# NCERT Solutions For Class 12 Maths Chapter 4

## NCERT Solutions Class 12 Maths Determinants Matrix Properties

### Ncert Solutions For Class 12 Maths Chapter 4 PDF Free Download

NCERT solutions for class 12 maths Chapter 4 – determinants matrix properties given here contain exercise wise solutions of the NCERT textbooks questions. Class 12 maths chapter 4 covers topics such as how to calculate determinants, different properties of determinants and more. Thus, for students of class 12 we have provided NCERT Solutions to help them test their knowledge with the intext questions and solutions based on which they can access their knowledge and learn what they do not know as well. NCERT class 12 maths solutions contains solutions for a variety of different problems related to the topic of determinants. There are a lot of different types of problems that can be asked in the examination.

### NCERT Solutions Class 12 Maths Chapter 4 Exercises

Exercise 4.1

Q1: Evaluate the determinants.

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

Ans:

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

Q2: Evaluate the determinant.

(i) $\begin{vmatrix} cos \Theta & -sin \Theta \\ sin \Theta & cos \Theta \end{vmatrix}\\$

(ii) $\\\begin{vmatrix} x^{ 2 } – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix}$

Ans:

(i) (cos Ɵ )(cos Ɵ ) – ( -sin Ɵ ) (sin Ɵ ) = cos2 Ɵ + sin2 Ɵ = 1

(ii) = ( x2 – x + 1 ) ( x + 1 ) – ( x – 1 ) ( x + 1 )

= x3 – x2 + x + x2 – x + 1 – ( x2 – 1 )

= x3 + 1 – x2 + 1

= x3 – x2 + 2

Q3: If  A = $\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$ , then show that |2A| = 4 |A|

Ans: The given matrix is A = $\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$

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

Therefore, L.H.S. = |2A| = $\begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix}$ = 2 × 4 – 4 × 8 = 8 – 32 = -24

Now, |A| = $\begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix}$ =

1 × 2 – 2 × 4 = 2 – 8 = -6

Therefore, R.H.S. = 4|A| = 4 × ( -6) = -24

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

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

Ans: The given matrix is A = $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.

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

Therefore, 27|A| = 27 (4) = 108                       …(i)

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

Therefore, |3A| = $3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} – 0 \begin{vmatrix} 0 & 3 \\ 0 & 12 \end{vmatrix} + 0 \begin{vmatrix} 0 & 3 \\ 3 & 6 \end{vmatrix}$

= 3 ( 36 – 0 ) = 3 ( 36 ) = 108                   …(ii)

From equations (i) and (ii), we have:

|3A| = 27|A|

Hence, the given result is proved.

Q5: Evaluate the determinants

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

(ii) $\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}\\$

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

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

Ans:

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

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculations.

|A| = $0 \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + 0 \begin{vmatrix} 3 & -2 \\ 3 & 0 \end{vmatrix} – (-1) \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}$ = ( -15 + 3 ) = -12

(ii) Let $\begin{bmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{bmatrix}$

By expanding along the first row, we have:

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

= 3 ( 1 + 6 ) + 4 ( 1 + 4 ) + 5 ( 3 – 2 )

= 3 ( 7 ) + 4 ( 5 ) + 5 ( 1 )

= 21 + 20 + 5 = 46

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

By expanding along the first row, we have:

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

= 0 – 1( 0 – 6 ) + 2 ( -3 – 0 )

= – 1 ( -6 ) + 2 ( -3 )

= 6 – 6 = 0

(iv) Let A = $\begin{bmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}$

By expanding along the first column, we have:

|A| = $2 \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} – 0 \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + 3 \begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix}$

= 2 ( 0 – 5 ) – 0 + 3 ( 1 + 4 )

= -10 + 15 = 5

Q6: If A = $\begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$ , find |A|.

Ans:

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

By expanding along the first row, we have:

|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 )

= 1 ( 3 ) – 1 ( -3 ) – 2 ( 3 )

= 3 + 3 – 6

= 6 – 6

= 0

Q7: Find values of x, if

(i) $\begin{vmatrix} 2 & 4 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\\$

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

Ans:

(i) $\begin{vmatrix} 2 & 4 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$

$\boldsymbol{\Rightarrow }$ 2 × 1 – 5 × 4 = 2x × x – 6 × 4

$\boldsymbol{\Rightarrow }$ 2 – 20 = 2x2 – 24

$\boldsymbol{\Rightarrow }$ 2x2 = 6

$\boldsymbol{\Rightarrow }$ x2 = 3

$\boldsymbol{\Rightarrow }$ x = ± $\sqrt{ 3 }$

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

$\boldsymbol{\Rightarrow }$ 2 × 5 – 3 × 4 = x × 5 – 3 × 2x

$\boldsymbol{\Rightarrow }$ 10 -12 = 5x – 6x

$\boldsymbol{\Rightarrow }$ -2 = -x

$\boldsymbol{\Rightarrow }$ x = 2

Q8: If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$ , then x is equal to

(A) 6

(B) ± 6

(C) – 6

(D) 0

Ans:

$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\\$

$\\\boldsymbol{\Rightarrow }$ x2 – 36 = 36 – 36

$\boldsymbol{\Rightarrow }$ x2 – 36 = 0

$\boldsymbol{\Rightarrow }$ x = ± 6

Hence, the correct answer is B

#### Exercise 4.2

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

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

Sol:

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

(Here, the two columns of the determinants are identical)

Q2: Using the property of determinants and without expanding, prove that:

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

Sol:

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

Applying $R_{1} \rightarrow R_{1} + R_{2}$, we have:

$\boldsymbol{\Rightarrow }$   $\Delta = \begin{vmatrix} a – b & b – a & c – b \\ b – c & c – a & a – b \\ -(a – c) & -(b – a) & -(c – b) \end{vmatrix} = 0\\$

$\\\boldsymbol{\Rightarrow }$   $- \begin{vmatrix} a – c & b – a & c – b \\ b – c & c – a & a – b \\ a – c & b – a & c – b \end{vmatrix} = 0$

Here, the two rows R1 and R3 are identical.

Therefore, $\bigtriangleup = 0$

Q3: Using the property of determinants and without expanding, prove that:

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

Sol:

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

$\\\boldsymbol{\Rightarrow }$   $\begin{vmatrix} 2 & 7 & 63 \\ 3 & 8 & 72 \\ 5 & 9 & 81 \end{vmatrix} + \begin{vmatrix} 2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5 \end{vmatrix}\\$

$\\\boldsymbol{\Rightarrow }$   $\begin{vmatrix} 2 & 7 & 9(7) \\ 3 & 8 & 9(8) \\ 5 & 9 & 9(9) \end{vmatrix} + 0$                 (Two columns are identical)

$\\9 \begin{vmatrix} 2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9 \end{vmatrix} + 0$                 (Two columns are identical)

= 0

Q4: Using the property of determinants and without expanding, prove that:

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

Sol:

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

By applying $C_{3} \rightarrow C_{3} + C_{2}$

$\\\boldsymbol{\Rightarrow }$    $\Delta = \begin{vmatrix} 1 & bc & ab + bc + ca \\ 1 & ca & ab + bc + ca \\ 1 & ab & ab + bc + ca \end{vmatrix} = 0\\$

$\\\boldsymbol{\Rightarrow }$   $(ab + bc + ca ) \begin{vmatrix} 1 & bc & 1 \\ 1 & ca & 1 \\ 1 & ab & 1 \end{vmatrix} = 0$

Here, two columns C1 and C3 are proportional.

$\boldsymbol{\Rightarrow }$   $\Delta = 0$

Q5: Using the property of determinants and without expanding, prove that:

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

Sol:

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

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

= $\Delta _{1} + \Delta _{2}$  ……………………….(i)

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

Applying $R_{2}\rightarrow R_{2} – R_{3}$, we have:

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

Applying $R_{1}\rightarrow R_{1} – R_{2}$, we have:

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

Applying $R_{1} \leftrightarrow R_{3} \;\; and R_{2} \leftrightarrow R_{3}$ we have:

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

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

Applying $R_{1} \rightarrow R_{1} – R_{3}$, we have:

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

Applying $R_{2} \rightarrow R_{2} – R_{1}$, we have:

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

Applying $R_{1} \leftrightarrow R_{2} \;\; and R_{2} \leftrightarrow R_{3}$ we have:

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

From (i), (ii) and (iii), we have:

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

Q6: By using properties of determinants, show that:

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

Sol:

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

Applying $R_{1} = c R_{1}$, we have:

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

Applying $R_{1} \rightarrow R_{1} – b R_{2}$, we have:

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

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

Here, the two rows $R_{1} \;\; and \;\; R_{3}$ are identical.

$\boldsymbol{\Rightarrow }$   $\Delta = 0$

Q7: By using the properties of determinants, show that:

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

Sol:

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

$\\\begin{vmatrix} -a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2} \end{vmatrix}$  (Taking out factors a, b, c from R1 , R2 and R3)

$\Delta = a^{2}b^{2}c^{2} \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}$  (Taking out factors a,b,c from C1, C2 and C3)

Applying $R_{2} \rightarrow R_{2} + R_{1} \;\; and \;\; R_{3} \rightarrow R_{3} + R_{1}$ we have:

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

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

$\\\boldsymbol{\Rightarrow }$   $-a^{2} b^{2} c^{2} (0 – 4) = 4a^{2} b^{2} c^{2}$

Q8: By using the properties of determinants, show that:

(i)  $\begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix} = (a – b) (b – c) (c – a)$

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

Sol:

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

Applying $R_{1} \rightarrow R_{1} – R_{3}$ and $R_{2} \rightarrow R_{2} – R_{3}$, we have:

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

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

Applying $R_{1}\rightarrow R_{1}+R_{2}$ we have:

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

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

Expanding along $C_{1}$ we have:

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

Hence, the given result is proved.

(ii) Let $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{vmatrix}$

Applying $C_{1}\rightarrow C_{1}-C_{3} \;\; and C_{2}\rightarrow C_{2}- C_{3}$ we have:

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

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

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

Applying $C_{1}\rightarrow C_{1}+C_{2}$ we have:

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

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

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

Expanding along $C_{1}$ we have:

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

= $\\(a-b)(b-c)(c-a)(c-a)(a+b+c)$

Hence, the given result is proved.

Q9: By using the properties of determinants, show that:

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

Sol:

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

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

$\\\boldsymbol{\Rightarrow }$   $\Delta = \begin{vmatrix} x & x^{2} & yz \\ y-x & y^{2}-x^{2} & zx-yz \\ z-x & z^{2}-x^{2} xy-yz & -y(z-x) \end{vmatrix}\\$

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

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

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

$\boldsymbol{\Rightarrow }$   $\Delta = (x-y)(z-x)\begin{vmatrix} x & x^{2} & yz \\ -1 & -x-y & z \\ 0 & z-y & z-y \end{vmatrix}\\$

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

Expanding along $R_{3}$ we have:

$\boldsymbol{\Rightarrow }$   $\Delta = \left [ (x-y)(z-x)(z-y) \right ] \left [ (-1) \begin{vmatrix} x & yz \\ -1 & z \end{vmatrix} + 1 \begin{vmatrix} x & x^{2} \\ -1 & -x-y \end{vmatrix} \right ]\\$

= (x-y)(z-x)(z-y)[(-xz-zy)$+(x^{2}-xy+x^{2})$]

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

Hence, the given result is proved.

Q10: By using properties of determinants, show that:

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

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

Sol:

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

Applying $R_{1}\rightarrow R_{1}+R_{2}+R_{3}$ we have:

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

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

Applying $C_{2}\rightarrow C_{2}-C_{1}, C_{3}\rightarrow C_{3}-C_{1}$ we have:

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

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

Expanding along $C_{3}$ we have:

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

$= (5x+4)(4-x)^{2}\\$

Hence, the given result is proved.

(ii) $\begin{vmatrix} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{vmatrix}$

Applying $R_{1}\rightarrow R_{1}+R_{2}+R_{3}$ we have:

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

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

Applying $C_{2}\rightarrow C_{2}-C_{1}, C_{3}\rightarrow C_{3}-C_{1}$ we have:

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

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

Expanding along $C_{3}$ we have:

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

Hence, the given result is proved.

Q.11: By using properties of determinants, show that:

(i) $\begin {vmatrix} a – b – c & 2a & 2a \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end {vmatrix} = \left ( a + b + c \right )^{3}\\$

(ii) $\\\begin {vmatrix} x + y + 2z & x & y \\ z & x + z + 2x & y \\ z & x & z + x + 2y \end {vmatrix} = 2 \left ( x + y + z \right )^{3}$

Sol:

$\Delta = \begin {vmatrix} a – b – c & 2a & 2a \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end {vmatrix}$

(i)

Applying $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$, we have:

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

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

Applying $C_{2} \rightarrow C_{2} – C_{1} , C_{3} \rightarrow C_{3} – C_{1}$, we have:

$\Delta = \left ( a + b + c \right ) \begin {vmatrix} 1 & 0 & 0 \\ 2b & – \left ( a + b + c \right ) & 0 \\ 2c & 0 & – \left ( a + b + c \right ) \end {vmatrix} \\$

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

Expanding along C3, we have:

∆ = (a + b + c)3 (-1) (-1) = (a + b + c)3

Hence, the given result is proved.

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

(ii) Applying C1 → C1 + C2 + C3, we have:

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

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

Applying R2 → R2 – R1 and R3 → R3 – R1, we have:

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

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

Expanding along R3, we have:

∆ = 2(x + y + z)3 (1) (1 – 0) = 2(x + y + z)3

Hence, the given results are proved.

Q.12: By using properties of determinants, show that:

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

Sol:

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

Applying R1 → R1 + R2 + R3, we have:

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

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

Applying C2 → C2 – C1 and C3 → C3 – C1, we have:

$\\ \Delta = \left ( 1 + x + x ^ {2} \right ) \begin {vmatrix} 1 & 0 & 0 \\ x ^ {2} & 1 – x ^ {2} & x – x ^ {2} \\ x & x ^ {2} – x & 1 – x \end {vmatrix}$

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

= $\\\left ( 1 – x ^ {3} \right ) \left ( 1 – x \right ) \begin {vmatrix} 1 & 0 & 0 \\ x ^ {2} & 1 + x & x \\ x & -x & 1 \end {vmatrix}\\$

Expanding along R1, we have:

$\\\Delta = \left ( 1 – x ^ {3} \right ) \left ( 1 – x \right ) \left ( x \right ) \begin {vmatrix} 1 + x & x \\ -x & 1 \end {vmatrix}\\$

= (1 – x3) (1 – x) (1 + x + x2)

= (1 – x3) (1 – x3)

= (1 – x3)2

Hence, the given result is proved.

Q.13: By using properties of determinants, show that:

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

Sol:

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

Applying R1 → R1 + bR3 and R2 → R2 – aR3, we have:

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

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

Expanding along R1, we have:

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

= (1 + a2 + b2)2 [1 – a2 – b2 + 2a2 –b (-2b)]

= (1 + a2 + b2)2  (1 + a2 + b2)

= (1 + a2 + b2)3

Q.14: By using properties of determinants, show that:

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

Sol:

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

Taking out common factors a, b, c from R1, R2 and R3 respectively, we have:

$\Delta = abc \begin {vmatrix} a + \frac {1} {b} & b & c \\ a & b + \frac {1} {b} & c \\ a & b & c + \frac {1} {c} \end {vmatrix}\\$

Applying R2 → R2 – R1 and R3 → R3 – R1, we have:

$\\ \Delta = abc \begin {vmatrix} a + \frac {1} {b} & b & c \\ – \frac {1} {a} & \frac {1} {b} & 0 \\ – \frac {1} {a} & 0 & \frac {1} {c} \end {vmatrix}\\$

Applying C1 → aC1, C2 → bC2, and C3 → cC3, we have:

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

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

Expanding along R3, we have:

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

= -1(-c2) + (a2 + 1 + b2) = 1 + a2 + b2 + c2

Hence, the given result is proved.

Let A be a square matrix of order 3 × 3, KA is equal to

(1).   k|A|

(2).   k2|A|

(3).   k3|A|

(4).   3k|A|

(3)

A is a square matrix of order 3 × 3

Let A = $\begin {vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end {vmatrix} \\$

Then, kA = $\begin{vmatrix} ka_{1} & kb_{1} & kc_{1} \\ ka_{2} & kb_{2} & kc_{2} \\ ka_{3} & kb_{3} & kc_{3} \end {vmatrix}$

Therefore, $\\|kA| = \begin {vmatrix} ka_{1} & kb_{1} & kc_{1} \\ ka_{2} & kb_{2} & kc_{2} \\ ka_{3} & kb_{3} & kc_{3} \end {vmatrix}\\$

$\\ k ^ {3} = \begin {vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end {vmatrix} \; \; \; \; \; \; \; \; \; \; \left ( Taking \; out \; common \; factors \; k \; from \; each \; row \right )$

= k3|A|

Therefore, |kA| = k3|A|

Hence, the correct answer is C.

Q.16: Which of the following is correct?

(1) Determinant is a square matrix.

(2) Determinant is a number associated to a matrix.

(3) Determinant is a number associated to a square matrix.

Sol:

(3)

We know that to every square matrix, A = [aij] of order n. we can associate a number called the determinant of a square matrix A, where aij = (i, j)th element of A.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is C.

#### Exercise – 4.3

Q.1: Find area of the triangle with vertices at the point given in each of the following:

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

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

(iii) (−2, −3), (3, 2), (−1, −8)

Sol:

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

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

$=\frac{1}{2}\left[1(0-3)-0(6-4)+1(18-0) \right ]\\$

$\\=\frac{1}{2}\left[-3+18 \right ]$ = $\frac{15}{2}$ unit2

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

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

$=\frac{1}{2}\left[2(1-8)-7(1-10)+1(8-0) \right ]\\$

= $\\\frac{1}{2}\left[-14+63-2 \right ]=\frac{1}{2}\left[-16+63 \right ]=\frac{47}{2}$ unit2

(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8) is given by the relation,

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

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

=$\\\frac{1}{2}\left[-20+12-22 \right ]\;=\;-\frac{30}{2}=-15$ unit2

Hence, the area of the triangle is $\begin{vmatrix} -15 \end{vmatrix}=15$ unit2

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

Sol:

Area of ∆ABC is given by the relation:

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

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

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

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

=0

Thus, the area of the triangle formed by points A, B, and C is zero. Hence, the points A, B, and C are collinear.

Q.3: Find values of k if area of triangle is 4 square units and vertices are

(i) (k, 0), (4, 0), (0, 2)

(ii) (−2, 0), (0, 4), (0, k)

Sol:

We know that the area of a triangle whose vertices are$(x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3})$ is the absolute value of the determinant (∆), where

$\Delta=\frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{vmatrix}$

It is given that the area of triangle is 4 square units.

Therefore $\Delta=\pm 4$

(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation:

$\Delta=\frac{1}{2}\begin{vmatrix} k & 0 & 1\\ 4 & 0 & 1\\ 0 & 2 & 1 \end{vmatrix}$

$=\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)]\\$

= $\\\frac{1}{2}[-2k+8]\;=\;-k+4 \;=\;-k+4=\pm4$

When −k + 4 = − 4, k = 8.
When −k + 4 = 4, k = 0.
Hence, k = 0, 8.
(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation:

$\Delta=\frac{1}{2}\begin{vmatrix} -2 & 0 & 1\\ 0 & 4 & 1\\ 0 & k & 1 \end{vmatrix}$

$=\frac{1}{2}[-2(4-k)]$

$=k-4\\$

i.e. $\\k-4 = \pm 4$
When k-4 = − 4, k = 0.
When k-4 = 4, k = 8.
Hence, k = 0, 8.

Q.4: (i) Find equation of line joining (1, 2) and (3, 6) using determinants

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants
Sol:
(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the
points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

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

$=\frac{1}{2}[1(6-y)-2(3-x)+1(3y-6x)]=0\\$

6-y-6+2x+3y-6x=0

2y-4x=0

y=2x
Hence, the equation of the line joining the given points is y = 2x.
(ii) Let P (x, y) be any point on the line joining points A (3, 1) and
B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP
will be zero.

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

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

9 – 3y – 9 + x + 9y – 3x = 0

6y – 2x = 0

x – 3y = 0

Hence, the equation of the line joining the given points is x − 3y = 0.

Q.5: If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is
A. 12 B. −2 C. −12, −2 D. 12, −2

Sol:

D

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation:

$\Delta=\frac{1}{2}\begin{vmatrix} 2 & -6 & 1\\ 5 & 4 & 1\\ k & 4 & 1 \end{vmatrix}=0$

$=\frac{1}{2}[2(4-4)+6(5-k)+1(20-4k)]$

$=\frac{1}{2}[30-6k+20-4k]$

$=\frac{1}{2}[50-10k]$ = 25 – 5k

It is given that the area of the triangle is ±35.

Therefore, we have:

$25-5k=\pm35$

$5(5-k)=\pm35$

$5-k=\pm7$

When 5-k = −7, k = 5+7 = 12.

When 5-k = 7, k = 5-7 = −2.

Hence, k = 12, −2.

The correct Sol: is D.

#### Exercise 4.4

Q1:  Write Minors and Cofactors of the elements of following determinants:

(i) $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$

(ii) $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$

Sol:

(i) The given determinant is $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$

Minor of element aij is Mij .

Therefore, M11 = minor of element a11 = 3

M12 = minor of element a12 = 0

M21 = minor of element a21 = -4

M22 = minor of element a22 = 2

Cofactor of aij is Aij = (-1) i+j Mij .

Therefore, A11 = (-1)1+1 M11 = (-1)2 (3) = 3

A12 = ( -1 )1 + 2 M12 = ( -1 )3 ( 0 ) = 0

A21 = ( – 1 )2 + 1 M21 = ( -1 )3 ( -4 ) = 4

A22 = ( -1 )2 + 2 M22 ( -1 )4 ( 2 ) = 2

(ii) The given determinant is $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$

Minor of element aij is Mij .

Therefore, M11 = minor of element a11 is Mij .

M12 = minor of element a12 = b

M21 = minor of element a21 = c

M22 = minor of element a22 = a

Cofactor of aij is Aij = (-1)I + j Mij .

Therefore, A11 = (-1)1 + 1 M11 = (-1)2 (d) = d

A12 = ( -1 )1 + 2 M12 = ( -1 )3 ( b ) = -b

A21 = ( -1 )2 + 1 M21 = ( -1 )3 ( c ) = -c

A22 = ( -1 ) 2 + 2 M22 = ( -1 )­4  ( a ) = a

Q2: (i) $\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$

(ii) $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$

Sol:

(i) The given determinant is $\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$

By the definition of minors and cofactors, we have :

M11 = minor of a11 = $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$ = 1

M12 = minor of a12 = $\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix}$ = 0

M13 = minor of a13 = $\begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix}$ = 0

M21 = minor of a21 = $\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix}$ = 0

M22 = minor of a22 = $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$ = 1

M23 = minor of a23 = $\begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix}$ = 0

M31 = minor of a31 = $\begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix}$ = 0

M32 = minor of a32 = $\begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix}$ = 0

M33 = minor of a33 = $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$ = 1

A11 = cofactor of a11 = ( -1 )1 + 1 M11 = 1

A12 = cofactor of a12 = ( -1 )1 + 2 M12 = 0

A13 = cofactor of a13 = ( -1 )1 + 3 M13 = 0

A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 0

A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 1

A23 = cofactor of a23 = ( -1 )2 + 3 M23 = 0

A31 = cofactor of a31 = ( -1 )3 + 1 M31 = 0

A32 = cofactor of a32 = ( -1 )3 + 2 M32 = 0

A33 = cofactor of a33 = ( -1 )3 + 3 M33 = 1

(ii) The given determinant is $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$

By definition of minors and cofactors , we have:

M11 = minor of a11 = $\begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix}$ = 10 + 1 = 11

M12 = minor of a12 = $\begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix}$ = 6 – 0 = 6

M13 = minor of a13 = $\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}$ = 3 – 0 = 3

M21 = minor of a21 = $\begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix}$ = 0 – 4 = -4

M22 = minor of a22 = $\begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix}$ = 2 – 0 = 2

M23 = minor of a23 = $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$ = 1 – 0  = 1

M31 = minor of a31 = $\begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix}$ = 0 – 20 = -20

M32 = minor of a32 = $\begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix}$ = -1 – 12 = -13

M33 = minor of a33 = $\begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix}$ = 5 – 0 = 5

A11 = cofactor of a11 = ( -1 )1 + 1 M11 = 11

A12 = cofactor of a12 = ( -1 )1 + 2 M12 = -6

A13 = cofactor of a13 = ( -1 )1 + 3 M13 = 3

A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 4

A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 2

A23 = cofactor of a23 = ( -1 )2 + 3 M23 = -1

A31 = cofactor of a31 = ( -1 )3 + 1 M31 = -20

A32 = cofactor of a32 = ( -1 )3 + 2 M32 = 13

A33 = cofactor of a33 = ( -1 )3 + 3 M33 = 5

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

Sol:

The given determinant is $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$

We have:

M21 = $\begin{vmatrix} 3 & 8\\ 2 & 3 \end{vmatrix}$ = 9 – 16 = -7

Therefore, A21 = cofactor of a21 = ( -1 )2 + 1 M21 = 7

M22 = $\begin{vmatrix} 5 & 8\\ 1 & 3 \end{vmatrix}$ = 15 – 8 = 7

Therefore, A22 = cofactor of a22 = ( -1 )2 + 2 M22 = 7

M23 = $\begin{vmatrix} 5 & 3\\ 1 & 2 \end{vmatrix}$ = 10 – 3 = 7

Therefore, A23 = cofactor of a23 = ( -1 )2 + 3 M23 = -7

We know that $\Delta$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

Therefore, $\Delta$ = a21A21 + a22A22 + a23A23 =

2 ( 7 ) + 0 ( 7 ) + 1 ( -7 ) = 14 – 7 = 7

Q4: Using Cofactors of elements of third column, evaluate $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$ .

Sol:

The given determinant is $\begin{vmatrix} 1 & x & yz\\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$ .

We have:

M13 = $\begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix}$ = z – y

M23 = $\begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix}$ = z – x

M33 = $\begin{vmatrix} 1 & x \\ 1 & \end{vmatrix}$ = y – x

Therefore, A13 = cofactor of a13 = ( -1 )1 + 3 M13 = ( z – y )

A23 = cofactor of a23 = ( -1 )2 + 3 M23 = – ( z – x ) = ( x – z )

A33 = cofactor of a33 = ( -1 )3 + 3 M33 = ( y – x )

We know that $\Delta$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

Therefore, $\Delta$ = a13A13 + a23A23 + a33A33

= yz ( z – y ) + zx ( x – z ) + xy ( y – x )

= yz2 – y2 z + x2 z + x2 z – xz2 + xy2 – x2 y

= ( x2 z – y2 z ) + ( yz2 – xz2 ) + ( xy2 – x2 y )

= z ( x2 – y2 ) + z2 ( y – x ) + xy ( y – x )

= z ( x – y ) ( x + y ) + z2 ( y – x ) + xy ( y – x )

= ( x – y ) [ zx + zy – z2 – xy ]

= ( x – y ) [ z ( x – z ) + y ( z – x ) ]

= ( x – y ) ( z – x ) [ -z + y ]

= ( x – y ) ( y – z ) ( z – x )

Hence, $\Delta$ = ( x – y ) ( y – z ) ( z – x )

Q5: For the matrices A and B, verify that (AB)’ = B’A’ where

(i) A = $\begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$ ,

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

(ii) A = = $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ ,

B = $\begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$

Sol:

(i) AB = $\begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$

Therefore, ( AB )’ = $\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Now, A’ = $\begin{bmatrix} -1 & -4 & 3 \end{bmatrix}$ ,

B’ = $\begin{bmatrix} -1\\ 2 \\ 1 \end{bmatrix}$

Therefore, B’A’= $\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Hence, we have verified that ( AB )’ = B’A’ .

(ii) AB = $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$

Therefore, ( AB )’ = $\begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

Now, A’ = $\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$ ,

B’ = $\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}$

Therefore, B’A’= $\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

Hence, we have verified that (AB)’ = B’A’

#### Exercise 4.5

Q.1: Find the adjoint of each of the matrices

$\begin{bmatrix} 2 & 4 \\ 5 & 7 \end{bmatrix}$

Sol:

Suppose A = $\begin{bmatrix} 2 & 4 \\ 5 & 7 \end{bmatrix}$

For X = $\begin{bmatrix} a & b \\ c & d \end{bmatrix}\\$

Then adj (X) = $\begin{bmatrix} d & – b \\ – c & a \end{bmatrix}$

We have,

A 11  = 7, A12  = – 4, A21  = -5, A22 = 2

Therefore, adj (A) = $\begin{bmatrix} 7 & – 4 \\ – 5 & 2 \end{bmatrix}$

Q.2: Find the adjoint of each of the matrices

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

Sol:

Suppose, D = $\begin{bmatrix} 1 & – 1 & 2 \\ 2 & 3 & 5 \\ – 2 & 0 & 1 \end{bmatrix}$

We have,

$D_{11} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3 – 0 = 3 \\ D_{12} = – \begin{vmatrix} 2 & 5 \\ – 2 & 1 \end{vmatrix} = – (2 + 10) = – 12 \\ D_{13} = \begin{vmatrix} 2 & 3 \\ – 2 & 0 \end{vmatrix} = 0 + 6 = 6 \\$

$D_{21} = – \begin{vmatrix} – 1 & 2 \\ 0 & 1 \end{vmatrix} = – (- 1 – 0) = 1 \\ D_{22} = \begin{vmatrix} 1 & 2 \\ – 2 & 1 \end{vmatrix} = 1 + 4 = 5 \\ D_{23} = – \begin{vmatrix} 1 & – 1 \\ – 2 & 0 \end{vmatrix} = – (0 – 2) = 2 \\$

$D_{31} = \begin{vmatrix} – 1 & 2 \\ 3 & 5 \end{vmatrix} = – 5 – 6 = – 11 \\ D_{32} = – \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = – (5 – 4) = – 1 \\ D_{33} = \begin{vmatrix} 1 & – 1 \\ 2 & 3 \end{vmatrix} = 3 + 2 = 5\\$

Hence, adj D = $\\\begin{bmatrix} D ^{11} & D ^{12} & D ^{13} \\ D ^{21} & D ^{22} & D ^{23} \\ D ^{31} & D ^{32} & D ^{33} \end{bmatrix} = \begin{bmatrix} 3 & 1 & – 11 \\ – 12 & 5 & – 1 \\ 6 & 2 & 5 \end{bmatrix}$

Q.3: Prove whether D (adj D) = (adj D) D = $\left | D \right | I$

$D = \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix}$

Sol:

$D = \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix}$

We have,

$\left | D \right | = – 12 – (- 12) = – 12 + 12 = 0 \\ \left | D \right | I = 0 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

D 11  = – 6, D 12  = 4, D 21  = – 3, D 22 = 2

adj D = $\begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix}$

Now,

D (adj D) = $\begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix} \begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix} \\ = \begin{bmatrix} – 12 + 12 & – 6 + 6 \\ 24 – 24 & 12 – 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

And,

(adj D) D = $\begin{bmatrix} – 6 & – 3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ – 4 & – 6 \end{bmatrix} \\ = \begin{bmatrix} – 12 + 12 & – 18 + 18 \\ 8 – 8 & 12 – 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Thus, D (adj D) = (adj D) D = $\left | D \right | I$

Q.4: Obtain the inverse of the matrices if it exists

$\begin{bmatrix} 2 & – 2 \\ 4 & 3 \end{bmatrix}$

Sol:

Suppose D = $\begin{bmatrix} 2 & – 2 \\ 4 & 3 \end{bmatrix}$

We have,

$\left | D \right | = 6 + 8 = 14$

Now,

D 11  = 3, D 12  = – 4, D 21  = 2, D 22 = 2

adj D = $\begin{bmatrix} 3 & 2 \\ – 4 & 2 \end{bmatrix} \\ D ^{- 1} = \frac{1}{\left | D \right |} = \frac{1}{14}\; \begin{bmatrix} 3 & 2 \\ – 4 & 2 \end{bmatrix}$

Q.5: Obtain the inverse of the matrices if it exists

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

Sol:

Suppose D = $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$

We have,

$\left | D \right | = 1 (10 – 0) – 2 (0 – 0) + 3 (0 – 0) = 10$

Now, D 11 = 10 – 0 = 10, D 12 = – (0 – 0) = 0, D 13 = 0 – 0 = 0,  D 21  = – (10 – 0) = – 10, D 22 = 5 – 0 = 5, D 23 = – (0 – 0) = 0,  D 31 = 8 – 6 = 2, D 32  = – (4 – 0) = – 4, D 33 = 2 – 0 = 2

adj D = $\begin{bmatrix} 10 & – 10 & 2 \\ 0 & 5 & – 4 \\ 0 & 0 & 2 \end{bmatrix} \\$

D-1 = $\\\frac{1}{\left | D \right |} adj D = \frac{1}{10} \begin{bmatrix} 10 & – 10 & 2 \\ 0 & 5 & – 4 \\ 0 & 0 & 2 \end{bmatrix}$

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

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

Sol:

Let A = $\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos \alpha & sin \alpha \\ 0 & sin \alpha & – cos \alpha \end{bmatrix}$

We have,

$\left |A \right | = 1 \left ( – cos ^{2} \alpha – sin ^{2} \alpha\right )$

$\left |A \right | = – \left ( cos ^{2} \alpha + sin ^{2} \alpha \right ) = – 1$

Now,

$A _{ 11 } = – cos ^{2 } \alpha – sin ^{2} \alpha = – 1 , A _{ 12 } = 0 , A _{ 13 } = 0$

$A _{21} = 0 , A _{ 22 } = – cos \alpha , A _{ 23 } = – sin \alpha \\ A _{ 31 } = 0 , A_{32} = – sin \alpha , A_{33} = cos \alpha$

Therefore,

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

Therefore,

$A ^{ -1 } = \frac{ 1 }{ \left | A \right |}.adj A$ = $\begin{bmatrix} -1 & 0 & 0 \\ 0 & – cos \alpha & – sin \alpha \\ 0 & – sin \alpha & cos \alpha \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos \alpha & sin \alpha \\ 0 & sin \alpha & – cos \alpha \end{bmatrix}$

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

Sol:

Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$

We have ,

$\left |A \right | = 15 – 14 = 1$

Now,

$A _{ 11} = 5 , A _{ 12} = – 2 , A _{ 21} = – 7 , A _{ 22} = 3$

Therefore,

$adj A = \begin{bmatrix} 5 & – 7 \\ – 2 & 3 \end{bmatrix}$

Therefore,

$A ^{ – 1 } = \frac{1}{ \left |A \right |}. adj A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$

Now, let $adj B = \begin{bmatrix} 6 & 8 \\ 7 & 6 \end{bmatrix}$

We have ,

$\left |B \right | = 54 – 56 = – 2$

Therefore,

$adj B = \begin{bmatrix} 9 & – 8 \\ – 7 & 6 \end{bmatrix}$

Therefore,

$B ^{ – 1 } = \frac{1}{ \left |B \right |}. adj B = 0.5 \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$

$\begin{bmatrix} \frac{ – 9}{2} & 4 \\ \frac{7}{2} & – 3 \end{bmatrix}$

Now,

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

$B^{-1}A^{-1} = \begin{bmatrix} \frac{-45}{2} – 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & \frac{-49}{2} – 9 \end{bmatrix}\begin{bmatrix} \frac{ 61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{ -67}{2} \end{bmatrix}$

Then,

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

$\\\boldsymbol{\Rightarrow }$ $AB = \begin{bmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{bmatrix}\\$

$\\\boldsymbol{\Rightarrow }$ $AB = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}\\$

Therefore, we have $AB = \begin{bmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{bmatrix}$

Also,

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

Therefore, $\left ( AB \right )^{ -1 } = \frac{1}{ \left |AB \right |} adj\left ( AB \right ) = \frac{-1}{2}\begin{bmatrix} 61 & – 87 \\ – 47 & 67 \end{bmatrix}$ . . . . . . . . . . . . . . . (2)

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

From (1) and (2), we have: (AB)−1 = B−1 A -1 Hence, the given result is proved.

Q.8: If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that A 2 – 5 A + 7 I = 0. Hence find A – 1

Sol:

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

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

Therefore,

A 2 – 5 A + 7 I = 0

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

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

= $\\\begin{bmatrix} – 7 & 0 \\ 0 & – 7 \end{bmatrix}$ –  $\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$ = $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\\$

Therefore, A 2 – 5 A + 7 I = 0

Therefore, A.A – 5 A = – 7I

$\boldsymbol{\Rightarrow }$ A.A ( A -1 ) – 5A A -1 = – 7I A -1         . . . . . . . . . . . . . . . . . [ post – multiplying by A -1 as $\left | A \right | \neq 0$ ]

$\boldsymbol{\Rightarrow }$ A. ( A A -1 ) – 5I = – 7 A -1

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

$\boldsymbol{\Rightarrow }$ A -1 $= – \frac{ 1 }{ 7 }\left ( A – 5I \right )$

$\boldsymbol{\Rightarrow }$ A -1 $= – \frac{ 1 }{ 7 }\left ( 5I – A \right )$

=$\\\frac{1}{7} \left ( \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} – \begin{bmatrix} 3 & 1 \\ – 1 & 2 \end{bmatrix} \right ) = \frac{1}{ 7 } \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

Therefore,

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

Q.9: Let A be a nonsingular square matrix of order 3 × 3. Then $\left | adj A \right |$ is equal to

(A) $\left | adj A \right |$

(B) $\left | adj A \right |^{2}$

(C) $\left | adj A \right |^{ 3 }$

(D) 3 $\left | adj A \right |$

Sol:

The correct option is B

$\left ( adj A \right )A = \left | A \right |I = \begin{bmatrix} \left | A \right | & 0 & 0 \\ 0 & \left | A \right | & 0 \\ 0 & 0 & \left | A \right | \end{bmatrix}$

$\boldsymbol{\Rightarrow }$ $\left ( adj A \right )A = \begin{bmatrix} \left | A \right | & 0 & 0 \\ 0 & \left | A \right | & 0 \\ 0 & 0 & \left | A \right | \end{bmatrix}$

$\boldsymbol{\Rightarrow }$ $\left ( adj A \right )\left |A \right | = \left | A \right |^{3} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \left | A \right |^{3} \left (I \right )$

Therefore, $\left |\left ( adj A \right ) \right | = \left | A \right |^{2}$

Hence, the correct Sol: is B.

Q.10: If A is an invertible matrix of order 2, then det (A −1 ) is equal to

(A) det (A)

(B) $\frac{ 1 }{ det \left ( A \right )}$

(C) 1

(D) 0

Sol:

Since A is an invertible matrix , A -1 exists and A -1 = $\frac{ 1 }{ \left | a \right |} adj A$

As matrix A is of order 2, let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

Then, $\left |A \right | = ad – bc and adj A = \begin{bmatrix} d & -b \\ – c & a \end{bmatrix}$

Now,

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

Therefore,

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

Therefore,

$det \left ( A \right )^{ -1} = \frac{ 1 }{ det \left ( A \right )}$

Hence, the correct Sol: is B.

Q.11: Suppose D = $\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}$, verify that D 3 – 6 D2 + 9D – 4I = 0 and hence find D – 1

Sol:

D = $\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}\\$

D 2 = $\\\begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix}\\$

$\\\begin{bmatrix} 4 + 1 + 1 & – 2 – 2 – 1 & 2 + 1 + 2 \\ – 2 – 2 – 1 & 1 + 4 + 1 & – 1 – 2 – 2 \\ 2 + 1 + 2 & – 1 – 2 – 2 & 1 + 1 + 4 \end{bmatrix} \\ = \begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix}\\$

D 3 = D 2 D = $\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} \\ = \begin{bmatrix} 12 + 5 + 5 & – 6 – 10 – 5 & 6 + 5 + 10 \\ – 10 – 6 – 5 & 5 + 12 + 5 & 5 – 6 – 10 \\ 10 + 5 + 6 & – 5 – 10 – 6 & 5 + 5 + 12 \end{bmatrix} \\ = \begin{bmatrix} 22 & – 21 & 21 \\ – 21 & 22 & – 21 \\ 21 & – 21 & 22 \end{bmatrix}\\$

Now,

D 3 – 6 D2 + 9D – 4I = 0

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

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

= $\\ \begin{bmatrix} 40 & – 30 & 30\\ – 30 & 40 & – 30 \\ 30 & – 30 & 40 \end{bmatrix} – \begin{bmatrix} 40 & – 30 & 30\\ – 30 & 40 & – 30 \\ 30 & – 30 & 40 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Hence, D 3 – 6 D2 + 9D – 4I = 0

Now,

D 3 – 6 D2 + 9D – 4I = 0

(DDD) D – 1 – 6 D + 9D – 4I = 0     [Multiplying by D – 1 on as $\left | D \right | I$ is not equal to 0]

DD (DD– 1) – 6D (DD– 1) + 9 (DD– 1) = 4 ID– 1

DDI – 6DI + 9I = 4 D– 1

D2 – 6D + 9I = 4 D– 1

D– 1 = $\frac{1}{4}$ (D2 – 6D + 9I)

D2 – 6D + 9I

= $\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} – 6 \begin{bmatrix} 2 & – 1 & 1 \\ – 1 & 2 & – 1 \\ 1 & – 1 & 2 \end{bmatrix} + 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\$

= $\\\begin{bmatrix} 6 & – 5 & 5 \\ – 5 & 6 & – 5 \\ 5 & – 5 & 6 \end{bmatrix} – \begin{bmatrix} 12 & – 6 & 6 \\ – 6 & 12 & – 6 \\ 6 & – 6 & 12 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \\ = \begin{bmatrix} 3 & 1 & – 1 \\ 1 & 3 & 1 \\ – 1 & 1 & 3 \end{bmatrix} From\; equation\; (1), we\; have, D ^{- 1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & – 1 \\ 1 & 3 & 1 \\ – 1 & 1 & 3 \end{bmatrix}$

#### Exercise- 4.6

Q-1: Check the consistency for the system of two equations given below:

a + 3b = 2

2a + 4b = 3

Sol:

As per the data given in the question,

The given system of the two equations is:

a + 3b = 2

2a + 4b = 3

The given system of equations will be written as in the form of MX = N, where

$M = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

Thus,

|M| = 1( 4 ) – 3( 2 ) = 3 – 4 = -1 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Therefore, the given system of two equations will be consistent.

Q-2. Check  the consistency of the system of two equations given below:

a + 4b = 5

2a + 7b = 8

Sol:

The given system of the two equations is:

a + 4b = 5

2a + 7b = 8

The given system of equations will be written as in the form of MX = N, where

$M = \begin{bmatrix} 1 & 4 \\ 2 & 7 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$

Thus,

|M| = 1(7) – 4(2) = 7 – 8 = -1 ≠ 0

Hence, M is a singular matrix.

Thus,

$\left ( adj M \right ) = \begin{bmatrix} 7 & -4 \\ -2 & 1 \end{bmatrix}$

$\left ( adj M \right )N = \begin{bmatrix} 7 & -4 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix}$

= $\begin{bmatrix} 35 – 32 \\ -10 + 8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} \neq 0$

Therefore, the solution for the given system of the equation does not exist. Thus, the given system of the equations will be inconsistent.

Q-3. Check the consistency of the system of three equations given below:

a + b + c = 1

2a + 3b + 2c = 2

pa + pb + 2pc = 4

Sol:

The given system of the two equations is:

a + b + c = 1

2a + 3b + 2c = 2

pa + pb + 2pc = 4

The given system of three equations will be written as in the form MX = N, where

$M = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ p & p & 2p \end{bmatrix}, X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$

|M| = 1( 6p – 2p ) – 1( 4p – 2p ) + 1( 2p – 3p )

= 1( 4p ) – 1( 2p ) + 1( -p )

= 4p – 2p – p

= p ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Therefore, the given system of two equations will be consistent.

Q-4. Check the consistency of the system of three equations given below:

3a – b – 2c = 2

2b – c = -1

3a – 5b = 3

Sol:

The given system of the two equations is:

3a – b – 2c = 2

2b – c = -1

3a – 5b = 3

The given system of three equations will be written as in the form MX = N, where

$M = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}, X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$

|M| = 3[ 2 × 0 – ( -1 ) × ( -5 ) ] – 0[ ( -1 ) × 0 –  ( -2 ) × ( -5 ) ]   + 3[ ( -1 ) × ( -1 ) – ( -1 ) × ( 4 ) ]

= 3( 0 – 5 ) – 0 + 3(1 + 4)

= 3( -5 ) – 0 + 3( 5 )

= -15 + 15 = 0

Thus,

M is a singular matrix.

Then,

$\left ( adj M \right ) = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}$

$\left ( adj M \right )N = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$

= $\begin{bmatrix} -10 – 10 + 15 \\ -6 -6 + 9 \\ -12 – 12 + 18 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \neq 0$

Therefore, the solution for the given system of the equation does not exist. Thus, the given system of the equations will be inconsistent.

Q-5.  Solve the following system of the linear equations by suing the matrix method:

5a + 2b = 4

7a + 3b = 5

Sol:

The given system of linear equations will be written as in the form of MX = N, where

$M = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

|M| = ( 5 × 3 ) – ( 7 × 2 ) = 15 – 14 = 1 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M-1 = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$

⟹ M-1 = $\frac{ 1 }{ 1 }\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$

Thus,

X =  M-1 N = $\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix}$

⟹  $\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 3 \times 4 -2 \times 5 \\ -7\times 4 + 5 \times 5 \end{bmatrix} = \begin{bmatrix} 12 – 10 \\ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$

⟹  $\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$

Hence, a = 2 and b = -3.

Q-6.  Solve the following system of the linear equations by suing the matrix method:

4a – 3b = 3

3a – 5b = 7

Sol:

The given system of linear equations will be written as in the form of MX = N, where

$M = \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}, X = \begin{bmatrix} a \\ b \end{bmatrix} \; and \; N = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$

|M| = [ 4 × (-5) ] – [ ( -3 ) × 3 ] = -20 + 9= -11 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M-1 = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$

⟹ M-1 = $- \frac{ 1 }{ 11 }\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}$ = $\frac{ 1 }{ 11 }\begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}$

Thus,

X =  M-1 N = $\frac{ 1 }{ 11 }\begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3 \\ 7 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3 \\ 7 \end{bmatrix}$

⟹  $\begin{bmatrix} a \\ b \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 5 \times 3 + \left( -3 \right )\times 7 \\ 3\times 3 + \left( -4 \right ) \times 7 \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} 15 – 21 \\ 9 – 28 \end{bmatrix} = \frac{ 1 }{ 11 } \begin{bmatrix} -6 \\ -19 \end{bmatrix} = \begin{bmatrix} \frac{ -6 }{ 11 } \\ -\frac{ 19 }{ 11 } \end{bmatrix}$

⟹  $\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \frac{ -6 }{ 11 } \\ -\frac{ 19 }{ 11 }\end{bmatrix}$

Hence, a = $- \frac{ 6 }{ 11 }$ and b = $- \frac{ 19 }{ 11 }$.

Q-7.  Solve the following system of the linear equations by suing the matrix method:

2a + b + c = 1

a – 2b – c = $\frac{ 3 }{ 2 }$

3b – 5c = 9

Sol:

The given system of linear equations will be written as in the form of MX = N, where

$M = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 1 \\ \frac{ 3 }{ 2 }\\ 9 \end{bmatrix}$

Thus,

|M| = 2[ (-2) × (-5) – (-1) × 3 ] – 1[ 1 × (-5) – 3 × 1 ] + 0[ (-2) × 1 – 1 × (-1) ]  = 2( 10 + 3 ) – 1( -5 – 3 ) + 0

= 2 × 13 – (-8)

= 26 + 8 = 34 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M11 = 13, M12 = 5, M13 = 3

M21 = 8, M22 = -10, M23 = -6

M31 = 1, M32 = 3, M33 = -5

Thus,

M-1 = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$ = $\frac{ 1 }{ 34 } \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}$

Thus,

X =  M-1 N = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$ = $\frac{ 1 }{ 34 } \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix} \begin{bmatrix} 1 \\ \frac{ 3 }{ 2 }\\ 9 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 13\times 1 + 8 \times \frac{ 3 }{ 2 } + 1 \times 9 \\ 5\times 1 + \left ( -10 \right ) \times \frac{ 3 }{ 2 } + 3 \times 9 \\ 3\times 1 + \left ( -6 \right ) \times \frac{ 3 }{ 2 } + \left ( -5 \right ) \times 9 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 13 + 12 + 9 \\ 5 – 15 + 27 \\ 3 – 9 – 45 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 34 } \begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{ 1 }{ 2 } \\ -\frac{ 3 }{ 2 } \end{bmatrix}$

Therefore, a = 1, b = $\frac{ 1 }{ 2 }$ and c = -$\frac{ 3 }{ 2 }$

Q-8.  Solve the following system of the linear equations by suing the matrix method:

a – b + c = 4

2a + b – 3c = 0

a + b + c = 2

Sol:

The given system of linear equations will be written as in the form of MX = N, where

$M = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

Thus,

|M| = 1[ 1 × 1 – (-3) × 1 ] – ( -1 )[ 2 × 1 – ( -3 ) × 1 ] + 1[ 2 × 1 – 1 × 1 ]  = 2( 1 + 3 ) + 1( 2 + 3 ) + 1( 2 – 1 )

= 1 × 4 + 1 × 5 + 1 × 1

= 4 + 5 + 1 = 10 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M11 = 4, M12 = -5, M13 = 1

M21 = 2, M22 = 0, M23 = -2

M31 = 2, M32 = 5, M33 = 3

Thus,

M-1 = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$ = $\frac{ 1 }{ 10 } \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$

Thus,

X =  M-1 N = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$ = $\frac{ 1 }{ 10 } \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 4\times 4 + 2 \times 0 + 2 \times 2 \\ -5\times 4 + 0 \times 0 + 5 \times 2 \\ 1 \times 4 + \left ( -2 \right ) \times 0 + 3 \times 2 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 10 } \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$

Therefore, a = 2, b = -1 and c = 1

Q-9. If from equations,

2a – 3b + 5c = 11, 3a + 2b – 4c = -5 and a + b – 2c = -3

M = $\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$

Find M-1. By using M-1, solve the system of the linear equations.

Sol:

M = $\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$

|M| = 2[ 2 × (-2) – 1 × (-4) ] + 3[ 3 × (-2) – 1 × (-4) ]  + 5[ 3 × 1 – 2 × 1 ]

= 2( -4 + 4 ) + 3( -6 + 4 ) + 5( 3 – 2 )

= 2 × 0 + 3 × (-2) + 5 × 1

= 0 – 6 + 5 = -1 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M11 = 0, M12 = 2, M13 = 1

M21 = -1, M22 = -9, M23 = -5

M31 = 2, M32 = 23, M33 = 13

Thus,

M-1 = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$ = $\frac{ 1 }{ -1 } \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$

The given system of linear equations will be written as in the form of MX = N, where

$M = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

The solution of the given system of linear equations will be given by X =  M-1 N.

⟹ X =  M-1 N = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$ = $\begin{bmatrix} 4 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0\times 11 + 1 \times \left( -5 \right ) + \left( -2 \right ) \times \left( -3 \right ) \\ -2\times 11 + 9 \times \left( -5 \right ) + \left( -23 \right ) \times \left( -3 \right ) \\ \left( -1 \right )\times 11 + 5 \times \left( -5 \right ) + \left( -13 \right ) \times \left( -3 \right ) \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0 – 5 + 6 \\ -22 – 45 + 69 \\ -11 – 25 + 39 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Therefore, a = 1, b = 2 and c = 3.

Q-10. The price of 4 kg of onion, 3 kg of wheat and 2 kg of rice is Rs 60. The price of 2 kg of onion, 4 kg of wheat and 6 kg of rice is Rs 90. The price of 6 kg of onion, 2 kg of wheat and 3 kg of rice is Rs 70. What is the price of each of the items (per kg)? Use matrix method to find the price.

Sol:

Let us consider the cost of onions, wheat, and rice per kg be given by Rs a, Rs b,and Rs c, respectively.

Thus, the given situation will be represented by the system of the equations as:

4a + 3b + 2c = 60

2a + 4b + 6c = 90

6a + 2b + 3c = 70

The given system of the equations will be written as in the form of MX = N, where

$M = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}, \; X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \; and \; N = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$

Thus,

|M| = 4[ 4 × 3 – 6 × 2 ] – 3[ 2 × 3 – 6 × 6 ] + 2[ 2 × 2 – 6 × 4 ]  = 2( 12 – 12 ) – 3( 6 – 36 ) + 2( 4 – 24 )

= 0 + 90 – 40 = 50 ≠ 0

Hence, M is non- singular.

Thus, M-1 exists.

Now,

M11 = 0, M12 = 30, M13 = -20

M21 = -5, M22 = 0, M23 = 10

M31 = 10, M32 = -20, M33 = 10

M-1 = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$ = $\frac{ 1 }{ 50 } \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$

Thus,

X =  M-1 N = $\frac{ 1 }{ \left | M \right | } \left( adj M \right )$ = $\frac{ 1 }{ 50 } \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \begin{bmatrix} 60 \\90 \\ 70 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 0\times 60 + \left( -5 \right ) \times 90 + 10 \times 70 \\ 30 \times 60 + 0 \times 90 + \left( -20 \right ) \times 70 \\ \left( -20 \right ) \times 60 + 10 \times 90 + 10 \times 70 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 0 – 450 + 700 \\ 1800 + 0 – 1400 \\ -1200 + 900 + 700 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \frac{ 1 }{ 50 } \begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix}$

⟹ $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$

Therefore, a = 5, b = 8 and c = 8

Thus, the price for onions is Rs. 5 per kg, the price for wheat is Rs. 8 per kg and the price for the rice is Rs. 8 per kg.

Understanding the NCERT solutions is important as it helps one understand the different solutions. Students can download the PDF containing all the solutions for easy learning and retention. Using NCERT books for class 12 maths, students can become incredibly proficient in learning all the solutions of the different problems and exercises. It can become quite easy to learn and understand the solutions to the majority of problems that one may encounter during the main exam. Preparing for the exam with the utmost dedication and preparation can ensure a student to understand the course material for the exam better and score the maximum amount of marks possible. A student can gain extreme competency and proficiency to prepare for the exams.