Exercise 2.3 [Pages 55 - 56] Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 2 Matrices.

Exercise 2.3 [Pages 55 - 56]

Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 2 Matrices Exercise 2.3 [Pages 55 - 56]

QUESTION

Exercise 2.3 | Q 1.1 | Page 55

Evaluate : \( \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 2 & -4 & 3 \end{bmatrix} \)

SOLUTION

\( \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 2 & -4 & 3 \end{bmatrix} \)

= \( \begin{bmatrix} 3(2) & 3(-4) & 3(3) \\ 2(2) & 2(-4) & 2(3) \\ 1(2) & 1(-4) & 1(3) \end{bmatrix} \)

= \( \begin{bmatrix} 6 & -12 & 9 \\ 4 & -8 & 6 \\ 2 & -4 & 3 \end{bmatrix} \).

QUESTION

Exercise 2.3 | Q 1.2 | Page 55

Evaluate : \( \begin{bmatrix} 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \\ 1 \end{bmatrix} \)

SOLUTION

\( \begin{bmatrix} 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \\ 1 \end{bmatrix} \)

= \( \begin{bmatrix} 2(4) + (-1)(3) + 3(1) \end{bmatrix} \)
= \( \begin{bmatrix} 8 - 3 + 3 \end{bmatrix} \)
= \( \begin{bmatrix} 8 \end{bmatrix} \).

QUESTION

Exercise 2.3 | Q 2 | Page 55

If A = \( \begin{bmatrix} -1 & 1 & 1 \\ 2 & 3 & 0 \\ 1 & -3 & 1 \end{bmatrix} \), B = \( \begin{bmatrix} 2 & 1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 & 1 \end{bmatrix} \), state whether AB = BA? Justify your answer.

SOLUTION

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

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

= \( \begin{bmatrix} -2+3+1 & -1+0+2 & -4+2+1 \\ 4+9+0 & 2+0+0 & 8+6+0 \\ 2-9+1 & 1+0+2 & 4-6+1 \end{bmatrix} \)

∴ AB = \( \begin{bmatrix} 2 & 1 & -1 \\ 13 & 2 & 14 \\ -6 & 3 & -1 \end{bmatrix} \)      ...(i)

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

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

= \( \begin{bmatrix} -2+2+4 & 2+3-12 & 2+0+4 \\ -3+0+2 & 3+0-6 & 3+0+2 \\ -1+4+1 & 1+6-3 & 1+0+1 \end{bmatrix} \)

∴ BA = \( \begin{bmatrix} 4 & -7 & 6 \\ -1 & -3 & 5 \\ 4 & 4 & 2 \end{bmatrix} \)       ...(ii)
From (i) and (ii), we get
AB ≠ BA.

QUESTION

Exercise 2.3 | Q 3 | Page 55

Show that AB = BA, where A = \( \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \), B = \( \begin{bmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{bmatrix} \).

SOLUTION

AB = \( \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \begin{bmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{bmatrix} \)

= \( \begin{bmatrix} -2(1)+3(2)+(-1)(3) & -2(3)+3(2)+(-1)(0) & -2(-1)+3(-1)+(-1)(-1) \\ -1(1)+2(2)+(-1)(3) & -1(3)+2(2)+(-1)(0) & -1(-1)+2(-1)+(-1)(-1) \\ -6(1)+9(2)+(-4)(3) & -6(3)+9(2)+(-4)(0) & -6(-1)+9(-1)+(-4)(-1) \end{bmatrix} \)

= \( \begin{bmatrix} -2+6-3 & -6+6-0 & 2-3+1 \\ -1+4-3 & -3+4-0 & 1-2+1 \\ -6+18-12 & -18+18+0 & 6-9+4 \end{bmatrix} \)

∴ AB = \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)          ...(i)

BA = \( \begin{bmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{bmatrix} \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \)

= \( \begin{bmatrix} 1(-2)+3(-1)+(-1)(-6) & 1(3)+3(2)+(-1)(9) & 1(-1)+3(-1)+(-1)(-4) \\ 2(-2)+2(-1)+(-1)(-6) & 2(3)+2(2)+(-1)(9) & 2(-1)+2(-1)+(-1)(-4) \\ 3(-2)+0(-1)+(-1)(-6) & 3(3)+0(2)+(-1)(9) & 3(-1)+0(-1)+(-1)(-4) \end{bmatrix} \)

= \( \begin{bmatrix} -2-3+6 & 3+6-9 & -1-3+4 \\ -4-2+6 & 6+4-9 & -2-2+4 \\ -6+0+6 & 9+0-9 & -3+0+4 \end{bmatrix} \)

∴ BA = \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)        ...(ii)

From (i) and (ii), we get
AB = BA.

QUESTION

Exercise 2.3 | Q 4 | Page 55

Verify A(BC) = (AB)C, if A = \( \begin{bmatrix} 1 & 0 & 1 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix} \), B = \( \begin{bmatrix} 2 & -2 \\ -1 & 1 \\ 0 & 3 \end{bmatrix} \), C = \( \begin{bmatrix} 3 & 2 & -1 \\ 2 & 0 & -2 \end{bmatrix} \)

SOLUTION

BC = \( \begin{bmatrix} 2 & -2 \\ -1 & 1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 3 & 2 & -1 \\ 2 & 0 & -2 \end{bmatrix} \)

= \( \begin{bmatrix} 2(3)+(-2)(2) & 2(2)+(-2)(0) & 2(-1)+(-2)(-2) \\ -1(3)+1(2) & -1(2)+1(0) & -1(-1)+1(-2) \\ 0(3)+3(2) & 0(2)+3(0) & 0(-1)+3(-2) \end{bmatrix} \)

= \( \begin{bmatrix} 6-4 & 4-0 & -2+4 \\ -3+2 & -2+0 & 1-2 \\ 0+6 & 0+0 & 0-6 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 2 \\ -1 & -2 & -1 \\ 6 & 0 & -6 \end{bmatrix} \)

∴ A(BC) = \( \begin{bmatrix} 1 & 0 & 1 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix} \begin{bmatrix} 2 & 4 & 2 \\ -1 & -2 & -1 \\ 6 & 0 & -6 \end{bmatrix} \)

= \( \begin{bmatrix} 1(2)+0(-1)+1(6) & 1(4)+0(-2)+1(0) & 1(2)+0(-1)+1(-6) \\ 2(2)+3(-1)+0(6) & 2(4)+3(-2)+0(0) & 2(2)+3(-1)+0(-6) \\ 0(2)+4(-1)+5(6) & 0(4)+4(-2)+5(0) & 0(2)+4(-1)+5(-6) \end{bmatrix} \)

= \( \begin{bmatrix} 2-0+6 & 4-0+0 & 2-0-6 \\ 4-3+0 & 8-6+0 & 4-3-0 \\ 0-4+30 & 0-8+0 & 0-4-30 \end{bmatrix} \)

∴ A(BC) = \( \begin{bmatrix} 8 & 4 & -4 \\ 1 & 2 & 1 \\ 26 & -8 & -34 \end{bmatrix} \)      ...(i)

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

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

= \( \begin{bmatrix} 2+0+0 & -2+0+3 \\ 4-3+0 & -4+3+0 \\ 0-4+0 & 0+4+15 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & -1 \\ -4 & 19 \end{bmatrix} \)

∴ (AB)C = \( \begin{bmatrix} 2 & 1 \\ 1 & -1 \\ -4 & 19 \end{bmatrix} \begin{bmatrix} 3 & 2 & -1 \\ 2 & 0 & -2 \end{bmatrix} \)

= \( \begin{bmatrix} 2(3)+1(2) & 2(2)+1(0) & 2(-1)+1(-2) \\ 1(3)+(-1)(2) & 1(2)+(-1)(0) & 1(-1)+(-1)(-2) \\ -4(3)+19(2) & -4(2)+19(0) & -4(-1)+19(-2) \end{bmatrix} \)

= \( \begin{bmatrix} 6+2 & 4+0 & -2-2 \\ 3-2 & 2-0 & -1+2 \\ -12+38 & -8+0 & 4-38 \end{bmatrix} \)

∴ (AB)C = \( \begin{bmatrix} 8 & 4 & -4 \\ 1 & 2 & 1 \\ 26 & -8 & -34 \end{bmatrix} \)        ...(ii)

From (i) and (ii), we get
A(BC) = (AB)C.

QUESTION

Exercise 2.3 | Q 5 | Page 55

Verify that A(B + C) = AB + AC, if A = \( \begin{bmatrix} 4 & -2 \\ 2 & 3 \end{bmatrix} \), B = \( \begin{bmatrix} -1 & 1 \\ 3 & -2 \end{bmatrix} \) and C = \( \begin{bmatrix} 4 & 1 \\ 2 & -1 \end{bmatrix} \).

SOLUTION

A(B + C) = \( \begin{bmatrix} 4 & -2 \\ 2 & 3 \end{bmatrix} \left\{ \begin{bmatrix} -1 & 1 \\ 3 & -2 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 2 & -1 \end{bmatrix} \right\} \)

= \( \begin{bmatrix} 4 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} -1+4 & 1+1 \\ 3+2 & -2-1 \end{bmatrix} \)

= \( \begin{bmatrix} 4 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & -3 \end{bmatrix} \)

= \( \begin{bmatrix} 4(3)+(-2)(5) & 4(2)+(-2)(-3) \\ 2(3)+3(5) & 2(2)+3(-3) \end{bmatrix} = \begin{bmatrix} 12-10 & 8+6 \\ 6+15 & 4-9 \end{bmatrix} \)

= \( \begin{bmatrix} 2 & 14 \\ 21 & -5 \end{bmatrix} \)           ...(i)

AB + AC = \( \begin{bmatrix} 4 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 3 & -2 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & -1 \end{bmatrix} \)

= \( \begin{bmatrix} 4(-1)+(-2)(3) & 4(1)+(-2)(-2) \\ 2(-1)+3(3) & 2(1)+3(-2) \end{bmatrix} + \begin{bmatrix} 4(4)+(-2)(2) & 4(1)+(-2)(-1) \\ 2(4)+3(2) & 2(1)+3(-1) \end{bmatrix} \)

= \( \begin{bmatrix} -4-6 & 4+4 \\ -2+9 & 2-6 \end{bmatrix} + \begin{bmatrix} 16-4 & 4+2 \\ 8+6 & 2-3 \end{bmatrix} \)

= \( \begin{bmatrix} -10 & 8 \\ 7 & -4 \end{bmatrix} + \begin{bmatrix} 12 & 6 \\ 14 & -1 \end{bmatrix} \)

= \( \begin{bmatrix} -10+12 & 8+6 \\ 7+14 & -4-1 \end{bmatrix} = \begin{bmatrix} 2 & 14 \\ 21 & -5 \end{bmatrix} \)        ...(ii)
From (i) and (ii),  we get
A(B + C) = AB + AC.

QUESTION

Exercise 2.3 | Q 6 | Page 56

If A = \( \begin{bmatrix} 4 & 3 & 2 \\ -1 & 2 & 0 \end{bmatrix} \), B = \( \begin{bmatrix} 1 & 2 \\ -1 & 0 \\ 1 & -2 \end{bmatrix} \) show that matrix AB is non singular.

SOLUTION

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

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

= \( \begin{bmatrix} 4-3+2 & 8+0-4 \\ -1-2+0 & -2+0+0 \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ -3 & -2 \end{bmatrix} \)

∴ |AB| = \( \begin{vmatrix} 3 & 4 \\ -3 & -2 \end{vmatrix} \)

= 3(-2) - 4(-3) = – 6 + 12
= 6 ≠ 0
∴ AB is a non-singular matrix.

QUESTION

Exercise 2.3 | Q 7 | Page 56

If A + I = \( \begin{bmatrix} 1 & 2 & 0 \\ 5 & 4 & 2 \\ 0 & 7 & -3 \end{bmatrix} \), find the product (A + I)(A − I).

SOLUTION

A − I = (A + I) − 2I

= \( \begin{bmatrix} 1 & 2 & 0 \\ 5 & 4 & 2 \\ 0 & 7 & -3 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 2 & 0 \\ 5 & 4 & 2 \\ 0 & 7 & -3 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \)

= \( \begin{bmatrix} 1-2 & 2-0 & 0-0 \\ 5-0 & 4-2 & 2-0 \\ 0-0 & 7-0 & -3-2 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 0 \\ 5 & 2 & 2 \\ 0 & 7 & -5 \end{bmatrix} \)

(A + I)(A − I) = \( \begin{bmatrix} 1 & 2 & 0 \\ 5 & 4 & 2 \\ 0 & 7 & -3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 0 \\ 5 & 2 & 2 \\ 0 & 7 & -5 \end{bmatrix} \)

= \( \begin{bmatrix} 1(-1)+2(5)+0(0) & 1(2)+2(2)+0(7) & 1(0)+2(2)+0(-5) \\ 5(-1)+4(5)+2(0) & 5(2)+4(2)+2(7) & 5(0)+4(2)+2(-5) \\ 0(-1)+7(5)+(-3)(0) & 0(2)+7(2)+(-3)(7) & 0(0)+7(2)+(-3)(-5) \end{bmatrix} \)

= \( \begin{bmatrix} -1+10+0 & 2+4+0 & 0+4+0 \\ -5+20+0 & 10+8+14 & 0+8-10 \\ 0+35-0 & 0+14-21 & 0+14+15 \end{bmatrix} = \begin{bmatrix} 9 & 6 & 4 \\ 15 & 32 & -2 \\ 35 & -7 & 29 \end{bmatrix} \).

QUESTION

Exercise 2.3 | Q 8 | Page 56

If A = \( \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \), show that A² – 4A is a scalar matrix.

SOLUTION

A² – 4A = A.A – 4A

= \( \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} - 4 \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \)

= \( \begin{bmatrix} 1(1)+2(2)+2(2) & 1(2)+2(1)+2(2) & 1(2)+2(2)+2(1) \\ 2(1)+1(2)+2(2) & 2(2)+1(1)+2(2) & 2(2)+1(2)+2(1) \\ 2(1)+2(2)+1(2) & 2(2)+2(1)+1(2) & 2(2)+2(2)+1(1) \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} \)

= \( \begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} \)

= \( \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} \)

= \( \begin{bmatrix} 9-4 & 8-8 & 8-8 \\ 8-8 & 9-4 & 8-8 \\ 8-8 & 8-8 & 9-4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \), which is a scalar matrix.

QUESTION

Exercise 2.3 | Q 9 | Page 56

If A = \( \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} \), find k so that A² – 8A – kI = O, where I is a 2 × 2 unit and O is null matrix of order 2.

SOLUTION

A² – 8A – kI = O
∴ A.A – 8A – kI = O

∴ \( \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} - 8 \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} - k \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

∴ \( \begin{bmatrix} 1(1)+0(-1) & 1(0)+0(7) \\ -1(1)+7(-1) & -1(0)+7(7) \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ -8 & 56 \end{bmatrix} - \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

∴ \( \begin{bmatrix} 1 & 0 \\ -8 & 49 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ -8 & 56 \end{bmatrix} - \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

∴ \( \begin{bmatrix} 1-8-k & 0-0-0 \\ -8-(-8)-0 & 49-56-k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

∴ \( \begin{bmatrix} -7-k & 0 \\ 0 & -7-k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

∴ By equality of matrices, we get
-7 – k = 0
∴ k = – 7.

QUESTION

Exercise 2.3 | Q 10 | Page 56

If A = \( \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \), prove that A² – 5A + 7I = O, where I is a 2 x 2 unit matrix.

SOLUTION

A² – 5A + 7I = A.A – 5A + 7I

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

= \( \begin{bmatrix} 3(3)+1(-1) & 3(1)+1(2) \\ -1(3)+2(-1) & -1(1)+2(2) \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \)

= \( \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \)

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

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

= \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

= O.

QUESTION

Exercise 2.3 | Q 11 | Page 56

If A = \( \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \), B = \( \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} \) and (A + B)² = A² + B², find the values of a and b.

SOLUTION

Given (A + B)² = A² + B²
∴ A² + AB + BA + B² = A² + B²
∴ AB + BA = O (null matrix)
∴ AB = – BA

∴ \( \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} = - \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \)

∴ \( \begin{bmatrix} 1(2)+2(-1) & 1(a)+2(b) \\ -1(2)+(-2)(-1) & -1(a)+(-2)(b) \end{bmatrix} = - \begin{bmatrix} 2(1)+a(-1) & 2(2)+a(-2) \\ -1(1)+b(-1) & -1(2)+b(-2) \end{bmatrix} \)

∴ \( \begin{bmatrix} 2-2 & a+2b \\ -2+2 & -a-2b \end{bmatrix} = - \begin{bmatrix} 2-a & 4-2a \\ -1-b & -2-2b \end{bmatrix} \)

∴ \( \begin{bmatrix} 0 & a+2b \\ 0 & -a-2b \end{bmatrix} = \begin{bmatrix} a-2 & 2a-4 \\ 1+b & 2+2b \end{bmatrix} \)

∴ By equality of matrices, we get
a – 2 = 0 => a = 2
and 1 + b = 0 => b = – 1.

QUESTION

Exercise 2.3 | Q 12 | Page 56

Find k, if A = \( \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \) and A² = kA – 2I.

SOLUTION

A² = kA – 2I
∴ A.A + 2I = kA

∴ \( \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \)

∴ \( \begin{bmatrix} 3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\ 4(3)+(-2)(4) & 4(-2)+(-2)(-2) \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} \)

∴ \( \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} \)

∴ \( \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} \)

∴ \( \begin{bmatrix} 1+2 & -2+0 \\ 4+0 & -4+2 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} \)

∴ \( \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} \)

∴ By equality of matrices, we get
3k = 3
∴ k = 1.

QUESTION

Exercise 2.3 | Q 13 | Page 56

Find x and y, if \( \left\{ 4 \begin{bmatrix} 2 & -1 & 3 \\ 1 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -3 & 4 \\ 2 & 1 & 1 \end{bmatrix} \right\} \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)

SOLUTION

\( \left\{ 4 \begin{bmatrix} 2 & -1 & 3 \\ 1 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -3 & 4 \\ 2 & 1 & 1 \end{bmatrix} \right\} \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)

∴ \( \left\{ \begin{bmatrix} 8 & -4 & 12 \\ 4 & 0 & 8 \end{bmatrix} - \begin{bmatrix} 3 & -3 & 4 \\ 2 & 1 & 1 \end{bmatrix} \right\} \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)

∴ \( \begin{bmatrix} 8-3 & -4-(-3) & 12-4 \\ 4-2 & 0-1 & 8-1 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)

∴ \( \begin{bmatrix} 5 & -1 & 8 \\ 2 & -1 & 7 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)

∴ \( \begin{bmatrix} 5(2)+(-1)(-1)+8(1) \\ 2(2)+(-1)(-1)+7(1) \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)

∴ \( \begin{bmatrix} 10+1+8 \\ 4+1+7 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)

∴ \( \begin{bmatrix} 19 \\ 12 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \)

∴ By equality of matrices, we get
x = 19 and y = 12.

QUESTION

Exercise 2.3 | Q 14 | Page 56

Find x, y, z, if \( \left\{ 3 \begin{bmatrix} 2 & 0 \\ 0 & 2 \\ 2 & 2 \end{bmatrix} - 4 \begin{bmatrix} 1 & 1 \\ -1 & 2 \\ 3 & 1 \end{bmatrix} \right\} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} x-3 \\ y-1 \\ 2z \end{bmatrix} \).

SOLUTION

\( \left\{ 3 \begin{bmatrix} 2 & 0 \\ 0 & 2 \\ 2 & 2 \end{bmatrix} - 4 \begin{bmatrix} 1 & 1 \\ -1 & 2 \\ 3 & 1 \end{bmatrix} \right\} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} x-3 \\ y-1 \\ 2z \end{bmatrix} \)

∴ \( \left\{ \begin{bmatrix} 6 & 0 \\ 0 & 6 \\ 6 & 6 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ -4 & 8 \\ 12 & 4 \end{bmatrix} \right\} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} x-3 \\ y-1 \\ 2z \end{bmatrix} \)

∴ \( \begin{bmatrix} 6-4 & 0-4 \\ 0-(-4) & 6-8 \\ 6-12 & 6-4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} x-3 \\ y-1 \\ 2z \end{bmatrix} \)

∴ \( \begin{bmatrix} 2 & -4 \\ 4 & -2 \\ -6 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} x-3 \\ y-1 \\ 2z \end{bmatrix} \)

∴ \( \begin{bmatrix} 2(1)+(-4)(2) \\ 4(1)+(-2)(2) \\ -6(1)+2(2) \end{bmatrix} = \begin{bmatrix} x-3 \\ y-1 \\ 2z \end{bmatrix} \)

∴ \( \begin{bmatrix} 2-8 \\ 4-4 \\ -6+4 \end{bmatrix} = \begin{bmatrix} x-3 \\ y-1 \\ 2z \end{bmatrix} \)

∴ \( \begin{bmatrix} -6 \\ 0 \\ -2 \end{bmatrix} = \begin{bmatrix} x-3 \\ y-1 \\ 2z \end{bmatrix} \)

∴ By equality of matrices, we get
x – 3 = – 6 => x = -6 + 3 => x = – 3
y – 1 = 0 => y = 1
2z = – 2 => z = – 1.

QUESTION

Exercise 2.3 | Q 15 | Page 56

Jay and Ram are two friends. Jay wants to buy 4 pens and 8 notebooks, Ram wants to buy 5 pens and 12 notebooks. The price of one pen and one notebook was ₹ 6 and ₹ 10 respectively. Using matrix multiplication, find the amount each one of them requires for buying the pens and notebooks.

SOLUTION

Let A be the matrix of pens and notebooks required by Jay and Ram, and B be the matrix of prices of one pen and one notebook.

Requirements matrix A:
\( A = \begin{bmatrix} 4 & 8 \\ 5 & 12 \end{bmatrix} \) (Row 1: Jay's items [Pens, Notebooks], Row 2: Ram's items [Pens, Notebooks])

Prices matrix B:
\( B = \begin{bmatrix} 6 \\ 10 \end{bmatrix} \) (Column 1: Price per Pen, Price per Notebook)

The total amount required for each one of them is obtained by the matrix product AB.

∴ AB = \( \begin{bmatrix} 4 & 8 \\ 5 & 12 \end{bmatrix} \begin{bmatrix} 6 \\ 10 \end{bmatrix} \)

= \( \begin{bmatrix} 4(6)+8(10) \\ 5(6)+12(10) \end{bmatrix} \)

= \( \begin{bmatrix} 24+80 \\ 30+120 \end{bmatrix} = \begin{bmatrix} 104 \\ 150 \end{bmatrix} \)

∴ Jay needs ₹ 104 and Ram needs ₹ 150.