# 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 :

#### SOLUTION

[321][2 -4  3]

= $\left[\begin{array}{ccc}3\left(2\right)& 3\left(-4\right)& 3\left(3\right)\\ 2\left(2\right)& 2\left(-4\right)& 2\left(3\right)\\ 1\left(2\right)& 1\left(-4\right)& 1\left(3\right)\end{array}\right]$

= $\left[\begin{array}{ccc}6& 12& 9\\ 4& 8& 6\\ 2& -4& 3\end{array}\right]$.

#### QUESTION

Exercise 2.3 | Q 1.2 | Page 55

Evaluate :

#### SOLUTION

[2-1  3][431]

= [2(4) –1(3) + 3(1)]
= [8 – 3 + 3]
= [8].

#### QUESTION

Exercise 2.3 | Q 2 | Page 55

If A = $\left[\begin{array}{ccc}-1& 1& 1\\ 2& 3& 0\\ 1& -3& 1\end{array}\right],\text{B}=\left[\begin{array}{ccc}2& 1& 4\\ 3& 0& 2\\ 1& 2& 1\end{array}\right]$, state whether AB = BA? Justify your answer.

#### SOLUTION

AB = $\left[\begin{array}{ccc}-1& 1& 1\\ 2& 3& 0\\ 1& -3& 1\end{array}\right]\left[\begin{array}{ccc}2& 1& 4\\ 3& 0& 2\\ 1& 2& 1\end{array}\right]$

= $\left[\begin{array}{ccc}-2+3+1& -1+0+2& -4+2+1\\ 4+9+0& 2+0+0& 8+6+0\\ 2-9+1& 1+0+2& 4-6+1\end{array}\right]$

∴ AB = $\left[\begin{array}{ccc}2& 1& -1\\ 13& 2& 14\\ -6& 3& -1\end{array}\right]$     ...(i)

BA = $\left[\begin{array}{ccc}2& 1& 4\\ 3& 0& 2\\ 1& 2& 1\end{array}\right]\left[\begin{array}{ccc}-1& 1& 1\\ 2& 3& 0\\ 1& -3& 1\end{array}\right]$

= $\left[\begin{array}{ccc}-2+2+4& 2+3-12& 2+0+4\\ -3+0+2& 3+0-6& 3+0+2\\ -1+4+1& 1+6-3& 1+0+1\end{array}\right]$

∴ BA = $\left[\begin{array}{ccc}4& -7& 6\\ -1& -3& 5\\ 4& 4& 2\end{array}\right]$      ...(ii)
From (i) and (ii), we get
AB ≠ BA.

#### QUESTION

Exercise 2.3 | Q 3 | Page 55

Show that AB = BA, where A = $\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right],\text{B}=\left[\begin{array}{ccc}1& 3& -1\\ 2& 2& -1\\ 3& 0& -1\end{array}\right]$.

#### SOLUTION

AB = $\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right]\left[\begin{array}{ccc}1& 3& -1\\ 2& 2& -1\\ 3& 0& -1\end{array}\right]$

= $\left[\begin{array}{ccc}-2+6-3& -6+6-0& 2-3+1\\ -1+4-3& -3+4-0& 1-2+1\\ -6+18-12& -18+18+0& 6-9+4\end{array}\right]$

∴ AB = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$         ...(i)

BA = $\left[\begin{array}{ccc}1& 3& -1\\ 2& 2& -1\\ 3& 0& -1\end{array}\right]\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right]$

= $\left[\begin{array}{ccc}2-3+6& 3+6-9& -1-3+4\\ -4-2+6& 6+4-9& -2-2+4\\ -6+0+6& 9+0-9& 3+0+4\end{array}\right]$

∴ BA = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$       ...(ii)

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

#### QUESTION

Exercise 2.3 | Q 4 | Page 55

Verify A(BC) = (AB)C, if A = $\left[\begin{array}{ccc}1& 0& 1\\ 2& 3& 0\\ 0& 4& 5\end{array}\right],\text{B}=\left[\begin{array}{cc}2& -2\\ -1& 1\\ 0& 3\end{array}\right]$

#### SOLUTION

BC = $\left[\begin{array}{cc}2& -2\\ -1& 1\\ 0& 3\end{array}\right]\left[\begin{array}{ccc}3& 2& -1\\ 2& 0& -2\end{array}\right]$

= $\left[\begin{array}{ccc}6-4& 4-0& -2+4\\ -3+2& -2+0& 1-2\\ 0+6& 0+0& 0-6\end{array}\right]$

= $\left[\begin{array}{ccc}2& 4& 2\\ 1& 2& -1\\ 6& 0& -6\end{array}\right]$

∴ A(BC) = $\left[\begin{array}{ccc}1& 0& 1\\ 2& 3& 0\\ 0& 4& 5\end{array}\right]\left[\begin{array}{ccc}2& 4& 2\\ -1& -2& -1\\ 6& 0& -6\end{array}\right]$

= $\left[\begin{array}{ccc}2-0+6& 4-0+0& 2-0-6\\ 4-3+0& 8-6+0& 4-3-0\\ 0-4+30& 0-8+0& 0-4-30\end{array}\right]$

∴ A(BC) = $\left[\begin{array}{ccc}8& 4& -4\\ 1& 2& 1\\ 26& -8& -34\end{array}\right]$     ...(i)

AB = $\left[\begin{array}{ccc}1& 0& 1\\ 2& 3& 0\\ 0& 4& 5\end{array}\right]\left[\begin{array}{cc}2& -2\\ -1& 1\\ 0& 3\end{array}\right]$

= $\left[\begin{array}{cc}2+0+0& -2+0+3\\ 4-3+0& -4+3+0\\ 0-4+0& 0+4+15\end{array}\right]$

= $\left[\begin{array}{cc}2& 1\\ 1& -1\\ -4& 19\end{array}\right]$

∴ (AB)C = $\left[\begin{array}{cc}2& 1\\ 1& -1\\ -4& 19\end{array}\right]\left[\begin{array}{ccc}3& 2& -1\\ 2& 0& -2\end{array}\right]$

= $\left[\begin{array}{ccc}6+2& 4+0& -2-2\\ 3-2& 2-0& -1+2\\ -12+38& -8+0& 4-38\end{array}\right]$

∴ (AB)C = $\left[\begin{array}{ccc}8& 4& -4\\ 1& 2& 1\\ 26& -8& 34\end{array}\right]$       ...(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 = $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right],\text{B}=\left[\begin{array}{cc}-1& 1\\ 3& -2\end{array}\right]\phantom{\rule{1ex}{0ex}}\text{and C}=\left[\begin{array}{cc}4& 1\\ 2& -1\end{array}\right]$.

#### SOLUTION

A(B + C) = $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left\{\left[\begin{array}{cc}-1& 1\\ 3& -2\end{array}\right]+\left[\begin{array}{cc}4& 1\\ 2& -1\end{array}\right]\right\}$

= $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}-1+4& 1+1\\ 3+2& -2-1\end{array}\right]$

= $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}3& 2\\ 5& -3\end{array}\right]$

= $\left[\begin{array}{cc}12-10& 8+6\\ 6+15& 4-9\end{array}\right]$

= $\left[\begin{array}{cc}2& 14\\ 21& -5\end{array}\right]$          ...(i)

AB + AC = $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 3& -2\end{array}\right]+\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}4& 1\\ 2& -1\end{array}\right]$

= $\left[\begin{array}{cc}-4-6& 4+4\\ -2+9& 2-6\end{array}\right]+\left[\begin{array}{cc}16-4& 4+2\\ 8+6& 2-3\end{array}\right]$

= $\left[\begin{array}{cc}-10& 8\\ 7& -4\end{array}\right]+\left[\begin{array}{cc}12& 6\\ 14& -1\end{array}\right]$

= $\left[\begin{array}{cc}-10+12& 8+6\\ 7+14& -4-1\end{array}\right]$

= $\left[\begin{array}{cc}2& 14\\ 21& -5\end{array}\right]$       ...(ii)
From (i) and (ii),  we get
A(B + C) = AB + AC.

#### QUESTION

Exercise 2.3 | Q 6 | Page 56

If  A = $\left[\begin{array}{ccc}4& 3& 2\\ -1& 2& 0\end{array}\right],\text{B}=\left[\begin{array}{cc}1& 2\\ -1& 0\\ 1& -2\end{array}\right]$ show that matrix AB is non singular.

#### SOLUTION

AB = $\left[\begin{array}{ccc}4& 3& 2\\ -1& 2& 0\end{array}\right]\left[\begin{array}{cc}1& 2\\ -1& 0\\ 1& -2\end{array}\right]$

= $\left[\begin{array}{cc}4-3+2& 8+0-4\\ -1-2+0& -2+0+0\end{array}\right]$

= $\left[\begin{array}{cc}3& 4\\ -3& -2\end{array}\right]$

∴ |AB| = $|\begin{array}{cc}3& 4\\ -3& -2\end{array}|$

= – 6 + 12
= 6 ≠ 0
∴ AB is a non-singular matrix.

#### QUESTION

Exercise 2.3 | Q 7 | Page 56

If A + I = $\left[\begin{array}{ccc}1& 2& 0\\ 5& 4& 2\\ 0& 7& -3\end{array}\right]$, find the product (A + I)(A − I).

#### SOLUTION

A − I = A I − 2I
= $\left[\begin{array}{ccc}1& 2& 0\\ 5& 4& 2\\ 0& 7& -3\end{array}\right]-2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

= $\left[\begin{array}{ccc}1& 2& 0\\ 5& 4& 2\\ 0& 7& -3\end{array}\right]\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$

= $\left[\begin{array}{ccc}1-2& 2-0& 0-0\\ 5-0& 4-2& 2-0\\ 0-0& 7-0& -3-2\end{array}\right]$

= $\left[\begin{array}{ccc}-1& 2& 0\\ 5& 2& 2\\ 0& 7& -5\end{array}\right]$

(A + I)(A − I) = $\left[\begin{array}{ccc}1& 2& 0\\ 5& 4& 2\\ 0& 7& -3\end{array}\right]\left[\begin{array}{ccc}-1& 2& 0\\ 5& 2& 2\\ 0& 7& -5\end{array}\right]$

= $\left[\begin{array}{ccc}-1+10+0& 2+4+0& 0+4+0\\ -5+20+0& 10+8+14& 0+8-10\\ 0+35-0& 0+14-21& 0+14+15\end{array}\right]$

$\left[\begin{array}{ccc}9& 6& 4\\ 15& 32& 2\\ 35& -7& 29\end{array}\right]$.

#### QUESTION

Exercise 2.3 | Q 8 | Page 56

If A = $\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$, show that A2 – 4A is a scalar matrix.

#### SOLUTION

A2 – 4A = A.A – 4A
= $\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]-4\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$

= $\left[\begin{array}{ccc}1+4+4& 2+2+4& 2+4+2\\ 2+2+4& 4+1+4& 4+2+2\\ 2+4+2& 4+2+2& 4+4+1\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]$

= $\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]$

= $\left[\begin{array}{ccc}9-4& 8-8& 8-8\\ 8-8& 9-4& 8-8\\ 8-8& 8-8& 9-4\end{array}\right]$

= $\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]$, which is a scalar martix.

#### QUESTION

Exercise 2.3 | Q 9 | Page 56

If A = $\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$, 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

$\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]-8\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]-\text{k}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]-\left[\begin{array}{cc}8& 0\\ -8& 56\end{array}\right]-\left[\begin{array}{cc}\text{k}& 0\\ 0& \text{k}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\left[\begin{array}{cc}1-8-\text{k}& 0-0-0\\ -8+8-0& 4-6-\text{k}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

∴ By equality of matrices, we get
1 – 8 – k = 0
∴ k = – 7.

#### QUESTION

Exercise 2.3 | Q 10 | Page 56

If A = $\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, prove that A2 – 5A + 7I = 0, where I is a 2 x 2 unit matrix.

#### SOLUTION

A2 – 5A + 7I = A.A – 5A + 7I

= $\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

=

= $\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$

= $\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right]$

= $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

= 0.

#### QUESTION

Exercise 2.3 | Q 11 | Page 56

If A = $\left[\begin{array}{cc}1& 2\\ -1& -2\end{array}\right],\text{B}=\left[\begin{array}{cc}2& \text{a}\\ -1& \text{b}\end{array}\right]$ 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 = 0
∴ AB = – BA

$\left[\begin{array}{cc}2-2& \text{a}+2\text{b}\\ -2+2& -\text{a}-2\text{b}\end{array}\right]=-\left[\begin{array}{cc}2-\text{a}& 4-2\text{a}\\ -1-\text{b}& -2-2\text{b}\end{array}\right]$

$\left[\begin{array}{cc}0& \text{a}+2\text{b}\\ 0& -\text{a}-2\text{b}\end{array}\right]=\left[\begin{array}{cc}-2+\text{a}& -4+2\text{a}\\ 1+\text{b}& 2+2\text{b}\end{array}\right]$

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

#### QUESTION

Exercise 2.3 | Q 12 | Page 56

Find k, if A = $\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$ and A2 = kA – 2I.

#### SOLUTION

A2 = kA – 2I
∴ A.A + 2I = kA

$\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]+2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\text{k}\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$

$\left[\begin{array}{cc}9-8& -6+4\\ 12-8& -8+4\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}3\text{k}& -2\text{k}\\ 4\text{k}& -2\text{k}\end{array}\right]$

$\left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}3\text{k}& -2\text{k}\\ 4\text{k}& -2\text{k}\end{array}\right]$

$\left[\begin{array}{cc}1+2& -2+0\\ 4+0& -4+2\end{array}\right]=\left[\begin{array}{cc}3\text{k}& -2\text{k}\\ 4\text{k}& -2\text{k}\end{array}\right]$

$\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]=\left[\begin{array}{cc}3\text{k}& -2\text{k}\\ 4\text{k}& -2\text{k}\end{array}\right]$

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

#### QUESTION

Exercise 2.3 | Q 13 | Page 56

Find x and y, if $\left\{4\left[\begin{array}{ccc}2& -1& 3\\ 1& 0& 2\end{array}\right]-\left[\begin{array}{ccc}3& -3& 4\\ 2& 1& 1\end{array}\right]\right\}\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

#### SOLUTION

{4[2-13102]-[3-34211]}[2-11]=[xy]

$\left\{\left[\begin{array}{ccc}8& -4& 12\\ 4& 0& 8\end{array}\right]-\left[\begin{array}{ccc}3& -3& 4\\ 2& 1& 1\end{array}\right]\right\}\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

$\left[\begin{array}{ccc}8-3& -4+3& 12-4\\ 4-2& 0-1& 8-1\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

$\left[\begin{array}{ccc}5& -1& 8\\ 2& -1& 7\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

$\left[\begin{array}{c}10+1+8\\ 4+1+7\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

$\left[\begin{array}{c}19\\ 12\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

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

#### QUESTION

Exercise 2.3 | Q 14 | Page 56

Find x, y, x, if $\left\{3\left[\begin{array}{cc}2& 0\\ 0& 2\\ 2& 2\end{array}\right]-4\left[\begin{array}{cc}1& 1\\ -1& 2\\ 3& 1\end{array}\right]\right\}\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$.

#### SOLUTION

{3[200222]-4[11-1231]}[12]=[x-3y-12z]

$\left\{\left[\begin{array}{cc}6& 0\\ 0& 6\\ 6& 6\end{array}\right]-\left[\begin{array}{cc}4& 4\\ -4& 8\\ 12& 4\end{array}\right]\right\}\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

$\left[\begin{array}{cc}6-4& 0-4\\ 0+4& 6-8\\ 6-12& 6-4\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

$\left[\begin{array}{cc}2& -4\\ 4& -2\\ -6& 2\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

$\left[\begin{array}{c}2-8\\ 4-4\\ -6+4\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

$\left[\begin{array}{c}-6\\ 0\\ -2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

∴ By equality of martices, we get
x – 3 = – 6, y – 1  = 0, 2z = – 2
∴ x = – 3, y = 1, 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 and B be the matrix od prices of one pen and one notebook.
Pens Notebooks
∴ A = $\left[\begin{array}{cc}4& 8\\ 5& 12\end{array}\right]\frac{\text{jay}}{\text{Ram}}$

and B = $\left[\begin{array}{c}6\\ 10\end{array}\right]\frac{\text{Pen}}{\text{Notebook}}$

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

∴ AB = $\left[\begin{array}{cc}4& 8\\ 5& 12\end{array}\right]\left[\begin{array}{c}6\\ 10\end{array}\right]$

= $\left[\begin{array}{c}24+80\\ 30+120\end{array}\right]$

= $\left[\begin{array}{c}104\\ 150\end{array}\right]$
∴ Jay needs ₹ 104 and Ram needs ₹ 150.