Ncert Solution of Math Class 12 Chapter 4

NCERT solutions for Class 12 Maths Chapter 4 Determinants - In this chapter, students will be able to understand the Class 12 Maths Chapter 4 NCERT solutions. If you multiply a matrix with coordinates of a point, it will give a new point in the space which is explained in NCERT solutions for Class 12 Maths Chapter 4 Determinants. In this sense, the matrix is a linear transformation. The determinant of the matrix is the factor by which its volume blows up. The important topics are determinants and their properties, finding the area of the triangle, minor and co-factors, adjoint and the inverse of the matrix, and applications of determinants like solving the system of linear equations etc are covered in NCERT solutions for Class 12 Maths Chapter 4 Determinants.

NCERT solutions for Class 12 Maths Chapter 4 Determinants

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What are the Determinants?

To every square matrix $A=\left [ a_{ij} \right ]$ of order n, we can associate a number (real or complex) called determinant of the square matrix A. Let's take a determinant (A) of order two-

If A is a then the determinant of A is written as |A|

matrix $A=\begin{bmatrix} a &b\\ c & d \end{bmatrix}$ , $|A| =\begin{vmatrix} a & b\\ c& d \end{vmatrix}=det(A)$

$det(A)=|A| =\Delta =\begin{vmatrix} a_{11} & a_{12}\\ a_{21}& a_{22} \end{vmatrix}=a_{11}a_{22}-a_{21}a_{12}$

The six exercises of this chapter determinants covers the properties of determinants, co-factors and applications like finding the area of triangle, solutions of linear equations in two or three variables, minors, consistency and inconsistency of system of linear equations, adjoint and inverse of a square matrix, and solution of linear equations in two or three variables using inverse of a matrix.

Topics and sub-topics of NCERT class 12 maths chapter 4 Determinants

4.1 Introduction

4.2 Determinant

4.2.1 Determinant of a matrix of order one

4.2.2 Determinant of a matrix of order two

4.2.3 Determinant of a matrix of order 3 × 3

4.3 Properties of Determinants

4.4 Area of a Triangle

4.5 Minors and Cofactors

4.6 Adjoint and Inverse of a Matrix

4.7 Applications of Determinants and Matrices

4.7.1 Solution of a system of linear equations using the inverse of a matrix

NCERT solutions for class 12 maths chapter-4 Determinants: Excercise- 4.1

Question:3 If $A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ , then show that $| 2 A |=4|A|$

Answer:

Given determinant $A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then we have to show that $| 2 A |=4|A|$ ,

So, $A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then, $2A =2 \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} = \begin{bmatrix} 2 & 4\\ 8 &4 \end{bmatrix}$

Hence we have $\left | 2A \right | = \begin{vmatrix} 2 &4 \\ 8& 4 \end{vmatrix} = 2(4) - 4(8) = -24$

So, L.H.S. = |2A| = -24

then calculating R.H.S. $4\left | A \right |$

We have,

$\left | A \right | = \begin{vmatrix} 1 &2 \\ 4& 2 \end{vmatrix} = 1(2) - 2(4) = -6$

hence R.H.S becomes $4\left | A \right | = 4\times(-6) = -24$

Therefore L.H.S. =R.H.S.

Hence proved.

Question:4 If $A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}$ then show that $|3A|=27|A|$

Answer:

Given Matrix $A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}$

Calculating $3A =3\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix} = \begin{bmatrix} 3 &0 &3 \\ 0& 3& 6\\ 0& 0 &12 \end{bmatrix}$

So, $\left | 3A \right | = 3(3(12) - 6(0) ) - 0(0(12)-0(6)) + 3(0-0) = 3(36) = 108$

calculating $27|A|$ ,

$|A| = \begin{vmatrix} 1 & 0 &1 \\ 0 & 1 & 2\\ 0& 0 &4 \end{vmatrix} = 1\begin{vmatrix} 1 &2 \\ 0 & 4 \end{vmatrix} - 0\begin{vmatrix} 0 &2 \\ 0& 4 \end{vmatrix} + 1\begin{vmatrix} 0 &1 \\ 0& 0 \end{vmatrix} = 4 -0 + 0 = 4$

So, $27|A| = 27(4) = 108$

Therefore $|3A|=27|A|$ .

Hence proved.

Question:6 If $A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix}$ , then find $|A|$ .

Answer:

Given the matrix $A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix}$ then,

Finding the determinant value of A;

$|A| = 1\begin{vmatrix} 1 &-3 \\ 4& -9 \end{vmatrix} -1\begin{vmatrix} 2 &-3 \\ 5& -9 \end{vmatrix}-2\begin{vmatrix} 2 &1 \\ 5& 4 \end{vmatrix}$

$= 1(-9+12)-1(-18+15)-2(8-5) =3+3-6 =0$

Question:7(i) Find values of x, if

$\begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}$

Answer:

Given that $\begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}$

First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S.,

$\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =2-20 = -18$ and $\begin{vmatrix}2x &4 \\6 &x \end{vmatrix} = 2x(x)-24 = 2x^2-24$

So, we have then,

$-18= 2x^2-24$ or $3= x^2$ or $x= \pm \sqrt{3}$

Question:7(ii) Find values of x, if

$\begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$

Answer:

Given $\begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$ ;

So, we here equate both sides after calculating each side's determinant values.

L.H.S. determinant value;

$\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}= 10 - 12 = -2$

Similarly R.H.S. determinant value;

$\begin{vmatrix}x &3 \\2x &5 \end{vmatrix} = 5(x) - 3(2x) = 5x - 6x =-x$

So, we have then;

$-2 = -x$ or $x =2$ .

NCERT solutions for class 12 maths chapter -4 Determinants: Excercise - 4.2

Question:1 Using the property of determinants and without expanding, prove that

$\begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}=0$

Answer:

We can split it in manner like;

$\begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}= \begin{vmatrix} x &a &x \\ y & b &y \\ z &c &z \end{vmatrix} + \begin{vmatrix} x &a & a\\ y &b &b \\ z&c & c \end{vmatrix}$

So, we know the identity that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Clearly, expanded determinants have identical columns.

$\therefore 0 + 0 = 0$

Hence the sum is zero.

Question:5 Using the property of determinants and without expanding, prove that

$\begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}=2\begin{vmatrix} a &p &x \\ b &q &y \\ c &r & z \end{vmatrix}$

Answer:

Given determinant :

$\triangle= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}$

Splitting the third row; we get,

$= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a &p & x \end{vmatrix} + \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ b &q & y \end{vmatrix} = \triangle_{1} + \triangle_{2}\ (assume\ that)$ .

Then we have,

$\triangle_{1} = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a &p & x \end{vmatrix}$

On Applying row transformation $R_{2} \rightarrow R_{2} - R_{3}$ and then $R_{1} \rightarrow R_{1} - R_{2}$ ;

we get, $\triangle_{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}$

Applying Rows exchange transformation $R_{1} \leftrightarrow R_{2}$ and $R_{2} \leftrightarrow R_{3}$ , we have:

$\triangle_{1} =(-1)^2 \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}= \begin{vmatrix} a & p & x\\ b & q&y \\ c& r & z \end{vmatrix}$

also $\triangle_{2} = \begin{vmatrix} b+c & q+r & y+z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$

On applying rows transformation, $R_{1} \rightarrow R_{1} - R_{3}$ and then $R_{2} \rightarrow R_{2} - R_{1}$

$\triangle_{2} = \begin{vmatrix} c & r & z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$ and then $\triangle_{2} = \begin{vmatrix} c & r & z \\ a&p &x \\ b & q & y \end{vmatrix}$

Then applying rows exchange transformation;

$R_{1} \leftrightarrow R_{2}$ and then $R_{2} \leftrightarrow R_{3}$ . we have then;

$\triangle_{2} =(-1)^2 \begin{vmatrix} a & p & x \\ b&q &y \\ c & r & z \end{vmatrix}$

So, we now calculate the sum = $\triangle_{1} + \triangle _{2}$

$\triangle_{1} + \triangle _{2} = 2 \begin{vmatrix} a &p &x \\ b& q& y\\ c & r& z \end{vmatrix}$

Hence proved.

Question:12 By using properties of determinants, show that:

$\begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}=(1-x^3)^2$

Answer:

Give determinant $\begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}$

Applying column transformation $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ we get;

$\triangle = \begin{vmatrix} 1+x+x^2 &x &x^2 \\ x^2+1+x &1 &x \\ x+x^2+1 &x^2 &1 \end{vmatrix}$

$= (1+x+x^2)\begin{vmatrix} 1 &x &x^2 \\ 1 &1 &x \\ 1 &x^2 &1 \end{vmatrix}$ [ after taking the (1+x+x ² ) factor common out.]

Now, applying row transformations, $R_{1} \rightarrow R_{1}-R_{2}$ and then $R_{2} \rightarrow R_{2}-R_{3}$ .

we have now,

$= (1+x+x^2)\begin{vmatrix} 0 &x-1 &x^2-x \\ 0 &1-x^2 &x-1 \\ 1 &x^2 &1 \end{vmatrix}$

$= (1+x+x^2)\begin{vmatrix} x-1 &x^2-x \\ 1-x^2 &x-1 \end{vmatrix}$

$= (1+x+x^2)((x-1)^2-x(x-1)(1-x^2))$

$= (1+x+x^2)(x-1)(x^3-1) = (x^3-1)^2$

As we know $\left [\because (1+x+x^2)(x-1) = (x^3-1) \right ]$

Hence proved.

Question:13 By using properties of determinants, show that:

$\begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}=(1+a^2+b^2)^3$

Answer:

We have determinant:

$\triangle = \begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

Applying row transformations, $R_{1} \rightarrow R_{1} +bR_{3}$ and $R_{2} \rightarrow R_{2} -aR_{3}$ then we have;

$= \begin{vmatrix} 1+a^2+b^2 &0 &-b(1+a^2+b^2) \\ 0 &1+a^2+b^2 &a(1+a^2+b^2) \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

taking common factor out of the determinant;

$= (1+a^2+b^2)^2\begin{vmatrix} 1 &0 &-b \\ 0 &1 &a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

Now expanding the remaining determinant we get;

$= (1+a^2+b^2)^2\left [ (1)\begin{vmatrix} 1& a\\ -2a&1-a^2-b^2 \end{vmatrix} - b\begin{vmatrix} 0&1 \\ 2b&-2a \end{vmatrix}\right ]$

$= (1+a^2+b^2)^2\left [ 1-a^2-b^2+2a^2-b(-2b)\right ]$

$= (1+a^2+b^2)^2\left [ 1+a^2+b^2\right ] = (1+a^2+b^2)^3$

Hence proved.

Question:14 By using properties of determinants, show that:

$\begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}=1+a^2+b^2+c^2$

Answer:

Given determinant:

$\dpi{100} \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}$

Let $\triangle = \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}$

Then we can clearly see that each column can be reduced by taking common factors like a,b, and c respectively from C _1, C _2, and C _3.

We then get;

$=abc \begin{vmatrix} \left ( a+\frac{1}{a} \right ) &a &a \\ b &(b+\frac{1}{b}) &b \\ c & c &(c+\frac{1}{c}) \end{vmatrix}$

Now, applying column transformations: $C_{1} \rightarrow C_{1} -C_{2}$ and $C_{2} \rightarrow C_{2} -C_{3}$

then we have;

$=abc \begin{vmatrix} \left ( \frac{1}{a} \right ) &0 &a \\ -\frac{1}{b} &(\frac{1}{b}) &b \\ 0 & -\frac{1}{c} &(c+\frac{1}{c}) \end{vmatrix}$

$=abc\times \frac{1}{abc} \begin{vmatrix} 1 &0 &a^2 \\ -1 &1 &b^2 \\ 0 & -1 &(c^2+1) \end{vmatrix}$

$= \begin{vmatrix} 1 &0 &a^2 \\ -1 &1 &b^2 \\ 0 & -1 &(c^2+1) \end{vmatrix}$

Now, expanding the remaining determinant:

$\triangle = 1\begin{vmatrix} 1&b^2 \\ -1&(c^2+1) \end{vmatrix} + a^2\begin{vmatrix} -1&1 \\ 0& -1 \end{vmatrix}$

$= 1[(c^2+1)+b^2] + a^2(1)=a^2+b^2+c^2+1$ .

Hence proved.

Question:2 Show that points $A (a, b+c), B (b,c+a), C (c,a+b)$ are collinear.

Answer:

If the area formed by the points is equal to zero then we can say that the points are collinear.

So, we have an area of a triangle given by,

$\triangle = \frac{1}{2} \begin{vmatrix} a &b+c &1 \\ b& c+a &1 \\ c& a+b & 1 \end{vmatrix}$

calculating the area:

$= \frac{1}{2}\left [ a\begin{vmatrix} c+a &1 \\ a+b& 1 \end{vmatrix} - (b+c)\begin{vmatrix} b & 1\\ c&1 \end{vmatrix}+1\begin{vmatrix} b &c+a \\ c&a+b \end{vmatrix} \right ]$

$= \frac{1}{2}\left [ a(c+a-a-b) - (b+c)(b-c)+1(b(a+b)-c(c+a)) \right ]$

$= \frac{1}{2}\left [ ac-ab - b^2+c^2+ab+b^2-c^2-ac \right ] = \frac{1}{2} \left [ 0 \right] = 0$

Hence the area of the triangle formed by the points is equal to zero.

Therefore given points $A (a, b+c), B (b,c+a), C (c,a+b)$ are collinear.

Question:4(ii) Find equation of line joining $\small (3,1)$ and $\small (9,3)$ using determinants.

Answer:

We can find the equation of the line by considering any arbitrary point $A(x,y)$ on line.

So, we have three points which are collinear and therefore area surrounded by them will be equal to zero .

$\triangle = \frac{1}{2}\begin{vmatrix} 3 &1 &1 \\ 9& 3 & 1\\ x& y &1 \end{vmatrix} = 0$

Calculating the determinant:

$=\frac{1}{2}\left [ 3\begin{vmatrix} 3 &1 \\ y& 1 \end{vmatrix}-1\begin{vmatrix} 9 &1 \\ x& 1 \end{vmatrix}+1\begin{vmatrix} 9 &3 \\ x &y \end{vmatrix} \right ]$

$=\frac{1}{2}\left [ 3(3-y)-1(9-x)+1(9y-3x) \right ] = 0$

$\frac{1}{2}\left [ 9-3y-9+x+9y-3x \right ] = \frac{1}{2}[6y-2x] = 0$

Hence we have the line equation:

$3y= x$ or $x-3y = 0$ .

NCERT solutions for class 12 maths chapter 4 Determinants-Excercise: 4.4

Question:1(i) Write Minors and Cofactors of the elements of following determinants:

$\small \begin{vmatrix}2 &-4 \\0 &3 \end{vmatrix}$

Answer:

GIven determinant: $\begin{vmatrix}2 &-4 \\0 &3 \end{vmatrix}$

Minor of element $a_{ij}$ is $M_{ij}$ .

Therefore we have

$M_{11}$ = minor of element $a_{11}$ = 3

$M_{12}$ = minor of element $a_{12}$ = 0

$M_{21}$ = minor of element $a_{21}$ = -4

$M_{22}$ = minor of element $a_{22}$ = 2

and finding cofactors of $a_{ij}$ is $A_{ij}$ = $(-1)^{i+j}M_{ij}$ .

Therefore we have:

$A_{11} = (-1)^{1+1}M_{11} = (-1)^2(3) = 3$

$A_{12} = (-1)^{1+2}M_{12} = (-1)^3(0) = 0$

$A_{21} = (-1)^{2+1}M_{21} = (-1)^3(-4) = 4$

$A_{22} = (-1)^{2+2}M_{22} = (-1)^4(2) = 2$

Question:2(i) Write Minors and Cofactors of the elements of following determinants:

$\small \begin{vmatrix} 1 & 0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{vmatrix}$

Answer:

Given determinant : $\begin{vmatrix} 1 & 0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{vmatrix}$

Finding Minors: by the definition,

$M_{11} =$ minor of $a_{11} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$ $M_{12} =$ minor of $a_{12} = \begin{vmatrix} 0 &0 \\ 0 &1 \end{vmatrix} = 0$

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 0 &1 \\ 0 &0 \end{vmatrix} = 0$ $M_{21} =$ minor of $a_{21} = \begin{vmatrix} 0 &0 \\ 0 &1 \end{vmatrix} = 0$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$ $M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &0 \\ 0 &0 \end{vmatrix} = 0$

$M_{31} =$ minor of $a_{31} = \begin{vmatrix} 0 &0 \\ 1 &0 \end{vmatrix} = 0$ $M_{32} =$ minor of $a_{32} = \begin{vmatrix} 1 &0 \\ 0 &0 \end{vmatrix} = 0$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$

Finding the cofactors:

$A_{11}=$ cofactor of $a_{11} = (-1)^{1+1}M_{11} = 1$

$A_{12}=$ cofactor of $a_{12} = (-1)^{1+2}M_{12} = 0$

$A_{13}=$ cofactor of $a_{13} = (-1)^{1+3}M_{13} = 0$

$A_{21}=$ cofactor of $a_{21} = (-1)^{2+1}M_{21} = 0$

$A_{22}=$ cofactor of $a_{22} = (-1)^{2+2}M_{22} = 1$

$A_{23}=$ cofactor of $a_{23} = (-1)^{2+3}M_{23} = 0$

$A_{31}=$ cofactor of $a_{31} = (-1)^{3+1}M_{31} = 0$

$A_{32}=$ cofactor of $a_{32} = (-1)^{3+2}M_{32} = 0$

$A_{33}=$ cofactor of $a_{33} = (-1)^{3+3}M_{33} = 1$ .

Question:2(ii) Write Minors and Cofactors of the elements of following determinants:

$\small \begin{vmatrix} 1 &0 &4 \\ 3 & 5 &-1 \\ 0 &1 &2 \end{vmatrix}$

Answer:

Given determinant : $\begin{vmatrix} 1 &0 &4 \\ 3 & 5 &-1 \\ 0 &1 &2 \end{vmatrix}$

Finding Minors: by the definition,

$M_{11} =$ minor of $a_{11} = \begin{vmatrix} 5 &-1 \\ 1 &2 \end{vmatrix} = 11$ $M_{12} =$ minor of $a_{12} = \begin{vmatrix} 3 &-1 \\ 0 &2 \end{vmatrix} = 6$

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 3 &5 \\ 0 &1 \end{vmatrix} = 3$ $M_{21} =$ minor of $a_{21} = \begin{vmatrix} 0 &4 \\ 1 &2 \end{vmatrix} = -4$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 1 &4 \\ 0 &2 \end{vmatrix} = 2$ $M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$

$M_{31} =$ minor of $a_{31} = \begin{vmatrix} 0 &4 \\ 5 &-1 \end{vmatrix} = -20$

$M_{32} =$ minor of $a_{32} = \begin{vmatrix} 1 &4 \\ 3 &-1 \end{vmatrix} = -1-12=-13$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &0 \\ 3 &5 \end{vmatrix} = 5$

Finding the cofactors:

$A_{11}=$ cofactor of $a_{11} = (-1)^{1+1}M_{11} = 11$

$A_{12}=$ cofactor of $a_{12} = (-1)^{1+2}M_{12} = -6$

$A_{13}=$ cofactor of $a_{13} = (-1)^{1+3}M_{13} = 3$

$A_{21}=$ cofactor of $a_{21} = (-1)^{2+1}M_{21} = 4$

$A_{22}=$ cofactor of $a_{22} = (-1)^{2+2}M_{22} = 2$

$A_{23}=$ cofactor of $a_{23} = (-1)^{2+3}M_{23} = -1$

$A_{31}=$ cofactor of $a_{31} = (-1)^{3+1}M_{31} = -20$

$A_{32}=$ cofactor of $a_{32} = (-1)^{3+2}M_{32} = 13$

$A_{33}=$ cofactor of $a_{33} = (-1)^{3+3}M_{33} = 5$ .

Question:3 Using Cofactors of elements of second row, evaluate . $\small \Delta =\begin{vmatrix} 5 &3 &8 \\ 2 & 0 & 1\\ 1 &2 &3 \end{vmatrix}$

Answer:

Given determinant : $\small \Delta =\begin{vmatrix} 5 &3 &8 \\ 2 & 0 & 1\\ 1 &2 &3 \end{vmatrix}$

First finding Minors of the second rows by the definition,

$M_{21} =$ minor of $a_{21} = \begin{vmatrix} 3 &8 \\ 2 &3 \end{vmatrix} =9-16 = -7$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 5 &8 \\ 1 &3 \end{vmatrix} = 15-8=7$

$M_{23} =$ minor of $a_{23} = \begin{vmatrix} 5 &3 \\ 1 &2 \end{vmatrix} = 10-3 =7$

Finding the Cofactors of the second row:

$A_{21}=$ Cofactor of $a_{21} = (-1)^{2+1}M_{21} = 7$

$A_{22}=$ Cofactor of $a_{22} = (-1)^{2+2}M_{22} = 7$

$A_{23}=$ Cofactor of $a_{23} = (-1)^{2+3}M_{23} = -7$

Therefore we can calculate $\triangle$ by sum of the product of the elements of the second row with their corresponding cofactors.

Therefore we have,

$\triangle = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2(7) +0(7) +1(-7) =14-7=7$

Question:4 Using Cofactors of elements of third column, evaluate $\small \Delta =\begin{vmatrix} 1 &x &yz \\ 1 &y &zx \\ 1 &z &xy \end{vmatrix}$

Answer:

Given determinant : $\small \Delta =\begin{vmatrix} 1 &x &yz \\ 1 &y &zx \\ 1 &z &xy \end{vmatrix}$

First finding Minors of the third column by the definition,

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 1 &y \\ 1 &z \end{vmatrix} =z-y$

$M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &x \\ 1 &z \end{vmatrix} = z-x$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &x \\ 1 &y \end{vmatrix} =y-x$

Finding the Cofactors of the second row:

$A_{13}=$ Cofactor of $a_{13} = (-1)^{1+3}M_{13} = z-y$

$A_{23}=$ Cofactor of $a_{23} = (-1)^{2+3}M_{23} = x-z$

$A_{33}=$ Cofactor of $a_{33} = (-1)^{3+3}M_{33} = y-x$

Therefore we can calculate $\triangle$ by sum of the product of the elements of the third column with their corresponding cofactors.

Therefore we have,

$\triangle = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$

$= (z-y)yz + (x-z)zx +(y-x)xy$

$=yz^2-y^2z + zx^2-xz^2 + xy^2-x^2y$

$=z(x^2-y^2) + z^2(y-x) +xy(y-x)$

$= (x-y) \left [ zx+zy-z^2-xy \right ]$

$=(x-y)\left [ z(x-z) +y(z-x) \right ]$

$= (x-y)(z-x)[-z+y]$

$= (x-y)(y-z)(z-x)$

Thus, we have value of $\triangle = (x-y)(y-z)(z-x)$ .

Question:3 Verify $\small A (adj A)=(adj A)A=|A|I$ .

$\small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

Answer:

Given the matrix: $\small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

Let $\small A = \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

Calculating the cofactors;

$\small A_{11} = (-1)^{1+1}(-6) = -6$

$\small A_{12} = (-1)^{1+2}(-4) = 4$

$\small A_{21} = (-1)^{2+1}(3) = -3$

$\small A_{22} = (-1)^{2+2}(2) = 2$

Hence, $\small adjA = \begin{bmatrix} -6 &-3 \\ 4& 2 \end{bmatrix}$

Now,

$\small A (adj A) = \begin{bmatrix} 2 &3 \\ -4&-6 \end{bmatrix}\left ( \begin{bmatrix} -6 &-3 \\ 4 &2 \end{bmatrix} \right )$

$\small \begin{bmatrix} -12+12 &-6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$

aslo,

$\small (adjA)A = \begin{bmatrix} -6 &-3 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 2 &3 \\ -4& -6 \end{bmatrix}$

$\small = \begin{bmatrix} -12+12 &-18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

Now, calculating |A|;

$\small |A| = -12-(-12) = -12+12 = 0$

So, $\small |A|I = 0\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

Hence we get

$\small A (adj A)=(adj A)A=|A|I$

Question:4 Verify $\small A (adj A)=(adjA)A=|A| I$ .

$\small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Answer:

Given matrix: $\small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Let $\small A= \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Calculating the cofactors;

$\small A_{11} = (-1)^{1+1} \begin{vmatrix} 0 &-2 \\ 0& 3 \end{vmatrix} = 0$

$\small A_{12} = (-1)^{1+2} \begin{vmatrix} 3 &-2 \\1& 3 \end{vmatrix} = -(9+2) =-11$

$\small A_{13} = (-1)^{1+3} \begin{vmatrix} 3 &0 \\ 1& 0 \end{vmatrix} = 0$

$\small A_{21} = (-1)^{2+1} \begin{vmatrix} -1 &2 \\ 0& 3 \end{vmatrix} = -(-3-0)= 3$

$\small A_{22} = (-1)^{2+2} \begin{vmatrix} 1 &2 \\ 1& 3 \end{vmatrix} = 3-2=1$

$\small A_{23} = (-1)^{2+3} \begin{vmatrix} 1 &-1 \\ 1& 0 \end{vmatrix} = -(0+1) = -1$

$\small A_{31} = (-1)^{3+1} \begin{vmatrix} -1 &2 \\ 0& -2 \end{vmatrix} = 2$

$\small A_{32} = (-1)^{3+2} \begin{vmatrix} 1 &2 \\ 3& -2 \end{vmatrix} = -(-2-6) = 8$

$\small A_{33} = (-1)^{3+3} \begin{vmatrix} 1 &-1 \\ 3& 0 \end{vmatrix} = 0+3 =3$

Hence, $\small adjA = \begin{bmatrix} 0 &3 &2 \\ -11 & 1& 8\\ 0 &-1 & 3 \end{bmatrix}$

Now,

$\small A (adj A) =\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}$

$\small =\begin{bmatrix} 0+11+0 &3-1-2 &2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 &3+0-3 & 2+0+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}$

also,

$\small A (adj A) =\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}$

$\small =\begin{bmatrix} 0+9+2 &0+0+0 &0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 &0+0+0 & 0+2+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}$

Now, calculating |A|;

$\small |A| = 1(0-0) +1(9+2) +2(0-0) = 11$

So, $\small |A|I = 11\begin{bmatrix} 1 &0&0 \\ 0& 1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 11 &0&0 \\ 0& 11&0\\ 0&0&11 \end{bmatrix}$

Hence we get,

$\small A (adj A)=(adj A)A=|A|I$ .

Question:5 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 2 &-2 \\ 4 & 3 \end{bmatrix}$

Answer:

Given matrix : $\small \begin{bmatrix} 2 &-2 \\ 4 & 3 \end{bmatrix}$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

|A| = (6+8) = 14

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (3) = 3$

$A_{12} = (-1)^{1+2} (4) = -4$

$A_{21} = (-1)^{2+1} (-2) = 2$

$A_{22} = (-1)^{2+2} (2) = 2$

So, we have $adjA = \begin{bmatrix} 3 &2 \\ -4& 2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{14}\begin{bmatrix} 3 &2 \\ -4& 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{14} &\frac{1}{7} \\ \\ \frac{-2}{7} & \frac{1}{7} \end{bmatrix}$

Question:6 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} -1 &5 \\ -3 &2 \end{bmatrix}$

Answer:

Given the matrix : $\small \begin{bmatrix} -1 &5 \\ -3 &2 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

|A| = (-2+15) = 13

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (2) = 2$

$A_{12} = (-1)^{1+2} (-3) = 3$

$A_{21} = (-1)^{2+1} (5) =-5$

$A_{22} = (-1)^{2+2} (-1) = -1$

So, we have $adjA = \begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{13}\begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix} = \begin{bmatrix} \frac{2}{13} &\frac{-5}{13} \\ \\ \frac{3}{13} & \frac{-1}{13} \end{bmatrix}$

Question:7 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 1 &2 &3 \\ 0 &2 &4 \\ 0 &0 &5 \end{bmatrix}$

Answer:

Given the matrix : $\small \begin{bmatrix} 1 &2 &3 \\ 0 &2 &4 \\ 0 &0 &5 \end{bmatrix}= A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(10-0)-2(0-0)+3(0-0) = 10$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (10) = 10$ $A_{12} = (-1)^{1+2} (0) = 0$

$A_{13} = (-1)^{1+3} (0) =0$ $A_{21} = (-1)^{2+1} (10) = -10$

$A_{22} = (-1)^{2+2} (5-0) = 5$ $A_{23} = (-1)^{2+1} (0-0) = 0$

$A_{31} = (-1)^{3+1} (8-6) = 2$ $A_{32} = (-1)^{3+2} (4-0) =-4$

$A_{33} = (-1)^{3+3} (2-0) = 2$

So, we have $adjA = \begin{bmatrix} 10 &-10 &2 \\ 0& 5 &-4 \\ 0& 0 &2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{10}\begin{bmatrix} 10 &-10 &2 \\ 0 & 5& -4\\ 0 &0 &2 \end{bmatrix}$

Question:8 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 1 &0 &0 \\ 3 &3 &0 \\ 5 &2 &-1 \end{bmatrix}$

Answer:

Given the matrix : $\small \begin{bmatrix} 1 &0 &0 \\ 3 &3 &0 \\ 5 &2 &-1 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(-3-0)-0(-3-0)+0(6-15) = -3$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-3-0) = -3$ $A_{12} = (-1)^{1+2} (-3-0) = 3$

$A_{13} = (-1)^{1+3} (6-15) =-9$ $A_{21} = (-1)^{2+1} (0-0) = 0$

$A_{22} = (-1)^{2+2} (-1-0) = -1$ $A_{23} = (-1)^{2+1} (2-0) = -2$

$A_{31} = (-1)^{3+1} (0-0) = 0$ $A_{32} = (-1)^{3+2} (0-0) =0$

$A_{33} = (-1)^{3+3} (3-0) = 3$

So, we have $adjA = \begin{bmatrix} -3 &0 &0 \\ 3& -1 &0 \\ -9& -2 &3 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{-1}{3}\begin{bmatrix} -3 &0 &0 \\ 3 & -1& 0\\ -9 &-2 &3 \end{bmatrix}$

Question:9 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix}$

Answer:

Given the matrix : $\small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix} =A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 2(-1-0)-1(4-0)+3(8-7) =-2-4+3 = -3$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-1-0) = -1$ $A_{12} = (-1)^{1+2} (4-0) = -4$

$A_{13} = (-1)^{1+3} (8-7) =1$ $A_{21} = (-1)^{2+1} (1-6) = 5$

$A_{22} = (-1)^{2+2} (2+21) = 23$ $A_{23} = (-1)^{2+1} (4+7) = -11$

$A_{31} = (-1)^{3+1} (0+3) = 3$ $A_{32} = (-1)^{3+2} (0-12) =12$

$A_{33} = (-1)^{3+3} (-2-4) = -6$

So, we have $adjA = \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}$

Question:10 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 1 & -1 & 2\\ 0 & 2 &-3 \\ 3 &-2 &4 \end{bmatrix}$

Answer:

Given the matrix : $\small \begin{bmatrix} 1 & -1 & 2\\ 0 & 2 &-3 \\ 3 &-2 &4 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(8-6)+1(0+9)+2(0-6) =2+9-12 = -1$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (8-6) = 2$ $A_{12} = (-1)^{1+2} (0+9) = -9$

$A_{13} = (-1)^{1+3} (0-6) =-6$ $A_{21} = (-1)^{2+1} (-4+4) = 0$

$A_{22} = (-1)^{2+2} (4-6) = -2$ $A_{23} = (-1)^{2+1} (-2+3) = -1$

$A_{31} = (-1)^{3+1} (3-4) = -1$ $A_{32} = (-1)^{3+2} (-3-0) =3$

$A_{33} = (-1)^{3+3} (2-0) = 2$

So, we have $adjA = \begin{bmatrix} 2 &0 &-1 \\ -9& -2 &3 \\ -6& -1 &2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 &0 &-1 \\ -9& -2 &3 \\ -6& -1 &2 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -2 &0 &1 \\ 9& 2 &-3 \\ 6& 1 &-2 \end{bmatrix}$

Question:11 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix}$

Answer:

Given the matrix : $\small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix} =A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(-\cos^2 \alpha-\sin^2 \alpha)+0(0-0)+0(0-0)$

$=-(\cos^2 \alpha + \sin^2 \alpha) = -1$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-\cos^2 \alpha - \sin^2 \alpha) = -1$ $A_{12} = (-1)^{1+2} (0-0) = 0$

$A_{13} = (-1)^{1+3} (0-0) =0$ $A_{21} = (-1)^{2+1} (0-0) = 0$

$A_{22} = (-1)^{2+2} (-\cos \alpha-0) = -\cos \alpha$ $A_{23} = (-1)^{2+1} (\sin \alpha-0) = -\sin \alpha$

$A_{31} = (-1)^{3+1} (0-0) = 0$ $A_{32} = (-1)^{3+2} (\sin \alpha-0) =-\sin \alpha$

$A_{33} = (-1)^{3+3} (\cos \alpha - 0) = \cos \alpha$

So, we have $adjA = \begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-1}\begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix} = \begin{bmatrix}1 &0 &0 \\ 0&\cos \alpha &\sin \alpha \\ 0& \sin \alpha &-\cos \alpha \end{bmatrix}$

Question:12 Let $\small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix}$ and $\small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}$ . Verify that $\small (AB)^-^1=B^{-1}A^{-1}$ .

Answer:

We have $\small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix}$ and $\small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}$ .

then calculating;

$AB = \begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}$

$=\begin{bmatrix} 18+49 &24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 &87 \\ 47& 61 \end{bmatrix}$

Finding the inverse of AB.

Calculating the cofactors fo AB:

$AB_{11}=(-1)^{1+1}(61) = 61$ $AB_{12}=(-1)^{1+2}(47) = -47$

$AB_{21}=(-1)^{2+1}(87) = -87$ $AB_{22}=(-1)^{2+2}(67) = 67$

Then we have adj(AB):

$adj(AB) = \begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}$

and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2

Therefore we have inverse:

$(AB)^{-1}=\frac{1}{|AB|}adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 &-87 \\ -47 & 67 \end{bmatrix}$

$= \begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}$ .....................................(1)

Now, calculating inverses of A and B.

|A| = 15-14 = 1 and |B| = 54- 56 = -2

$adjA = \begin{bmatrix} 5 &-7 \\ -2 & 3 \end{bmatrix}$ and $adjB = \begin{bmatrix} 9 &-8 \\ -7 & 6 \end{bmatrix}$

therefore we have

$A^{-1} = \frac{1}{|A|}adjA= \frac{1}{1} \begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}$ and $B^{-1} = \frac{1}{|B|}adjB= \frac{1}{-2} \begin{bmatrix} 9&-8 \\ -7& 6 \end{bmatrix}= \begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}$

Now calculating $B^{-1}A^{-1}$ .

$B^{-1}A^{-1} =\begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}\begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}$

$=\begin{bmatrix} \frac{-45}{2}-8 && \frac{63}{2}+12 \\ \\ \frac{35}{2}+6 && \frac{-49}{2}-9 \end{bmatrix} = \begin{bmatrix} \frac{-61}{2} && \frac{87}{2} \\ \\ \frac{47}{2} && \frac{-67}{2} \end{bmatrix}$ ........................(2)

From (1) and (2) we get

$\small (AB)^-^1=B^{-1}A^{-1}$

Hence proved.

Question:13 If $\small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}$ ? , show that $A^2-5A+7I=O$ . Hence find $\small A^-^1$

Answer:

Given $\small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}$ then we have to show the relation $A^2-5A+7I=0$

So, calculating each term;

$A^2 = \begin{bmatrix} 3& 1\\ -1& 2 \end{bmatrix}\begin{bmatrix} 3&1 \\ -1& 2 \end{bmatrix} = \begin{bmatrix} 9-1 &3+2 \\ -3-2&-1+4 \end{bmatrix} = \begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix}$

therefore $A^2-5A+7I$ ;

$=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - 5\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix} + 7 \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$

$=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - \begin{bmatrix} 15 &5 \\ -5& 10 \end{bmatrix} + \begin{bmatrix} 7 &0 \\ 0 & 7 \end{bmatrix}$

$\begin{bmatrix} 8-15+7 &&5-5+0 \\ -5+5+0 && 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 &&0 \\ 0 && 0 \end{bmatrix}$

Hence $A^2-5A+7I = 0$ .

$\therefore A.A -5A = -7I$

$\Rightarrow A.A(A^{-1}) - 5AA^{-1} = -7IA^{-1}$

[ Post multiplying by $A^{-1}$ , also $|A| \neq 0$ ]

$\Rightarrow A(AA^{-1}) - 5I = -7A^{-1}$

$\Rightarrow AI - 5I = -7A^{-1}$

$\Rightarrow -\frac{1}{7}(AI - 5I)= \frac{1}{7}(5I-A)$

$\therefore A^{-1} = \frac{1}{7}(5\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}-\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}) = \frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}$

Question:15 For the matrix $\small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}$ Show that $\small A^3-6A^2+5A+11I=O$ Hence, find $\small A^-^1$ .

Answer:

Given matrix: $\small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}$ ;

To show: $\small A^3-6A^2+5A+11I=O$

Finding each term:

$A^{2} = \begin{bmatrix} 1 & 1& 1\\ 1 & 2& -3\\ 2& -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1& 1\\ 1 & 2& -3\\ 2& -1 & 3 \end{bmatrix}$

$= \begin{bmatrix} 1+1+2 &&1+2-1 &&1-3+3 \\ 1+2-6 &&1+4+3 &&1-6-9 \\ 2-1+6 &&2-2-3 && 2+3+9 \end{bmatrix}$

$= \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}$

$A^{3} = \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}$

$= \begin{bmatrix} 4+2+2 &4+4-1 &4-6+3 \\ -3+8-28 &-3+16+14 & -3-24-42 \\ 7-3+28&7-6-14 &7+9+42 \end{bmatrix}$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}$

So now we have, $\small A^3-6A^2+5A+11I$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-6\begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}+5\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+11\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-\begin{bmatrix} 24 &&12 &&6 \\ -18 &&48 &&-84 \\ 42 &&-18 && 84 \end{bmatrix}+\begin{bmatrix} 5 &5 &5 \\ 5 &10 &-15 \\ 10 &-5 &15 \end{bmatrix}+\begin{bmatrix} 11 &0 &0 \\ 0 &11 & 0\\ 0& 0& 11 \end{bmatrix}$

$= \begin{bmatrix} 8-24+5+11 &7-12+5 &1-6+5 \\ -23+18+5&27-48+10+11 &-69+84-15 \\ 32-42+10&-13+18-5 & 58-84+15+11 \end{bmatrix}$

$= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = 0$

Now finding the inverse of A;

Post-multiplying by $A^{-1}$ as, $|A| \neq 0$

$\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +5AA^{-1}+11IA^{-1} = 0$

$\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +5(AA^{-1})=- 11IA^{-1}$

$\Rightarrow A^{2}-6A +5I=- 11A^{-1}$

$A^{-1} = \frac{-1}{11}(A^{2}-6A+5I)$ ...................(1)

Now,

From equation (1) we get;

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}-6\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+5\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})$

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4-6+5 &&2-6 &&1-6 \\ -3-6 &&8-12+5 &&-14+18 \\ 7-12 &&-3+6 && 14-18+5 \end{bmatrix}$

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 3 &&-4 &&-5 \\ -9 &&1 &&4 \\ -5 &&3 && 1 \end{bmatrix}$

Question:16 If $\small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$ , verify that $\small A^3-6A^2+9A-4I=O$ . Hence find $\small A^-^1$ .

Answer:

Given matrix: $\small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$ ;

To show: $\small A^3-6A^2+9A-4I$

Finding each term:

$A^{2} = \begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$

$= \begin{bmatrix} 4+1+1 &&-2-2-1 &&2+1+2 \\ -2-2-1 &&1+4+1 &&-1-2-2 \\ 2+1+2 &&-1-2-2 && 1+1+4 \end{bmatrix}$

$= \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}$

$A^{3} =\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$

$= \begin{bmatrix} 12+5+5 &-6-10-5 &6+5+10 \\ -10-6-5 &5+12+5 & -5-6-10 \\ 10+5+6&-5-10-6 &5+5+12 \end{bmatrix}$

$= \begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}$

So now we have, $\small A^3-6A^2+9A-4I$

$=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}-6 \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}+9\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}-4\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}$

$=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}- \begin{bmatrix} 36 &&-30 &&30 \\ -30 &&36 &&-30 \\30 &&-30 && 36 \end{bmatrix}+\begin{bmatrix} 18 &-9 &9 \\ -9 &18 &-9 \\ 9 &-9 &18 \end{bmatrix}-\begin{bmatrix} 4 &0 &0 \\ 0 &4 & 0\\ 0& 0& 4 \end{bmatrix}$

$= \begin{bmatrix} 22-36+18-4 &-21+30-9 &21-30+9 \\ -21+30-9&22-36+18-4 &-21+30-9 \\ 21-30+9&-21+30-9 & 22-36+18-4 \end{bmatrix}$

$= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = O$

Now finding the inverse of A;

Post-multiplying by $A^{-1}$ as, $|A| \neq 0$

$\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +9AA^{-1}-4IA^{-1} = 0$

$\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +9(AA^{-1})=4IA^{-1}$

$\Rightarrow A^{2}-6A +9I=4A^{-1}$

$A^{-1} = \frac{1}{4}(A^{2}-6A+9I)$ ...................(1)

Now,

From equation (1) we get;

$A^{-1} = \frac{1}{4}(\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}-6\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}+9\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})$

$A^{-1} = \frac{1}{4} \begin{bmatrix} 6-12+9 &&-5+6 &&5-6 \\ -5+6 &&6-12+9 &&-5+6 \\ 5-6 &&-5+6 && 6-12+9 \end{bmatrix}$

Hence inverse of A is :

$A^{-1} = \frac{1}{4} \begin{bmatrix} 3 &&1 &&-1 \\ 1 &&3 &&1 \\ -1 &&1 && 3 \end{bmatrix}$

NCERT solutions for class 12 chapter 4 Determinants: Excercise- 4.6

Question:1 Examine the consistency of the system of equations.

$\small x+2y=2$

$\small 2x+3y=3$

Answer:

We have given the system of equations:18967

$\small x+2y=2$

$\small 2x+3y=3$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &2 \\ 2&3 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 2\\3 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(3) -2(2) = -1 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:2 Examine the consistency of the system of equations

$\small 2x-y=5$

$\small x+y=4$

Answer:

We have given the system of equations:

$\small 2x-y=5$

$\small x+y=4$

The given system of equations can be written in the form of matrix; $AX =B$

where $A= \begin{bmatrix} 2 &-1 \\ 1&1 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\4 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 2(1) -1(-1) = 3 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:3 Examine the consistency of the system of equations.

$\small x+3y=5$

$\small 2x+6y=8$

Answer:

We have given the system of equations:

$\small x+3y=5$

$\small 2x+6y=8$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &3 \\ 2&6 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\8 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(6) -2(3) = 0$

Here A is singular matrix therefore now we will check whether the $(adjA)B$ is zero or non-zero.

$adjA= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}$

So, $(adjA)B= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}\begin{bmatrix} 5\\8 \end{bmatrix} = \begin{bmatrix} 30-24\\-10+8 \end{bmatrix}=\begin{bmatrix} 6\\-2 \end{bmatrix} \neq 0$

As, $(adjA)B \neq 0$ , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

Question:4 Examine the consistency of the system of equations.

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

Answer:

We have given the system of equations:

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 1& 1&1 \\ 2& 3& 2\\ a& a &2a \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 1\\2 \\ 4 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(6a-2a) -1(4a-2a)+1(2a-3a)$

$= 4a -2a-a = 4a -3a =a \neq 0$

[ If zero then it won't satisfy the third equation ]

Here A is non- singular matrix therefore there exist $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:5 Examine the consistency of the system of equations.

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

Answer:

We have given the system of equations:

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

The given system of equations can be written in the form of matrix; $AX =B$

where $A = \begin{bmatrix} 3& -1&-2 \\ 0& 2& -1\\ 3& -5 &0 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 3(0-5) -(-1)(0+3)-2(0-6)$

$= -15 +3+12 = 0$

Therefore matrix A is a singular matrix.

So, we will then check $(adjA)B,$

$(adjA) = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}$

$\therefore (adjA)B = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}\begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix} = \begin{bmatrix} -10-10+15\\ -6-6+9 \\ -12-12+18 \end{bmatrix} = \begin{bmatrix} -5\\-3 \\ -6 \end{bmatrix} \neq 0$

As, $(adjA)B$ is non-zero thus the solution of the given system of the equation does not exist. Hence, the given system of equations is inconsistent.

Question:6 Examine the consistency of the system of equations.

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

Answer:

We have given the system of equations:

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 5& -1&4 \\ 2& 3& 5\\ 5& -2 &6 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 5\\2 \\ -1 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 5(18+10) +1(12-25)+4(-4-15)$

$= 140-13-76 = 51 \neq 0$

Here A is non- singular matrix therefore there exist $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:7 Solve system of linear equations, using matrix method.

$\small 5x+2y=4$

$\small 7x+3y=5$

Answer:

The given system of equations

$\small 5x+2y=4$

$\small 7x+3y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &4 \\ 7& 3 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 4\\5 \end{bmatrix}$

we have,

$|A| = 15-14=1 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = (adjA) = \begin{bmatrix} 3 &-2 \\ -7& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 3 &-2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 12-10\\ -28+25 \end{bmatrix} = \begin{bmatrix} 2\\-3 \end{bmatrix}$

Hence the solutions of the given system of equations;

x = 2 and y =-3 .

Question:8 Solve system of linear equations, using matrix method.

$2x-y=-2$

$3x+4y=3$

Answer:

The given system of equations

$2x-y=-2$

$3x+4y=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &-1 \\ 3& 4 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} -2\\3 \end{bmatrix}$

we have,

$|A| = 8+3=11 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3& 2 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2\\3 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -8+3\\ 6+6 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -5\\12 \end{bmatrix}= \begin{bmatrix} -\frac{5}{11}\\ \\-\frac{12}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-5}{11} \ and\ y =\frac{12}{11}.$

Question:9 Solve system of linear equations, using matrix method.

$\small 4x-3y=3$

$\small 3x-5y=7$

Answer:

The given system of equations

$\small 4x-3y=3$

$\small 3x-5y=7$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 4 &-3 \\ 3& -5 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\7 \end{bmatrix}$

we have,

$|A| = -20+9=-11 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{-1}{11}\begin{bmatrix} -5 &3 \\ -3& 4 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 5 &-3 \\ 3& -4 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 5 &-3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3\\7 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 15-21\\ 9-28 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -6\\-19 \end{bmatrix}= \begin{bmatrix} -\frac{6}{11}\\ \\-\frac{19}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-6}{11} \ and\ y =\frac{-19}{11}.$

Question:10 Solve system of linear equations, using matrix method.

$\small 5x+2y=3$

$\small 3x+2y=5$

Answer:

The given system of equations

$\small 5x+2y=3$

$\small 3x+2y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &2 \\ 3& 2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\5 \end{bmatrix}$

we have,

$|A| = 10-6=4 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3 & 5 \end{bmatrix}\begin{bmatrix} 3\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 6-10\\ -9+25 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4\\16 \end{bmatrix}= \begin{bmatrix} -1\\4 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =-1 \ and\ y =4.$

Question:11 Solve system of linear equations, using matrix method.

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

Answer:

The given system of equations

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &1 &1 \\ 1 & -2 &-1 \\ 0& 3 &-5 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ and $B =\begin{bmatrix} 1\\ \\ \frac{3}{2} \\ \\ 9 \end{bmatrix}$

we have,

$|A| =2(10+3)-1(-5-0)+1(3-0) = 26+5+3 = 34 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(10+3) = 13$ $A_{12} =(-1)^{1+2}(-5-0) = 5$

$A_{13} =(-1)^{1+3}(3-0) = 3$ $A_{21} =(-1)^{2+1}(-5-3) = 8$

$A_{22} =(-1)^{2+2}(-10-0) = -10$ $A_{23} =(-1)^{2+3}(6-0) = -6$

$A_{31} =(-1)^{3+1}(-1+2) = 1$ $A_{32} =(-1)^{3+2}(-2-1) = 3$

$A_{33} =(-1)^{3+3}(-4-1) = -5$

$(adjA) =\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}\begin{bmatrix} 1\\\frac{3}{2} \\ 9 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 13+12+9\\5-15+27 \\ 3-9-45 \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 34\\17 \\ -51 \end{bmatrix}= \begin{bmatrix} 1\\\frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =\frac{1}{2},\ and\ \ z=-\frac{3}{2}.$

Question:12 Solve system of linear equations, using matrix method.

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

Answer:

The given system of equations

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &1 \\ 2 & 1 &-3 \\ 1& 1 &1 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 4\\ 0 \\ 2 \end{bmatrix}.$

we have,

$|A| =1(1+3)+1(2+3)+1(2-1) = 4+5+1= 10 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(1+3) = 4$ $A_{12} =(-1)^{1+2}(2+3) = -5$

$A_{13} =(-1)^{1+3}(2-1) = 1$ $A_{21} =(-1)^{2+1}(-1-1) = 2$

$A_{22} =(-1)^{2+2}(1-1) = 0$ $A_{23} =(-1)^{2+3}(1+1) = -2$

$A_{31} =(-1)^{3+1}(3-1) = 2$ $A_{32} =(-1)^{3+2}(-3-2) = 5$

$A_{33} =(-1)^{3+3}(1+2) = 3$

$(adjA) =\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}\begin{bmatrix} 4\\0 \\ 2 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 16+0+4\\-20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 20\\-10 \\ 10 \end{bmatrix}= \begin{bmatrix} 2\\-1 \\ 1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =-1,\ and\ \ z=1.$

Question:13 Solve system of linear equations, using matrix method.

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

Answer:

The given system of equations

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &3 &3 \\ 1 & -2 &1 \\ 3& -1 &-2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 5\\ -4 \\ 3 \end{bmatrix}.$

we have,

$|A| =2(4+1) -3(-2-3)+3(-1+6) = 10+15+15 = 40$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(4+1) = 5$ $A_{12} =(-1)^{1+2}(-2-3) = 5$

$A_{13} =(-1)^{1+3}(-1+6) = 5$ $A_{21} =(-1)^{2+1}(-6+3) = 3$

$A_{22} =(-1)^{2+2}(-4-9) = -13$ $A_{23} =(-1)^{2+3}(-2-9) = 11$

$A_{31} =(-1)^{3+1}(3+6) = 9$ $A_{32} =(-1)^{3+2}(2-3) = 1$

$A_{33} =(-1)^{3+3}(-4-3) = -7$

$(adjA) =\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5&11 & -7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}\begin{bmatrix} 5\\-4 \\ 3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 25-12+27\\25+52+3 \\ 25-44-21 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 40\\80 \\ -40 \end{bmatrix}= \begin{bmatrix} 1\\2 \\ -1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=-1.$

Question:14 Solve system of linear equations, using matrix method.

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

Answer:

The given system of equations

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &2 \\ 3 & 4 &-5 \\ 2& -1 &3 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 7\\ -5 \\ 12 \end{bmatrix}.$

we have,

$|A| =1(12-5) +1(9+10)+2(-3-8) = 7+19-22 = 4 \neq0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{12-5} = 7$ $A_{12} =(-1)^{1+2}(9+10) = -19$

$A_{13} =(-1)^{1+3}(-3-8) = -11$ $A_{21} =(-1)^{2+1}(-3+2) = 1$

$A_{22} =(-1)^{2+2}(3-4) = -1$ $A_{23} =(-1)^{2+3}(-1+2) = -1$

$A_{31} =(-1)^{3+1}(5-8) = -3$ $A_{32} =(-1)^{3+2}(-5-6) = 11$

$A_{33} =(-1)^{3+3}(4+3) = 7$

$(adjA) =\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}\begin{bmatrix} 7\\-5 \\ 12 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 49-5-36\\-133+5+132 \\ -77+5+84 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8\\4 \\ 12 \end{bmatrix}= \begin{bmatrix} 2\\1 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =1,\ and\ \ z=3.$

Question:15 If $A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$ , find $A^-^1$ . Using $A^-^1$ solve the system of equations

$2x-3y+5z=11$

$3x+2y-4z=-5$

$x+y-2z=-3$

Answer:

The given system of equations

$2x-3y+5z=11$

$3x+2y-4z=-5$

$x+y-2z=-3$

can be written in the matrix form of AX =B, where

$A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}.$

we have,

$|A| =2(-4+4) +3(-6+4)+5(3-2) = 0-6+5 = -1 \neq0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{-4+4} = 0$ $A_{12} =(-1)^{1+2}(-6+4) = 2$

$A_{13} =(-1)^{1+3}(3-2) = 1$ $A_{21} =(-1)^{2+1}(6-5) = -1$

$A_{22} =(-1)^{2+2}(-4-5) = -9$ $A_{23} =(-1)^{2+3}(2+3) = -5$

$A_{31} =(-1)^{3+1}(12-10) = 2$ $A_{32} =(-1)^{3+2}(-8-15) = 23$

$A_{33} =(-1)^{3+3}(4+9) = 13$

$(adjA) =\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = -1\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix} = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}\begin{bmatrix} 11\\-5 \\ -3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-5+6\\-22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1\\2 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=3.$

Question:16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

Answer:

So, let us assume the cost of onion, wheat, and rice be x , y and z respectively.

Then we have the equations for the given situation :

$4x+3y+2z = 60$

$2x+4y+6z = 90$

$6x+2y+3y = 70$

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix $A$ .

We have;

$A = \begin{bmatrix} 4 &3 &2 \\ 2& 4 &6 \\ 6 & 2 & 3 \end{bmatrix},$ $X= \begin{bmatrix} x\\y \\ z \end{bmatrix}$ $and\ B = \begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}.$

$|A| = 4(12-12) -3(6-36)+2(4-24) = 0 +90-40 = 50 \neq 0$

Now, we will find the cofactors of A;

$A_{11} = (-1)^{1+1}(12-12) = 0$ $A_{12} = (-1)^{1+2}(6-36) = 30$

$A_{13} = (-1)^{1+3}(4-24) = -20$ $A_{21} = (-1)^{2+1}(9-4) = -5$

$A_{22} = (-1)^{2+2}(12-12) = 0$ $A_{23} = (-1)^{2+3}(8-18) = 10$

$A_{31} = (-1)^{3+1}(18-8) = 10$ $A_{32} = (-1)^{3+2}(24-4) = -20$

$A_{33} = (-1)^{3+3}(16-6) = 10$

Now we have adjA;

$adjA = \begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$ s

So, the solutions can be found by $X = A^{-1}B = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-450+700\\1800+0-1400 \\ -1200+900+700 \end{bmatrix} =\frac{1}{50} \begin{bmatrix} 250\\400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5\\8 \\ 8 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =5,\ y =8,\ and\ \ z=8.$

Therefore, we have the cost of onions is Rs. 5 per Kg, the cost of wheat is Rs. 8 per Kg, and the cost of rice is Rs. 8 per kg.

NCERT solutions for class 12 maths chapter 4 Determinants: Miscellaneous exercise

Question:1 Prove that the determinant $\begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}$ is independent of $\theta$ .

Answer:

Calculating the determinant value of $\begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}$ ;

$= x\begin{bmatrix} -x &1 \\ 1& x \end{bmatrix}-\sin \Theta\begin{bmatrix} -\sin \Theta &1 \\ \cos \Theta& x \end{bmatrix} + \cos \Theta \begin{bmatrix} -\sin \Theta &-x \\ \cos \Theta& 1 \end{bmatrix}$

$= x(-x^2-1)-\sin \Theta (-x\sin \Theta-\cos \Theta)+\cos\Theta(-\sin \Theta+x\cos\Theta)$

$= -x^3-x+x\sin^2 \Theta+ \sin \Theta\cos \Theta-\cos\Theta\sin \Theta+x\cos^2\Theta$

$= -x^3-x+x(\sin^2 \Theta+\cos^2\Theta)$

$= -x^3-x+x = -x^3$

Clearly, the determinant is independent of $\Theta$ .

Question:3 Evaluate $\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta &-\sin \alpha \\ -\sin \beta & \cos \beta &0 \\ \sin \alpha \cos \beta &\sin \alpha \sin \beta & \cos \alpha \end{vmatrix}$ .

Answer:

Given determinant $\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta &-\sin \alpha \\ -\sin \beta & \cos \beta &0 \\ \sin \alpha \cos \beta &\sin \alpha \sin \beta & \cos \alpha \end{vmatrix}$ ;

$= \cos \alpha \cos \beta \begin{vmatrix} \cos \beta &0 \\ \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} - \cos \alpha \sin \beta \begin{vmatrix} -\sin \beta & 0 \\ \sin \alpha \cos \beta & \cos \alpha \end{vmatrix} -\sin \alpha \begin{vmatrix} -\sin \beta &\cos \beta \\ \sin \alpha \cos \beta& \sin \alpha \sin \beta \end{vmatrix}$ $= \cos \alpha \cos \beta (\cos \beta \cos \alpha -0 )- \cos \alpha \sin \beta (-\cos \alpha\sin \beta- 0) -\sin \alpha (-\sin \alpha\sin^2\beta - \sin \alpha \cos^2 \beta)$

$= \cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta +\sin^2 \alpha \sin^2\beta + \sin^2 \alpha \cos^2 \beta$

$= \cos^2 \alpha(\cos^2 \beta+\sin^2 \beta) +\sin^2 \alpha(\sin^2\beta+\cos^2 \beta)$

$= \cos^2 \alpha(1) +\sin^2 \alpha(1) = 1$ .

Question:4 If $a,b$ and $c$ are real numbers, and

$\Delta =\begin{vmatrix} b+c & c+a &a+b \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}=0$

Show that either $a+b+c=0$ or $a=b=c$

Answer:

We have given $\Delta =\begin{vmatrix} b+c & c+a &a+b \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}=0$

Applying the row transformations; $R_{1} \rightarrow R_{1} +R_{2} +R_{3}$ we have;

$\Delta =\begin{vmatrix} 2(a+b+c) & 2(a+b+c) &2(a+b+c) \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}$

Taking out common factor 2(a+b+c) from the first row;

$\Delta =2(a+b+c)\begin{vmatrix} 1 & 1 &1 \\ c+a &a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}$

Now, applying the column transformations; $C_{1}\rightarrow C_{1} - C_{2}\ and\ C_{2} \rightarrow C_{2}- C_{3}$

we have;

$=2(a+b+c)\begin{vmatrix} 0 & 0 &1 \\ c-b &a-c &b+c \\ a-c & b-a & c+a \end{vmatrix}$

$=2(a+b+c)[(c-b)(b-a)-(a-c)^2]$

$=2(a+b+c)[ab+bc+ca-a^2-b^2-c^2]$

and given that the determinant is equal to zero. i.e., $\triangle = 0$ ;

$(a+b+c)[ab+bc+ca-a^2-b^2-c^2] = 0$

So, either $(a+b+c) = 0$ or $[ab+bc+ca-a^2-b^2-c^2] = 0$ .

we can write $[ab+bc+ca-a^2-b^2-c^2] = 0$ as;

$\Rightarrow -2ab-2bc-2ca+2a^2+2b^2+2c^2 =0$

$\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2 =0$

$\because (a-b)^2,(b-c)^2,\ and\ (c-a)^2$ are non-negative.

Hence $(a-b)^2= (b-c)^2=(c-a)^2 = 0$ .

we get then $a=b=c$

Therefore, if given $\triangle$ = 0 then either $(a+b+c) = 0$ or $a=b=c$ .

Question:5 Solve the equation

$\begin{vmatrix} x+a & x &x \\ x &x+a &x\\ x & x & x+a \end{vmatrix}=0; a\neq 0$

Answer:

Given determinant $\begin{vmatrix} x+a & x &x \\ x &x+a &x\\ x & x & x+a \end{vmatrix}=0; a\neq 0$

Applying the row transformation; $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ we have;

$\begin{vmatrix} 3x+a & 3x+a &3x+a \\ x &x+a &x\\ x & x & x+a \end{vmatrix} =0$

Taking common factor (3x+a) out from first row.

$(3x+a)\begin{vmatrix} 1 & 1 &1 \\ x &x+a &x\\ x & x & x+a \end{vmatrix} =0$

Now applying the column transformations; $C_{1} \rightarrow C_{1}-C_{2}$ and $C_{2} \rightarrow C_{2}-C_{3}$ .

we get;

$(3x+a)\begin{vmatrix} 0 & 0 &1 \\ -a &a &x\\ 0 & -a & x+a \end{vmatrix} =0$

$=(3x+a)(a^2)=0$ as $a^2 \neq 0$ ,

or $3x+a=0$ or $x= -\frac{a}{3}$

Question:6 Prove that $\begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}=4a^2b^2c^2$ .

Answer:

Given matrix $\begin{vmatrix} a^2 &bc &ac+a^2 \\ a^2+ab & b^2 & ac\\ ab &b^2+bc &c^2 \end{vmatrix}$

Taking common factors a,b and c from the column $C_{1}, C_{2}, and\ C_{3}$ respectively.

we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b & a\\ b &b+c &c \end{vmatrix}$

Applying $R_{2} \rightarrow R_{2}-R_{1}\ and\ R_{3} \rightarrow R_{3} - R_{1}$ , we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ b & b-c &-c\\ b-a &b &-a \end{vmatrix}$

Then applying $R_{2} \rightarrow R_{2}+R_{1}$ , we get;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ b-a &b &-a \end{vmatrix}$

Applying $R_{3} \rightarrow R_{3}+R_{2}$ , we have;

$\triangle = abc\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 2b &2b &0 \end{vmatrix} = 2ab^2c\begin{vmatrix} a &c &a+c \\ a+b & b &a\\ 1 &1 &0 \end{vmatrix}$

Now, applying column transformation; $C_{2} \rightarrow C_{2 }-C_{1}$ , we have

$\triangle = 2ab^2c\begin{vmatrix} a &c-a &a+c \\ a+b & -a &a\\ 1 &0 &0 \end{vmatrix}$

So we can now expand the remaining determinant along $R_{3}$ we have;

$\triangle = 2ab^2c\left [ a(c-a)+a(a+c) \right ]$

$= 2ab^2c\left [ ac-a^2+a^2+ac) \right ] = 2ab^2c\left [ 2ac \right ]$

$= 4a^2b^2c^2$

Hence proved.

Question:7 If $A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$ and $B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$ , find $(AB)^-^1$ .

Answer:

We know from the identity that;

$(AB)^{-1} = B^{-1}A^{-1}$ .

Then we can find easily,

Given $A^-^1=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$ and $B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

Then we have to basically find the $B^{-1}$ matrix.

So, Given matrix $B=\begin{bmatrix} 1 &2 &-2 \\ -1&3 &0 \\ 0 &-2 &1 \end{bmatrix}$

$|B| = 1(3-0) -2(-1-0)-2(2-0) = 3+2-4 = 1 \neq 0$

Hence its inverse $B^{-1}$ exists;

Now, as we know that

$B^{-1} = \frac{1}{|B|} adjB$

So, calculating cofactors of B,

$B_{11} = (-1)^{1+1}(3-0) = 3$ $B_{12} = (-1)^{1+2}(-1-0) = 1$

$B_{13} = (-1)^{1+3}(2-0) = 2$ $B_{21} = (-1)^{2+1}(2-4) = 2$

$B_{22} = (-1)^{2+2}(1-0) = 1$ $B_{23} = (-1)^{2+3}(-2-0) = 2$

$B_{31} = (-1)^{3+1}(0+6) = 6$ $B_{32} = (-1)^{3+2}(0-2) = 2$

$B_{33} = (-1)^{3+3}(3+2) = 5$

$adjB = \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|} adjB = \frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}$

Now, We have both $A^{-1}$ as well as $B^{-1}$ ;

Putting in the relation we know; $(AB)^{-1} = B^{-1}A^{-1}$

$(AB)^{-1}=\frac{1}{1} \begin{bmatrix} 3 &2 &6 \\ 1& 1& 2\\ 2& 2& 5 \end{bmatrix}\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}$

$= \begin{bmatrix} 9-30+30 &-3+12-12 &3-10+12 \\ 3-15+10&-1+6-4 &1-5+4 \\ 6-30+25 &-2+12-10 &2-10+10 \end{bmatrix}$

$= \begin{bmatrix} 9 &-3 &5 \\ -2&1 &0 \\ 1 &0 &2\end{bmatrix}$

Question:8(i) Let $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ . Verify that,

$\dpi{100} [adj A]^-^1 = adj (A^-^1)$

Answer:

Given that $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ ;

So, let us assume that $A^{-1} = B$ matrix and $adjA = C$ then;

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$ or $B = \frac{1}{|A|}C$ ;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$ $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$ $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$ $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$ $A_{32}= (-1)^{3+2}(1-2) = 1$

$A_{33}= (-1)^{3+3}(3-4) = -1$

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

Finding the inverse of C;

$|C| = 14(-4-1)+9(9+1)-1(-9+4) = -70+90+5 = 25 \neq 0$

Hence its inverse exists;

$C^{-1} = \frac{1}{|C|}adj C$

Now, finding the $adjC$ ;

$C_{11}= (-1)^{1+1}(-4-1) = -5$ $C_{12}= (-1)^{1+2}(9+1) = -10$

$C_{13}= (-1)^{1+3}(-9+4) = -5$ $C_{21}= (-1)^{2+1}(9+1) = -10$

$C_{22}= (-1)^{2+2}(-14-1) = -15$ $C_{23}= (-1)^{2+3}(14-9) = -5$

$C_{31}= (-1)^{3+1}(-9+4) = -5$ $C_{32}= (-1)^{3+2}(14-9) = -5$

$C_{33}= (-1)^{3+3}(56-81) = -25$

$adjC = \begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix}$

$C^{-1} = \frac{1}{|C|}adjC = \frac{1}{25}\begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

or $L.H.S. = C^{-1} = [adjA]^{-1} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Now, finding the R.H.S.

$adj (A^{-1}) = adj B$

$A^{-1} =B= \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5}&& \frac{-4}{5}&& \frac{-1}{5}\\ \\ \frac{1}{5}&& \frac{-1}{5} &&\frac{1}{5}\end{bmatrix}$

Cofactors of B;

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$

$B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) =- \frac{2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$

$B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = -1$

$R.H.S. = adjB = adj(A^{-1}) =\begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Hence L.H.S. = R.H.S. proved.

Question:8(ii) Let $A=\begin{bmatrix} 1 &2 &1 \\ 2 & 3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ , Verify that

$(A^-^1)^-^1=A$

Answer:

Given that $A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$ ;

So, let us assume that $A^{-1} = B$

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$ or $B = \frac{1}{|A|}C$ ;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$ $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$ $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$ $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$ $A_{32}= (-1)^{3+2}(1-2) = 1$

$A_{33}= (-1)^{3+3}(3-4) = -1$

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix} = \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5} && \frac{-4}{5} &&\frac{-1}{5} \\ \\ \frac{1}{5} &&\frac{-1}{5} &&\frac{1}{5} \end{bmatrix}$

Finding the inverse of B ;

$|B| = \frac{-14}{5}(\frac{-4}{25}-\frac{1}{25})-\frac{9}{5}(\frac{9}{25}+\frac{1}{25})+\frac{1}{5}(\frac{-9}{25}+\frac{4}{25})$

$= \frac{70}{125}-\frac{90}{125}-\frac{5}{125} = \frac{-25}{125} = \frac{-1}{5} \neq 0$

Hence its inverse exists;

$B^{-1} = \frac{1}{|B|}adj B$

Now, finding the $adjB$ ;

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$ $B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) = \frac{-2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$ $B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$ $B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = \frac{-25}{25} =-1$

$adjB = \begin{bmatrix} \frac{-1}{5} &&\frac{-2}{5} &&\frac{-1}{5} \\ \\\frac{-2}{5}&& \frac{-3}{5} && \frac{-1}{5}\\ \\ \frac{-1}{5}&& \frac{-1}{5}&& -1 \end{bmatrix}$

$B^{-1} = \frac{1}{|B|}adjB = \frac{-5}{1}\begin{bmatrix} \frac{-1}{5} &&\frac{-2}{5} &&\frac{-1}{5} \\ \\\frac{-2}{5}&& \frac{-3}{5} && \frac{-1}{5}\\ \\ \frac{-1}{5}&& \frac{-1}{5}&& -1 \end{bmatrix}= \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1 & 1 &5 \end{bmatrix}$

$L.H.S. = B^{-1} = (A^{-1})^{-1} = \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1 & 1 &5 \end{bmatrix}$

$R.H.S. = A = \begin{bmatrix} 1 &2 &1 \\ 2& 3& 1\\ 1& 1 &5 \end{bmatrix}$

Hence proved L.H.S. =R.H.S. .

Question:9 Evaluate $\begin{vmatrix} x & y &x+y \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Answer:

We have determinant $\triangle = \begin{vmatrix} x & y &x+y \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Applying row transformations; $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ , we have then;

$\triangle = \begin{vmatrix} 2(x+y) & 2(x+y) &2(x+y) \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Taking out the common factor 2(x+y) from the row first.

$= 2(x+y)\begin{vmatrix} 1 & 1 &1 \\ y & x+y &x \\ x+y & x & y \end{vmatrix}$

Now, applying the column transformation; $C_{1} \rightarrow C_{1} - C_{2}$ and $C_{2} \rightarrow C_{2} - C_{1}$ we have ;

$= 2(x+y)\begin{vmatrix} 0 & 0 &1 \\ -x & y &x \\ y & x-y & y \end{vmatrix}$

Expanding the remaining determinant;

$= 2(x+y)(-x(x-y)-y^2) = 2(x+y)[-x^2+xy-y^2]$

$= -2(x+y)[x^2-xy+y^2] = -2(x^3+y^3)$ .

Question:10 Evaluate $\begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}$

Answer:

We have determinant $\triangle = \begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}$

Applying row transformations; $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$ then we have then;

$\triangle = \begin{vmatrix} 0 & -y &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}$

Taking out the common factor -y from the row first.

$\triangle = -y\begin{vmatrix} 0 & 1 &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}$

Expanding the remaining determinant;

$-y[1(-x-o)] = xy$

Question:13 Using properties of determinants, prove that

$\begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}=3(a+b+c)(ab+bc+ca)$

Answer:

Given determinant $\triangle = \begin{vmatrix} 3a &-a+b &-a+c \\ -b+c &3b &-b+c \\ -c+a &-c+b &3c \end{vmatrix}$

Applying the column transformation, $C_{1} \rightarrow C_{1} +C_{2}+C_{3}$ we have then;

$\triangle = \begin{vmatrix} a+b+c &-a+b &-a+c \\ a+b+c &3b &-b+c \\ a+b+c &-c+b &3c \end{vmatrix}$

Taking common factor (a+b+c) out from the column first;

$=(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 1 &3b &-b+c \\1 &-c+b &3c \end{vmatrix}$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ , we have then;

$\triangle=(a+b+c) \begin{vmatrix} 1 &-a+b &-a+c \\ 0 &2b+a &a-b \\0 &a-c &2c+a \end{vmatrix}$

Now we can expand the remaining determinant along $C_{1}$ we have;

$\triangle=(a+b+c) [(2b+a)(2c+a)-(a-b)(a-c)]$

$=(a+b+c) [4bc+2ab+2ac+a^2-a^2+ac+ba-bc]$

$=(a+b+c)(3ab+3bc+3ac)$

$=3(a+b+c)(ab+bc+ac)$

Hence proved.

Question:14 Using properties of determinants, prove that

$\begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}=1$

Answer:

Given determinant $\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}$

Applying the row transformation; $R_{2} \rightarrow R_{2}-2R_{1}$ and $R_{3} \rightarrow R_{3}-3R_{1}$ we have then;

$\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&3 & 7+3p \end{vmatrix}$

Now, applying another row transformation $R_{3}\rightarrow R_{3}-3R_{2}$ we have;

$\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&0 & 1 \end{vmatrix}$

We can expand the remaining determinant along $C_{1}$ , we have;

$\triangle = 1\begin{vmatrix} 1 & 2+p\\ 0 &1 \end{vmatrix} = 1(1-0) =1$

Hence the result is proved.

Question:15 Using properties of determinants, prove that

$\begin{vmatrix} \sin \alpha &\cos \alpha &\cos (\alpha +\delta ) \\ \sin \beta & \cos \beta & \cos (\beta +\delta )\\ \sin \gamma &\cos \gamma & \cos (\gamma +\delta ) \end{vmatrix}=0$

Answer:

Given determinant $\triangle = \begin{vmatrix} \sin \alpha &\cos \alpha &\cos (\alpha +\delta ) \\ \sin \beta & \cos \beta & \cos (\beta +\delta )\\ \sin \gamma &\cos \gamma & \cos (\gamma +\delta ) \end{vmatrix}$

Multiplying the first column by $\sin \delta$ and the second column by $\cos \delta$ , and expanding the third column, we get

$\triangle =\frac{1}{\sin \delta \cos \delta} \begin{vmatrix} \sin \alpha \sin \delta &\cos \alpha\cos \delta &\cos \alpha \cos \delta -\sin \alpha \sin \delta \\ \sin \beta\sin \delta & \cos \beta\cos \delta & \cos \beta \cos \delta - \sin \beta\sin \delta \\ \sin \gamma\sin \delta &\cos \gamma\cos \delta & \cos \gamma \cos\delta- \sin \gamma\sin \delta \end{vmatrix}$

Applying column transformation, $C_{1} \rightarrow C_{1}+C_{3}$ we have then;

$\triangle =\frac{1}{\sin \delta \cos \delta} \begin{vmatrix} \cos \alpha \cos \delta &\cos \alpha\cos \delta &\cos \alpha \cos \delta -\sin \alpha \sin \delta \\ \cos \beta\cos \delta & \cos \beta\cos \delta & \cos \beta \cos \delta - \sin \beta\sin \delta \\ \cos \gamma\cos \delta &\cos \gamma\cos \delta & \cos \gamma \cos\delta- \sin \gamma\sin \delta \end{vmatrix}$

Here we can see that two columns $C_{1}\ and\ C_{2}$ are identical.

The determinant value is equal to zero. $\therefore \triangle = 0$

Hence proved.

Question:16 Solve the system of equations

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

Answer:

We have a system of equations;

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

So, we will convert the given system of equations in a simple form to solve the problem by the matrix method;

Let us take, $\frac{1}{x} = a$ , $\frac{1}{y} = b\ and\ \frac{1}{z} = c$

Then we have the equations;

$2a +3b+10c = 4$

$4a-6b+5c =1$

$6a+9b-20c = 2$

We can write it in the matrix form as $AX =B$ , where

$A= \begin{bmatrix} 2 &3 &10 \\ 4& -6 & 5\\ 6 & 9 & -20 \end{bmatrix} , X = \begin{bmatrix} a\\b \\c \end{bmatrix}\ and\ B = \begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}.$

Now, Finding the determinant value of A;

$|A| = 2(120-45)-3(-80-30)+10(36+36)$

$=150+330+720$

$=1200 \neq 0$

Hence we can say that A is non-singular $\therefore$ its invers exists;

Finding cofactors of A;

$A_{11} = 75$ , $A_{12} = 110$ , $A_{13} = 72$

$A_{21} = 150$ , $A_{22} = -100$ , $A_{23} = 0$

$A_{31} =75$ , $A_{31} =30$ , $A_{33} =-24$

$\therefore$ as we know $A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}$

Now we will find the solutions by relation $X = A^{-1}B$ .

$\Rightarrow \begin{bmatrix} a\\b \\ c \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 &150 &75 \\ 110& -100& 30\\ 72&0 &-24 \end{bmatrix}\begin{bmatrix} 4\\1 \\ 2 \end{bmatrix}$

$= \frac{1}{1200}\begin{bmatrix} 300+150+150\\440-100+60 \\ 288+0-48 \end{bmatrix}$

$= \frac{1}{1200}\begin{bmatrix} 600\\400\\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\\ \\ \frac{1}{3} \\ \\ \frac{1}{5} \end{bmatrix}$

Therefore we have the solutions $a = \frac{1}{2},\ b= \frac{1}{3},\ and\ c = \frac{1}{5}.$

Or in terms of x, y, and z;

$x =2,\ y =3,\ and\ z = 5$

An insight to the NCERT solutions for Class 12 Maths Chapter 4 Determinants:

The six exercises of NCERT Class 12 Maths solutions chapter 4 Determinants covers the properties of determinants, co-factors and applications like finding the area of triangle, solutions of linear equations in two or three variables, minors, consistency and inconsistency of system of linear equations, adjoint and inverse of a square matrix, and solution of linear equations in two or three variables using inverse of a matrix.

Topics and sub-topics of NCERT class 12 maths chapter 4 Determinants

4.1 Introduction

4.2 Determinant

4.2.1 Determinant of a matrix of order one

4.2.2 Determinant of a matrix of order two

4.2.3 Determinant of a matrix of order 3 × 3

4.3 Properties of Determinants

4.4 Area of a Triangle

4.5 Minors and Cofactors

4.6 Adjoint and Inverse of a Matrix

4.7 Applications of Determinants and Matrices

4.7.1 Solution of a system of linear equations using the inverse of a matrix

NCERT solutions for class 12 Maths - Chapter wise

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Benefits of NCERT solutions for Class 12 Maths Chapter 4 Determinants:

NCERT Class 12 Maths solutions chapter 4 will assist the students in the exam preparation in a strategic way.
Class 12 Maths Chapter 4 NCERT solutions are prepared by the experts, therefore, students can rely upon the same without any second thought .
NCERT solutions for Class 12 Maths Chapter 4 provides the detailed solution for all the questions. This will help the students in analysing and understanding the questions in a better way.

Solutions

Frequently Asked Question (FAQs) - NCERT solutions for Class 12 Maths Chapter 4 Determinants

Question: What are the important topics in chapter determinants?

Answer:

Determinant of a matrix of order upto three, properties of determinants, area of a triangle, minors and co-factors, adjoint and inverse of a matrix, applications of determinants and matrices, and solution of a system of linear equations using the inverse of a matrix are the important topics from this chapter.

Question: What is the weightage of the chapter determinants for CBSE board exam?

Answer:

The topic algebra which contains two topics matrices and determinants which has 13 % weightage in the maths CBSE 12th board final examination.

Question: How are the NCERT solutions helpful in the board exam?

Answer:

Only knowing the answer does not guarantee to score good marks in the exam. One should know how to answer in order to get good marks. NCERT solutions are provided by the experts who know how best to write answers in the board exam in order to get good marks in the board exam.

Question: Which is the best book for CBSE class 12 Maths ?

Answer:

NCERT textbook is the best book for CBSE class 12 maths. Most of the questions in CBSE class 12 board exam are directly asked from NCERT textbook. So you don't need to buy any supplementary books for CBSE class 12 maths.

Question: Does CBSE provide the solutions of NCERT class 12 Maths ?

Answer:

No, CBSE does not provide NCERT solutions for any class or subject.

Question: Where can I find the complete solutions of NCERT class 12 Maths ?

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NCERT solutions for Class 12 Maths Chapter 4 Determinants

Ncert Solution of Math Class 12 Chapter 4

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Ncert Solution of Math Class 12 Chapter 4

NCERT solutions for class 12 maths chapter-4 Determinants: Excercise- 4.1

NCERT solutions for class 12 maths chapter -4 Determinants: Excercise - 4.2

NCERT solutions for class 12 maths chapter 4 Determinants-Excercise: 4.4

NCERT solutions for class 12 chapter 4 Determinants: Excercise- 4.6

NCERT solutions for class 12 maths chapter 4 Determinants: Miscellaneous exercise

An insight to the NCERT solutions for Class 12 Maths Chapter 4 Determinants:

NCERT solutions for class 12 Maths - Chapter wise

Benefits of NCERT solutions for Class 12 Maths Chapter 4 Determinants:

Frequently Asked Question (FAQs) - NCERT solutions for Class 12 Maths Chapter 4 Determinants

Question: What are the important topics in chapter determinants?

Question: What is the weightage of the chapter determinants for CBSE board exam?

Question: How are the NCERT solutions helpful in the board exam?

Question: Which is the best book for CBSE class 12 Maths ?

Question: Does CBSE provide the solutions of NCERT class 12 Maths ?

Question: Where can I find the complete solutions of NCERT class 12 Maths ?

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