# NCERT Solutions for Class 12 Maths Chapter 4 Determinants

NCERT Solutions for Class 12 Maths Chapter 4 Determinants: In our previous chapter, you have already learnt about matrices and properties of matrices. In this article, you will find NCERT solutions for class 12 maths chapter 4 determinants. You know that if you multiply a matrix with coordinates of a point, it will give a new point in the space. In this sense, the matrix is a linear transformation. The determinant of the matrix is the factor by which its volume blow up. For example, if the determinant is 1 which means its volume is unchanged, if the determinant is 2 means the volume is doubled after transformation. What does the physical meaning of the determinant is negative or zero mean? Why is the inverse of the matrix is not possible if its determinant is zero? You will get all these answers in the CBSE NCERT solutions for class 12 maths chapter 4 determinants article. The Important topics are determinants and their properties, finding the area of the triangle, minor and cofactors, adjoint and the inverse of the matrix, and applications of determinants like solving the system of linear equations etc are covered in this chapter. In the solutions of NCERT for class 12 maths chapter 4 determinants article, you will get detailed explanations to all these above topics.  The practice of NCERT questions is very important to get a command on this chapter otherwise you will get confused with the formulas of this chapter. You should solve every problem on your own, if you are finding difficulties, you can take help with these solutions of NCERT for class 12 maths chapter 4 determinants. Check all NCERT solutions at a single place which will help you to learn CBSE maths.

The topic algebra which contains two topics matrices and determinants has 13 % weightage in the CBSE 12th board final examination, which means 10 marks questions out of 80 marks will be asked from these two chapters matrices and determinants in the final examination. The determinant is an important part of matrices. In the solutions of NCERT class 12 maths chapter 4 determinants, you will be dealing with determinants of order up to three only. In this chapter, there are 6 exercises with 68 questions. All these questions are prepared and explained in this NCERT solutions for class 12 maths chapter 4 determinants article.

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, cofactors 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

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

The determinant is evaluated as follows

$\begin{vmatrix} 2 & 4\\ -5 & -1\end{vmatrix} = 2(-1) - 4(-5) = -2 + 20 = 18$

The given two by two determinant is calculated as follows

$\dpi{100} \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix} = cos \theta (\cos \theta) - (-\sin \theta)\sin \theta = \cos^2\theta + \sin ^2 \theta = 1$

We have determinant $\dpi{100} \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}$

So, $\dpi{100} \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix} = (x^2-x+1)(x+1) - (x-1)(x+1)$

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

$=x^3-2x^2+2x +x^2-2x+2$

$= x^3-x^2+2$

Given determinant $\dpi{100} A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then we have to show that $\dpi{100} | 2 A |=4|A|$,

So, $\dpi{100} A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then, $\dpi{100} 2A =2 \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} = \begin{bmatrix} 2 & 4\\ 8 &4 \end{bmatrix}$

Hence we have $\dpi{100} \left | 2A \right | = \begin{vmatrix} 2 &4 \\ 8& 4 \end{vmatrix} = 2(4) - 4(8) = -24$

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

then calculating R.H.S. $\dpi{100} 4\left | A \right |$

We have,

$\dpi{100} \left | A \right | = \begin{vmatrix} 1 &2 \\ 4& 2 \end{vmatrix} = 1(2) - 2(4) = -6$

hence R.H.S becomes $\dpi{100} 4\left | A \right | = 4\times(-6) = -24$

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

Hence proved.

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 $\dpi{100} 27|A|$,

$\dpi{100} |A| = \begin{vmatrix} 1 & 0 &1 \\ 0 & 1 & 2\\ 0& 0 &4 \end{vmatrix} = 1\begin{vmatrix} 1 &2 \\ 0 & 4 \end{vmatrix} - 0\begin{vmatrix} 0 &2 \\ 0& 4 \end{vmatrix} + 1\begin{vmatrix} 0 &1 \\ 0& 0 \end{vmatrix} = 4 -0 + 0 = 4$

So, $\dpi{100} 27|A| = 27(4) = 108$

Therefore $\dpi{100} |3A|=27|A|$.

Hence proved.

Question:5(i) Evaluate the determinants.

$\dpi{100} \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix}$

Given the determinant $\dpi{100} \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix}$;

now, calculating its determinant value,

$\dpi{100} \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix} = 3\begin{vmatrix} 0 &-1 \\ -5& 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 &-1 \\ 3& 0 \end{vmatrix} +(-2)\begin{vmatrix} 0 &0 \\ 3& -5 \end{vmatrix}$

$\dpi{100} = 3(0-5)+1(0+3) -2(0-0) = -15+3-0 = -12$.

Question:5(ii) Evaluate the determinants.

$\dpi{100} \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix}$

Given determinant $\dpi{100} \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix}$;

Now calculating the determinant value;

$\dpi{100} \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix} = 3\begin{vmatrix} 1 &-2 \\ 3&1 \end{vmatrix} -(-4)\begin{vmatrix} 1 &-2 \\ 2& 1 \end{vmatrix}+5\begin{vmatrix} 1 & 1\\ 2& 3 \end{vmatrix}$

$= 3(1+6) +4(1+4) +5(3-2) = 21+20+5 = 46$.

Question:5(iii) Evaluate the determinants.

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

Given determinant $\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix}$;

Now calculating the determinant value;

$\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix} = 0\begin{vmatrix} 0 &-1 \\ 3& 0 \end{vmatrix} -1\begin{vmatrix} -1 &-3 \\ -2& 0 \end{vmatrix}+2\begin{vmatrix} -1 &0 \\ -2& 3 \end{vmatrix}$

$= 0 - 1(0-6)+2(-3-0) = 6 -6 =0$

Question:5(iv) Evaluate the determinants.

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

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

We now calculate determinant value:

$\begin{vmatrix}2 &-1 &-2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix} =2\begin{vmatrix} 2 &-1 \\ -5 & 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 &-1 \\ 3 & 0 \end{vmatrix}+(-2)\begin{vmatrix} 0 &2 \\ 3&-5 \end{vmatrix}$

$=2(0-5)+1(0+3)-2(0-6) = -10+3+12 = 5$

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

$\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}$

Given that $\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}$

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

So, we have then,

$\dpi{100} -18= 2x^2-24$      or    $\dpi{100} 3= x^2$       or   $\dpi{100} x= \pm \sqrt{3}$

Question:7(ii) Find values of x, if

$\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$

Given  $\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$;

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;

$\dpi{100} \begin{vmatrix}x &3 \\2x &5 \end{vmatrix} = 5(x) - 3(2x) = 5x - 6x =-x$

So, we have then;

$\dpi{100} -2 = -x$    or    $\dpi{100} x =2$.

(A) $\dpi{100} 6$        (B) $\dpi{100} \pm 6$       (C) $\dpi{100} -6$     (D) $\dpi{100} 0$

Solving the L.H.S. determinant ;

$\dpi{100} \begin{vmatrix}x &2 \\18 &x \end{vmatrix}= x^2 - 36$

and solving R.H.S determinant;

$\dpi{100} \begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix} = 36-36 = 0$

So equating both sides;

$\dpi{100} x^2 - 36 =0$        or     $\dpi{100} x^2 = 36$        or     $\dpi{100} x = \pm 6$

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

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

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.

$\dpi{100} \begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0$

Given determinant $\dpi{100} \triangle =\begin{vmatrix}a-b &b-c &c-a \\b-c &c-a &a-b \\c-a &a-b &b-c \end{vmatrix}=0$

Applying the rows addition $R_{1} \rightarrow R_{1}+R_{2}$   then we have;

$\dpi{100} \triangle =\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\-(a-c) &-(b-a) &-(c-b) \end{vmatrix}=0$

$\dpi{100} =-\begin{vmatrix}a-c &b-a &c-b \\b-c &c-a &a-b \\(a-c) &(b-a) &(c-b) \end{vmatrix}=0$

So, we have two rows $\dpi{100} R_{1}$ and $\dpi{100} R_{2}$ identical hence we can say that the value of determinant = 0

Therefore $\dpi{100} \triangle = 0$.

$\dpi{100} \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix}=0$

Given determinant $\dpi{100} \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix}$

So, we can split it in two addition determinants:

$\dpi{100} \begin{vmatrix}2 & 7 &65 \\3 &8 &75 \\5 &9 &86 \end{vmatrix} = \begin{vmatrix} 2 &7 &63+2 \\ 3& 8 &72+3 \\ 5& 9 & 81+5 \end{vmatrix}$

$\dpi{100} \begin{vmatrix} 2 &7 &63+2 \\ 3& 8 &72+3 \\ 5& 9 & 81+5 \end{vmatrix} = \begin{vmatrix} 2 & 7 &2 \\ 3& 8& 3\\ 5 & 9 & 5 \end{vmatrix} + \begin{vmatrix} 2 & 7 &63 \\ 3& 8 &72 \\ 5 & 9 & 81 \end{vmatrix}$

$\dpi{100} \begin{vmatrix} 2 & 7 &2 \\ 3& 8& 3\\ 5 & 9 & 5 \end{vmatrix} = 0$     [$\dpi{100} \because$ Here two columns are identical ]

and $\dpi{100} \begin{vmatrix} 2 & 7 &63 \\ 3& 8 &72 \\ 5 & 9 & 81 \end{vmatrix} = \begin{vmatrix} 2 & 7 &9(7) \\ 3& 8 &9(8) \\ 5 &9 & 9(9) \end{vmatrix} = 9 \begin{vmatrix} 2 & 7 &7 \\ 3& 8& 8\\ 5& 9&9 \end{vmatrix}$    [$\dpi{100} \because$ Here two columns are identical ]

$\dpi{100} = 0$

Therefore we have the value of determinant = 0.

$\dpi{100} \begin{vmatrix}1 &bc &a(b+c) \\1 &ca &b(c+a) \\1 &ab & c(a+b) \end{vmatrix}=0$

We have determinant:

$\triangle = \begin{vmatrix} 1 &bc &a(b+c) \\ 1& ca &b(c+a) \\ 1& ab &c(a+b) \end{vmatrix}$

Applying $C_{3} \rightarrow C_{2} + C_{3}$ we have then;

$\triangle = \begin{vmatrix} 1 &bc & ab+bc+ca \\ 1& ca &ab+bc+ca \\ 1& ab &ab+bc+ca \end{vmatrix}$

So, here column 3 and column 1 are proportional.

Therefore, $\triangle = 0$.

$\dpi{100} \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}=2\begin{vmatrix} a &p &x \\ b &q &y \\ c &r & z \end{vmatrix}$

Given determinant :

$\dpi{100} \triangle= \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a+b &p+q & x+y \end{vmatrix}$

Splitting the third row; we get,

$\dpi{100} = \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ a &p & x \end{vmatrix} + \begin{vmatrix}b+c &q+r &y+z \\ c+a & r+p &z+x \\ b &q & y \end{vmatrix} = \triangle_{1} + \triangle_{2}\ (assume\ that)$.

Then we have,

$\dpi{100} \triangle_{1} = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a &p & x \end{vmatrix}$

On Applying row transformation $\dpi{100} R_{2} \rightarrow R_{2} - R_{3}$   and then  $\dpi{100} R_{1} \rightarrow R_{1} - R_{2}$;

we get, $\dpi{100} \triangle_{1} = \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}$

Applying Rows exchange transformation $\dpi{100} R_{1} \leftrightarrow R_{2}$   and   $\dpi{100} R_{2} \leftrightarrow R_{3}$, we have:

$\dpi{100} \triangle_{1} =(-1)^2 \begin{vmatrix} b & q & y \\ c & r & z \\ a &p & x \end{vmatrix}= \begin{vmatrix} a & p & x\\ b & q&y \\ c& r & z \end{vmatrix}$

also $\dpi{100} \triangle_{2} = \begin{vmatrix} b+c & q+r & y+z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$

On applying rows transformation, $\dpi{100} R_{1} \rightarrow R_{1} - R_{3}$ and then $\dpi{100} R_{2} \rightarrow R_{2} - R_{1}$

$\dpi{100} \triangle_{2} = \begin{vmatrix} c & r & z \\ c+a&r+p &z+x \\ b & q & y \end{vmatrix}$  and then  $\dpi{100} \triangle_{2} = \begin{vmatrix} c & r & z \\ a&p &x \\ b & q & y \end{vmatrix}$

Then applying rows exchange transformation;

$\dpi{100} R_{1} \leftrightarrow R_{2}$   and then $\dpi{100} R_{2} \leftrightarrow R_{3}$. we have then;

$\dpi{100} \triangle_{2} =(-1)^2 \begin{vmatrix} a & p & x \\ b&q &y \\ c & r & z \end{vmatrix}$

So, we now calculate the sum = $\dpi{100} \triangle_{1} + \triangle _{2}$

$\dpi{100} \triangle_{1} + \triangle _{2} = 2 \begin{vmatrix} a &p &x \\ b& q& y\\ c & r& z \end{vmatrix}$

Hence proved.

$\dpi{100} \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}=0$

We have given determinant

$\dpi{100} \triangle = \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}$

Applying transformation, $\dpi{100} R_{1} \rightarrow cR_{1}$ we have then,

$\dpi{100} \triangle = \frac{1}{c}\begin{vmatrix} 0 &ac &-bc \\-a &0 & -c\\b &c &0 \end{vmatrix}$

We can make the first row identical to the third row so,

Taking another row transformation: $\dpi{100} R_{1} \rightarrow R_{1}-bR_{2}$ we have,

$\dpi{100} \triangle = \frac{1}{c}\begin{vmatrix} ab &ac &0 \\-a &0 & -c\\b &c &0 \end{vmatrix} = \frac{a}{c} \begin{vmatrix} b &c &0 \\-a &0 & -c\\b &c &0 \end{vmatrix}$

So, determinant has two rows $\dpi{100} R_{1}\ and\ R_{3}$ identical.

Hence $\dpi{100} \triangle = 0$.

$\dpi{100} \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}=4a^2b^2c^2$

Given determinant : $\dpi{100} \begin{vmatrix} -a^2 &ab &ac \\ ba &-b^2 &bc \\ ca & cb & -c^2 \end{vmatrix}$

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

As we can easily take out the common factors a,b,c from rows $R_{1},R_{2},R_{3}$ respectively.

So, get then:

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

Now, taking common factors a,b,c from the columns $C_{1},C_{2},C_{3}$ respectively.

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

Now, applying rows transformations $R_{1} \rightarrow R_{1} + R_{2}$   and then $R_{3} \rightarrow R_{2} + R_{3}$ we have;

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

Expanding to get R.H.S.

$\triangle = a^2b^2c^2 \left ( 2\begin{vmatrix} 1 &-1 \\ 2& 0 \end{vmatrix} \right ) = 2a^2b^2c^2(0+2) =4a^2b^2c^2$

Question:8(i) By using properties of determinants, show that:

We have the determinant $\dpi{100} \begin{vmatrix} 1 &a &a^2 \\ 1 &b &b^2 \\ 1 &c &c^2 \end{vmatrix}$

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

$\dpi{100} \triangle = \begin{vmatrix} 0 &a-b &a^2-b^2 \\ 0 &b-c &b^2-c^2 \\ 1 &c &c^2 \end{vmatrix}$

$\dpi{100} = \begin{vmatrix} 0 &a-b &(a-b)(a+b) \\ 0 &b-c &(b-c)(b+c) \\ 1 &c &c^2 \end{vmatrix} = (a-b)(b-c)\begin{vmatrix} 0 &1 &(a+b) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}$

Now, applying $R_{1} \rightarrow R_{1} -R_{2}$ we have:

$\dpi{100} = (a-b)(b-c)\begin{vmatrix} 0 &0 &(a-c) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}$    or    $\dpi{100} = (a-b)(b-c)(a-c)\begin{vmatrix} 0 &0 &1 \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix} =(a-b)(b-c)(a-c)\begin{vmatrix} 0 &1 \\ 1 & c \end{vmatrix}$

$\dpi{100} = (a-b)(b-c)(c-a)$

Hence proved.

Question:8(ii) By using properties of determinants, show that:

$\dpi{100} \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)$

Given determinant :

$\dpi{100} \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}$,

Applying column transformation $\dpi{100} C_{1} \rightarrow C_{1}-C_{3}$ and then $\dpi{100} C_{2} \rightarrow C_{2}-C_{3}$

We get,

$\dpi{100} \triangle =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ a^3-c^3 &b^3-c^3 & c^3 \end{vmatrix}$

$\dpi{100} =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ (a-c)(a^2+ac+c^2) &(b-c)(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 1& 1 & c \\ (a^2+ac+c^2) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

Now, applying column transformation $\dpi{100} C_{1} \rightarrow C_{1} - C_{2}$, we have:

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a^2-b^2+ac-bc) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a-b)(a+b+c) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)(a-b)(a+b+c)\begin{vmatrix} 0&1 \\ 1& c \end{vmatrix}$

$\dpi{100} =-(a-c)(b-c)(a-b)(a+b+c) = (a-b)(b-c)(c-a)(a+b+c)$

Hence proved.

$\dpi{100} \begin{vmatrix} x & x^2 & yz\\ y & y^2 &zx \\ z & z^2 & xy \end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$

We have the determinant:

$\triangle = \begin{vmatrix} x & x^2 & yz\\ y & y^2 &zx \\ z & z^2 & xy \end{vmatrix}$

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

$\triangle = \begin{vmatrix} x-z & x^2-z^2 & yz-xy\\ y-z & y^2-z^2 &zx-xy \\ z & z^2 & xy \end{vmatrix}$

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

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

Now, applying $R_{1} \rightarrow R_{1} - R_{2}$; we have

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

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

Now, expanding the remaining determinant;

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

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

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

Hence proved.

Question:10(i) By using properties of determinants, show that:

$\dpi{100} \begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}=(5x+4)(4-x)$

Given determinant:

$\begin{vmatrix} x+4 &2x &2x \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}$

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

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

Taking a common factor: 5x+4

$= (5x+4)\begin{vmatrix} 1 &1 &1 \\ 2x & x+4 & 2x\\ 2x & 2x & x+4 \end{vmatrix}$

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

$= (5x+4)\begin{vmatrix} 0 &0 &1 \\ x-4 & 4-x & 2x\\ 0 & x-4 & x+4 \end{vmatrix}$

$= (5x+4)(4-x)(4-x)\begin{vmatrix} 0 &0 &1 \\ 1 & 1 & 2x\\ 0 & 1 & x+4 \end{vmatrix}$

$= (5x+4)(4-x)^2$

Question:10(ii) By using properties of determinants, show that:

Given determinant:

$\triangle = \begin{vmatrix} y+k & y & y\\ y & y+k &y \\ y & y & y+k \end{vmatrix}$

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

$= \begin{vmatrix} 3y+k & 3y+k & 3y+k\\ y & y+k &y \\ y & y & y+k \end{vmatrix}$

$=(3y+k) \begin{vmatrix}1 & 1 & 1\\ y & y+k &y \\ y & y & y+k \end{vmatrix}$                        [taking common (3y + k) factor]

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

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

$=(3y+k)(k^2) \begin{vmatrix}0 & 0 & 1\\ -1 & 1 &y \\ 0 & -1 & y+k \end{vmatrix}$

$=k^2 (3y+k)$

Hence proved.

Question:11(i) By using properties of determinants, show that:

$\dpi{100} \begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}=(a+b+c)^3$

Given determinant:

$\triangle = \begin{vmatrix} a-b-c &2a &2a \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}$

We apply row transformation: $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ we have;

$= \begin{vmatrix} a+b+c &a+b+c &a+b+c \\ 2b &b-c-a &2b \\ 2c &2c &c-a-b \end{vmatrix}$

Taking common factor (a+b+c) out.

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

Now, applying column tranformation $C_{1} \rightarrow C_{1}- C_{2}$    and then  $C_{2} \rightarrow C_{2}- C_{3}$

We have;

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

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

$=(a+b+c)(a+b+c)(a+b+c) = (a+b+c)^3$

Hence Proved.

Question:11(ii) By using properties of determinants, show that:

$\dpi{100} \begin{vmatrix} x+y+2z &x &y \\ z & y+z+2x & y\\ z & x &z+x+2y \end{vmatrix}=2(x+y+z)^3$

Given determinant

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

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

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

Taking 2(x+y+z) factor out, we get;

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

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

we get;

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

$=2(x+y+z)^3\begin{vmatrix} 0 &-1 &0 \\ 0 & 1 & -1\\ 1 & x &z+x+2y \end{vmatrix}$

$=2(x+y+z)^3\begin{vmatrix} -1 &0 \\ 1& -1 \end{vmatrix} = 2(x+y+z)^3$

Hence proved.

$\dpi{100} \begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}=(1-x^3)^2$

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) 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.

$\dpi{100} \begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}=(1+a^2+b^2)^3$

We have determinant:

$\dpi{100} \triangle = \begin{vmatrix} 1+a^2-b^2 &2ab &-2b \\ 2ab &1-a^2+b^2 &2a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

Applying row transformations, $\dpi{100} R_{1} \rightarrow R_{1} +bR_{3}$    and   $\dpi{100} R_{2} \rightarrow R_{2} -aR_{3}$ then we have;

$\dpi{100} = \begin{vmatrix} 1+a^2+b^2 &0 &-b(1+a^2+b^2) \\ 0 &1+a^2+b^2 &a(1+a^2+b^2) \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

taking common factor out of the determinant;

$\dpi{100} = (1+a^2+b^2)^2\begin{vmatrix} 1 &0 &-b \\ 0 &1 &a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}$

Now expanding the remaining determinant we get;

$\dpi{100} = (1+a^2+b^2)^2\left [ (1)\begin{vmatrix} 1& a\\ -2a&1-a^2-b^2 \end{vmatrix} - b\begin{vmatrix} 0&1 \\ 2b&-2a \end{vmatrix}\right ]$

$\dpi{100} = (1+a^2+b^2)^2\left [ 1-a^2-b^2+2a^2-b(-2b)\right ]$

$\dpi{100} = (1+a^2+b^2)^2\left [ 1+a^2+b^2\right ] = (1+a^2+b^2)^3$

Hence proved.

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

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 C1,C2,and C3.

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.

(A) $k|A|$          (B) $k^2|A|$        (C) $k^3|A|$        (D)  $3k|A|$

Assume a square matrix A of order of $3\times3$.

$A = \begin{bmatrix} a_1 & b_1&c_1 \\ a_2& b_2& c_2\\ a_3& b_3 & c_3 \end{bmatrix}$

Then we have;

$kA = \begin{bmatrix} ka_1 & kb_1&kc_1 \\ ka_2& kb_2& kc_2\\ ka_3& kb_3 & kc_3 \end{bmatrix}$

(Taking the common factors k from each row.)

$|kA| = \begin{vmatrix} ka_1 & kb_1&kc_1 \\ ka_2& kb_2& kc_2\\ ka_3& kb_3 & kc_3 \end{vmatrix} = k^3 \begin{vmatrix} a_1 & b_1&c_1 \\a_2& b_2& c_2\\ a_3& b_3 & c_3 \end{vmatrix}$

$= k^3 |A|$

Therefore correct option is (C).

Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these

The answer is (C) Determinant is a number associated to a square matrix.

As we know that To every square matrix $A = [a_{ij}]$of order n, we can associate a number (real or complex) called determinant of the square matrix A, where $a_{ij} = (i, j)^{th}$ element of A.

## NCERT solutions for class 12 maths chapter 4 Determinants: Excercise-4.3

$(1,0), (6,0), (4,3)$

We can find the area of the triangle with vertices $(1,0), (6,0), (4,3)$ by the following determinant relation:

$\triangle =\frac{1}{2} \begin{vmatrix} 1& 0 &1 \\ 6 & 0 &1 \\ 4& 3& 1 \end{vmatrix}$

Expanding using second column

$=\frac{1}{2} (-3) \begin{vmatrix} 1 &1 & \\ 6& 1 & \end{vmatrix}$

$= \frac{15}{2}\ square\ units.$

$(2,7), (1,1), (10,8)$

We can find the area of the triangle with given coordinates by the following method:

$\triangle = \begin{vmatrix} 2 &7 &1 \\ 1 & 1& 1\\ 10& 8 &1 \end{vmatrix}$

$=\frac{1}{2} \begin{vmatrix} 2 &7 &1 \\ 1 & 1& 1\\ 10& 8 &1 \end{vmatrix} = \frac{1}{2}\left [ 2(1-8)-7(1-10)+1(8-10) \right ]$

$= \frac{1}{2}\left [ 2(-7)-7(-9)+1(-2) \right ] = \frac{1}{2}\left [ -14+63-2 \right ] = \frac{47}{2}\ square\ units.$

$(-2,-3), (3,2), (-1,-8)$

Area of the triangle by the determinant method:

$Area\ \triangle = \frac{1}{2} \begin{vmatrix} -2 &-3 &1 \\ 3& 2 & 1\\ -1& -8 & 1 \end{vmatrix}$

$=\frac{1}{2}\left [ -2(2+8)+3(3+1)+1(-24+2) \right ]$

$=\frac{1}{2}\left [ -20+12-22 \right ] = \frac{1}{2}[-30]= -15$

Hence the area is equal to $|-15| = 15\ square\ units.$

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.

$(k,0), (4,0), (0,2)$

We can easily calculate the area by the formula :

$\triangle = \frac{1}{2} \begin{vmatrix} k &0 &1 \\ 4& 0& 1\\ 0 &2 & 1 \end{vmatrix} = 4\ sq.\ units$

$= \frac{1}{2}\left [ k\begin{vmatrix} 0 &1 \\ 2& 1 \end{vmatrix} -0\begin{vmatrix} 4 &1 \\ 0 & 1 \end{vmatrix}+1\begin{vmatrix} 4 &0 \\ 0& 2 \end{vmatrix} \right ]= 4\ sq.\ units$

$=\frac{1}{2}\left [ k(0-2)-0+1(8-0) \right ] = \frac{1}{2}\left [ -2k+8 \right ] = 4\ sq.\ units$

$\left [ -2k+8 \right ] = 8\ sq.\ units$    or   $-2k +8 = \pm 8\ sq.\ units$

or  $k = 0$    or  $k = 8$

Hence two values are possible for k.

$(-2,0), (0,4), (0,k)$

The area of the triangle is given by the formula:

$\triangle = \frac{1}{2} \begin{vmatrix} -2 &0 &1 \\ 0 & 4 & 1\\ 0& k & 1 \end{vmatrix} = 4\ sq.\ units.$

Now, calculating the area:

$= \frac{1}{2} \left | -2(4-k)-0(0-0)+1(0-0) \right | = \frac{1}{2} \left | -8+2k \right | = 4$

or  $-8+2k =\pm 8$

Therefore we have two possible values of 'k' i.e.,  $k = 8$  or  $k = 0$.

As we know the line joining $\small (1,2)$ ,$\small (3,6)$  and let say a point on line $A\left ( x,y \right )$ will be collinear.

Therefore area formed by them will be equal to zero.

$\triangle = \frac{1}{2}\begin{vmatrix} 1 &2 &1 \\ 3& 6 &1 \\ x & y &1 \end{vmatrix} = 0$

So, we have:

$=1(6-y)-2(3-x)+1(3y-6x) = 0$

or $6-y-6+2x+3y-6x = 0 \Rightarrow 2y-4x=0$

Hence, we have the equation of line $\Rightarrow y=2x$.

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  $x-3y = 0$.

Area of triangle is given by:

$\triangle = \frac{1}{2} \begin{vmatrix} 2 &-6 &1 \\ 5& 4 & 1\\ k& 4& 1 \end{vmatrix} = 35\ sq.\ units.$

or $\begin{vmatrix} 2 &-6 &1 \\ 5& 4 & 1\\ k& 4& 1 \end{vmatrix} = 70\ sq.\ units.$

$2\begin{vmatrix} 4 &1 \\ 4& 1 \end{vmatrix}-(-6)\begin{vmatrix} 5 &1 \\ k &1 \end{vmatrix}+1\begin{vmatrix} 5 &4 \\ k&4 \end{vmatrix} = 70$

$2(4-4) +6(5-k)+(20-4k) = \pm70$

$50-10k = \pm70$

$k = 12$ or  $k = -2$

Hence the possible values of k are 12 and -2.

Therefore option (D) is correct.

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

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

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:

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

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

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

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

$\small \begin{vmatrix} a &c \\ b &d \end{vmatrix}$

GIven determinant: $\begin{vmatrix} a &c \\ b &d \end{vmatrix}$

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

Therefore we have

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

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

$M_{21}$ = minor of element $a_{21}$ = c

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

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

Therefore we have:

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

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

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

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

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

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 $a_{33} = (-1)^{3+3}M_{33} = 1$.

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

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 $a_{33} = (-1)^{3+3}M_{33} = 5$.

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,

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

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 $\triangle = (x-y)(y-z)(z-x)$.

## CBSE NCERT solutions for class 12 maths chapter 4 Determinants- Excercise: 4.5

Question:1 Find adjoint of each of the matrices.

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

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

Then we have,

$A_{11} = 4, A_{12}=-(1)3, A_{21} = -(1)2,\ and\ A_{22}= 1$

Hence we get:

$adjA = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} &A_{22} \end{bmatrix}^T = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} &A_{22} \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -3 &1 \end{bmatrix}$

Question:2  Find adjoint of each of the matrices

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

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

Then we have,

$A_{11} = (-1)^{1+1}\begin{vmatrix} 3 &5 \\ 0& 1 \end{vmatrix} =(3-0)= 3$

$A_{12} = (-1)^{1+2}\begin{vmatrix} 2 &5 \\ -2& 1 \end{vmatrix} =-(2+10)= -12$

$A_{13} = (-1)^{1+3}\begin{vmatrix} 2 &3 \\ -2& 0 \end{vmatrix} =0+6= 6$

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

$A_{22} = (-1)^{2+2}\begin{vmatrix} 1 &2 \\ -2& 1 \end{vmatrix} =(1+4)= 5$

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

$A_{31} = (-1)^{3+1}\begin{vmatrix} -1 &2 \\ 3& 5 \end{vmatrix} =(-5-6)= -11$

$A_{32} = (-1)^{3+2}\begin{vmatrix} 1 &2 \\2& 5\end{vmatrix} =-(5-4)= -1$

$A_{33} = (-1)^{3+3}\begin{vmatrix} 1 &-1 \\ 2& 3 \end{vmatrix} =(3+2)= 5$

Hence we get:

$adjA = \begin{bmatrix} A_{11} &A_{21} &A_{31} \\ A_{12}&A_{22} &A_{32} \\ A_{13}&A_{23} &A_{33} \end{bmatrix} = \begin{bmatrix} 3 &1 &-11 \\ -12&5 &-1 \\ 6&2 &5 \end{bmatrix}$

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

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

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

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

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,

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

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

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:

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

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

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

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:

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

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

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

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:

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

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

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

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:

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

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

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

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:

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

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

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

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:

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

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

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

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

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:

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

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

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:

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

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

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

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.

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 $|A| \neq 0$]

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

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

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

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

Given $\small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix}$ then we have the relation $A^2+aA+bI=O$

So, calculating each term;

$A^2 = \begin{bmatrix} 3& 2\\ 1& 1 \end{bmatrix}\begin{bmatrix} 3&2 \\ 1& 1 \end{bmatrix} = \begin{bmatrix} 9+2 &6+2 \\ 3+1&2+1 \end{bmatrix} = \begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}$

therefore  $A^2+aA+bI=O$;

$=\begin{bmatrix}11 &8 \\ 4& 3 \end{bmatrix} + a\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + b \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

$\begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$

So, we have equations;

$11+3a+b = 0,\ 8+2a = 0$   and $4+a = 0,and\ \ 3+a+b = 0$

We get $a = -4\ and\ b= 1$.

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;

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

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

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

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 :

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

(A) $\small |A|$      (B) $\small |A|^2$      (C) $\small |A|^3$      (D) $\small 3|A|$

We know the identity $(adjA)A = |A| I$

Hence we can determine the value of $|(adjA)|$.

Taking both sides determinant value we get,

$|(adjA)A| = ||A| I|$     or   $|(adjA)||A| = ||A||| I|$

or taking R.H.S.,

$||A||| I| = \begin{vmatrix} |A| & 0&0 \\ 0&|A| &0 \\ 0&0 &|A| \end{vmatrix}$

$= |A| (|A|^2) = |A|^3$

or, we have then $|(adjA)||A| = |A|^3$

Therefore $|(adjA)| = |A|^2$

Hence the correct answer is B.

(A) $\small det(A)$       (B)  $\small \frac{1}{det (A)}$       (C)  $\small 1$       (D) $\small 0$

Given that the matrix is invertible hence $A^{-1}$ exists and $A^{-1} = \frac{1}{|A|}adjA$

Let us assume a matrix of the order of 2;

$A = \begin{bmatrix} a &b \\ c &d \end{bmatrix}$.

Then $|A| = ad-bc$.

$adjA = \begin{bmatrix} d &-b \\ -c & a \end{bmatrix}$    and  $|adjA| = ad-bc$

Now,

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

Taking determinant both sides;

$|A^{-1}| = |\frac{1}{|A|}adjA| = \begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}$

$\therefore|A^{-1}| = \begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix} = \frac{1}{|A|^2}\begin{vmatrix} d &-b \\ -c& a \end{vmatrix} = \frac{1}{|A|^2}(ad-bc) =\frac{1}{|A|^2}.|A| = \frac{1}{|A|}$

Therefore we get;

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

Hence the correct answer is B.

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

$\small x+2y=2$

$\small 2x+3y=3$

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.

$\small 2x-y=5$

$\small x+y=4$

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.

$\small x+3y=5$

$\small 2x+6y=8$

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.

$\small x+y+z=1$

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

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

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.

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

$\small 2y-z=-1$

$\small 3x-5y=3$

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.

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

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

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

$\small 5x-y+4z=5$
$\small 2x+3y+5z=2$
$\small 5x-2y+6z=-1$