Chapter 2 - Matrices EXERCISE 2.2 [PAGES 46 - 47] Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board

Chapter 2 - Matrices EXERCISE 2.2 [PAGES 46 - 47]

Exercise 2.2 | Q 1.1 | Page 46

QUESTION

If A = ${\displaystyle \left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right],\text{B}=\left[\begin{array}{cc}-1& 2\\ 2& 2\\ 0& 3\end{array}\right]\text{and C}=\left[\begin{array}{cc}4& 3\\ -1& 4\\ -2& 1\end{array}\right]}$, Show that A + B = B + A

SOLUTION

A + B = ${\displaystyle \left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right]+\left[\begin{array}{cc}-1& 2\\ 2& 2\\ 0& 3\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}2-1& -3+2\\ 5+2& -4+2\\ -6+0& 1+3\end{array}\right]}$

∴ A + B = ${\displaystyle \left[\begin{array}{cc}1& -1\\ 7& -2\\ -6& 4\end{array}\right]}$       ....(i)

B + A = ${\displaystyle \left[\begin{array}{cc}-1& 2\\ 2& 2\\ 0& 3\end{array}\right]+\left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}-1+2& 2-3\\ 2+5& 2-4\\ 0-6& 3+1\end{array}\right]}$

∴ B + A = ${\displaystyle \left[\begin{array}{cc}1& -1\\ 7& -2\\ -6& 4\end{array}\right]}$       ....(ii)

From (i) and (ii), we get
A + B = B + A.

Exercise 2.2 | Q 1.2 | Page 46

QUESTION

If A = ${\displaystyle \left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right],\text{B}=\left[\begin{array}{cc}-1& 2\\ 2& 2\\ 0& 3\end{array}\right]\text{and C}=\left[\begin{array}{cc}4& 3\\ -1& 4\\ -2& 1\end{array}\right]}$, Show that (A + B) + C = A + (B + C)

SOLUTION

(A + B) + C = ${\displaystyle \left\{\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right]+\left[\begin{array}{cc}-1& 2\\ 2& 2\\ 0& 3\end{array}\right]\}+\left[\begin{array}{cc}4& 3\\ -1& 4\\ -2& 1\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}2-1& -3+2\\ 5+2& -4+2\\ -6+0& 1+3\end{array}\right]+\left[\begin{array}{cc}4& 3\\ -1& 4\\ -2& 1\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}1& -1\\ 7& -2\\ -6& 4\end{array}\right]+\left[\begin{array}{cc}4& 3\\ -1& 4\\ -2& 1\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}1+4& -1+3\\ 7-1& -2+4\\ -6-2& 4+1\end{array}\right]}$

∴ (A+ B) + C = ${\displaystyle \left[\begin{array}{cc}5& 2\\ 6& 2\\ -8& 5\end{array}\right]}$      ....(i)

A + (B + C) = ${\displaystyle \left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right]+\{\left[\begin{array}{cc}-1& 2\\ 2& 2\\ 0& \end{array}\right]+\left[\begin{array}{cc}4& 3\\ -1& 4\\ -2& 1\end{array}\right]\}}$

= ${\displaystyle \left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right]+\left[\begin{array}{cc}-1+4& 2+3\\ 2-1& 2+4\\ 0-2& 3+1\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}2& -3\\ 5& -4\\ -6& 1\end{array}\right]\left[\begin{array}{cc}3& 5\\ 1& 6\\ -2& 4\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}2+3& -3+5\\ 5+1& -4+6\\ -6-2& 1+4\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}5& 2\\ 6& 2\\ -8& 5\end{array}\right]}$         ....(ii)

From (i) and (ii), we get
(A + B) + C = A + (B + C).

Exercise 2.2 | Q 2 | Page 46

QUESTION

If A = ${\displaystyle \left[\begin{array}{cc}1& -2\\ 5& 3\end{array}\right],\text{B}=\left[\begin{array}{cc}1& -3\\ 4& -7\end{array}\right]}$ , then find the matrix A − 2B + 6I, where I is the unit matrix of order 2.

SOLUTION

A – 2B + 6I = ${\displaystyle \left[\begin{array}{cc}1& -2\\ 5& 3\end{array}\right] 2\left[\begin{array}{cc}1& -3\\ 4& -7\end{array}\right]+6\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}1& -2\\ 5& 3\end{array}\right]-\left[\begin{array}{cc}2& -6\\ 8& -14\end{array}\right]+\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}1-2+6& -2+6+0\\ 5-8+0& +14+6\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}5& 4\\ -3& 23\end{array}\right]}$.

Exercise 2.2 | Q 3 | Page 46

QUESTION

If A = ${\displaystyle \left[\begin{array}{ccc}1& 2& -3\\ -3& 7& -8\\ 0& -6& 1\end{array}\right],\text{B}=\left[\begin{array}{ccc}9& -1& 2\\ -4& 2& 5\\ 4& 0& -3\end{array}\right]}$ then find the matrix C such that A + B + C is a zero matrix.

SOLUTION

A + B + C is a zero martix.
∴ A + B + C = 0
∴ C = – (A + B)

= ${\displaystyle -\{\left[\begin{array}{ccc}1& 2& 3\\ -3& 7& -8\\ 0& -6& 1\end{array}\right]+\left[\begin{array}{ccc}9& -1& 2\\ -4& 2& 5\\ 4& 0& -3\end{array}\right]\}}$

= ${\displaystyle -\left[\begin{array}{ccc}1+9& 2-1& -3+2\\ -3-4& 7+2& -8+5\\ 0+4& -6+0& 1-3\end{array}\right]}$

= ${\displaystyle -\left[\begin{array}{ccc}10& 1& -1\\ -7& 9& -3\\ 4& -6& -2\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{ccc}10& -1& 1\\ -7& -9& 3\\ -4& 6& -2\end{array}\right]}$.

Exercise 2.2 | Q 4 | Page 46 

QUESTION

If A = ${\displaystyle \left[\begin{array}{cc}1& -2\\ 3& -5\\ -6& 0\end{array}\right],\text{B}=\left[\begin{array}{cc}-1& -2\\ 4& 2\\ 1& 5\end{array}\right]\text{and C}=\left[\begin{array}{cc}2& 4\\ -1& -4\\ -3& 6\end{array}\right]}$, find the matrix X such that 3A – 4B + 5X = C.

SOLUTION

3A – 4B + 5X = C
∴ 5X =C + 4B – 3A

= ${\displaystyle \left[\begin{array}{cc}2& 4\\ -1& -4\\ -3& 6\end{array}\right]+4\left[\begin{array}{cc}-1& -2\\ 4& 2\\ 1& 5\end{array}\right]-3\left[\begin{array}{cc}1& -2\\ 3& -5\\ -6& 0\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}2& 4\\ -1& -4\\ -3& 6\end{array}\right]+\left[\begin{array}{cc}-4& -8\\ 16& 8\\ 4& 20\end{array}\right]-\left[\begin{array}{cc}3& -6\\ 9& -15\\ -18& 0\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}2-4- 3& 4-8+6\\ -1+16-9& -4+8+5\\ -3+4+ 18& 6+20-0\end{array}\right]}$

= 5X = ${\displaystyle \left[\begin{array}{cc}-5& 2\\ 6& 19\\ 19& 26\end{array}\right]}$

∴ X = ${\displaystyle \frac{1}{5}\left[\begin{array}{cc}-5& 2\\ 6& 19\\ 19& 26\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}-1& \frac{2}{5}\\ \frac{6}{5}& \frac{19}{5}\\ \frac{19}{5}& \frac{26}{5}\end{array}\right]}$.

Exercise 2.2 | Q 5 | Page 46

QUESTION

If A = ${\displaystyle \left[\begin{array}{ccc}5& 1& -4\\ 3& 2& 0\end{array}\right]}$, find (A^T)^T.

SOLUTION

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

∴ A^T = ${\displaystyle \left[\begin{array}{cc}5& 3\\ 1& 2\\ -4& 0\end{array}\right]}$

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

 Exercise 2.2 | Q 6 | Page 46

QUESTION

If A = ${\displaystyle \left[\begin{array}{ccc}7& 3& 1\\ -2& -4& 1\\ 5& 9& 1\end{array}\right]}$, find (A^T)^T.

SOLUTION

A = ${\displaystyle \left[\begin{array}{ccc}7& 3& 1\\ -2& -4& 1\\ 5& 9& 1\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}7& -2& 5\\ 3& -4& 9\\ 1& 1& 1\end{array}\right]}$

∴ (A^T)^T = ${\displaystyle \left[\begin{array}{ccc}7& 3& 1\\ -2& -4& 1\\ 5& 9& 1\end{array}\right]}$ = A.

Exercise 2.2 | Q 7 | Page 47

QUESTION

Find a, b, c, if ${\displaystyle \left[\begin{array}{ccc}1& \frac{3}{5}& \text{a}\\ \text{b}& -5& -7\\ -4& \text{c}& 0\end{array}\right]}$ is a symmetric matrix.

SOLUTION

Let A = ${\displaystyle \left[\begin{array}{ccc}1& \frac{3}{5}& \text{a}\\ \text{b}& -5& -7\\ -4& \text{c}& 0\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}1& \text{b}& 4\\ \frac{3}{5}& -5& \text{c}\\ \text{a}& -7& 0\end{array}\right]}$

Since A is a symmetric matrix,
A = A^T

∴ ${\displaystyle \left[\begin{array}{ccc}1& \frac{3}{5}& \text{a}\\ \text{b}& -5& -7\\ -4& \text{c}& 0\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{ccc}1& \text{b}& -4\\ \frac{3}{5}& -5& \text{c}\\ \text{a}& -7& 0\end{array}\right]}$

∴ By equality of matrices, we get

a = – 4, b = ${\displaystyle \frac{3}{5}}$, c = – 7.

Exercise 2.2 | Q 8 | Page 47

QUESTION

Find x, y, z if ${\displaystyle \left[\begin{array}{ccc}0& -5i& x\\ y& 0& z\\ \frac{3}{2}& -\sqrt{2}& 0\end{array}\right]}$ is a skew symmetric matrix.

SOLUTION

Let A = ${\displaystyle \left[\begin{array}{ccc}0& -5i& x\\ y& 0& z\\ \frac{3}{2}& -\sqrt{2}& 0\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}0& -5\text{i}& x\\ y& 0& z\\ \frac{3}{2}& -\sqrt{2}& 0\end{array}\right]}$

Since A is a skew-symmetric matrix,
A = A^T

∴ ${\displaystyle \left[\begin{array}{ccc}0& -5\text{i}& x\\ y& 0& z\\ \frac{3}{2}& -\sqrt{2}& 0\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{ccc}0& -y& \frac{-3}{2}\\ 5\text{i}& 0& \sqrt{2}\\ -x& -z& 0\end{array}\right]}$

∴ By equality of matrices, we get

x = ${\displaystyle \frac{-3}{2},y=5\text{i},z=\sqrt{2}}$

Exercise 2.2 | Q 9.1 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew- symmetric or neither.

${\displaystyle \left[\begin{array}{ccc}1& 2& -5\\ 2& -3& 4\\ -5& 4& 9\end{array}\right]}$

${\displaystyle }$

SOLUTION

Let A = ${\displaystyle \left[\begin{array}{ccc}1& 2& -5\\ 2& -3& 4\\ -5& 4& 9\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}1& 2& -5\\ 2& -3& 4\\ -5& 4& 9\end{array}\right]}$

∴ A^T = A i.e., A = A^T
∴ A is a symmetric matrix.

Exercise 2.2 | Q 9.2 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew- symmetric or neither.

${\displaystyle \left[\begin{array}{ccc}2& 5& 1\\ -5& 4& 6\\ -1& -6& 3\end{array}\right]}$

${\displaystyle }$

SOLUTION

Let A = ${\displaystyle \left[\begin{array}{ccc}2& 5& 1\\ -5& 4& 6\\ -1& -6& 3\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}2& -5& -1\\ -5& 4& -6\\ 1& 6& 3\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}-2& 5& 1\\ -5& -4& 6\\ -1& -6& -3\end{array}\right]}$

∴ A ≠ A^T and A ≠ – A^T
∴ A is neither symmetric nor skew – symmetric matrix.

Exercise 2.2 | Q 9.3 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew- symmetric or neither.

${\displaystyle \left[\begin{array}{ccc}0& 1+2\text{i}& \text{i}-2\\ -1-2\text{i}& 0& -7\\ 2-\text{i}& 7& 0\end{array}\right]}$

${\displaystyle }$

SOLUTION

Let A = ${\displaystyle \left[\begin{array}{ccc}0& 1+2\text{i}& \text{i}-2\\ -1-2\text{i}& 0& -7\\ 2-\text{i}& 7& 0\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}0& 1-2\text{i}& 2-\text{i}\\ -1+2\text{i}& 0& 7\\ \text{i}-2& -7& 0\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}0& 1+2\text{i}& \text{i}-2\\ -1-2\text{i}& 0& -7\\ 2-\text{i}& 7& 0\end{array}\right]}$

^∴ A^T = – A i.e. A = – A^T
∴ A is a skew-symmetric matrix.

Exercise 2.2 | Q 10 | Page 47

QUESTION

Construct the matrix A = [a_ij]_3×3 where a_ij = i − j. State whether A is symmetric or skew-symmetric.

SOLUTION

A = [a_ij]_3x3  

∴ A = ${\displaystyle \left[\begin{array}{ccc}{\text{a}}_{11}& {\text{a}}_{12}& {\text{a}}_{13}\\ {\text{a}}_{21}& {\text{a}}_{22}& {\text{a}}_{23}\\ {\text{a}}_{31}& {\text{a}}_{32}& {\text{a}}_{33}\end{array}\right]}$

Given, a_ij = i – j
∴ a₁₁ = 1 – 1 = 0, a₁₂ = 1 – 2 = – 1, a₁₃ = 1 – 3 = – 2
a₂₁ = 2 – 1= 1, a₂₂ = 2 – 2 = 0, a₂₃ = 2 – 3 = – 1,
a₃₁ = 3 – 1 = 2, a₃₂ = 3 - 2 = 1, a₃₃ = 3 – 3 = 0

∴ A = ${\displaystyle \left[\begin{array}{ccc}0& -1& -2\\ 1& 0& -1\\ 2& 1& 0\end{array}\right]}$

∴ A^T = ${\displaystyle \left[\begin{array}{ccc}0& 1& 2\\ -1& 0& 1\\ -2& -1& 0\end{array}\right]}$

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

∴ A^T = – A i.e., A = – A^T
∴ A is a skew-symmetric matrix.

Exercise 2.2 | Q 11 | Page 47

QUESTION

Solve the following equations for X and Y, if 3X − Y = ${\displaystyle \left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]}$  and X – 3Y = ${\displaystyle \left[\begin{array}{cc}0& -1\\ 0& -1\end{array}\right]}$.

SOLUTION

Given equations are

3X – Y = ${\displaystyle \left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]}$          ...(i)

and X – 3Y = ${\displaystyle \left[\begin{array}{cc}0& -1\\ 0& -1\end{array}\right]}$     ...(ii)

By (i) x 3 – (ii), we get

8X = ${\displaystyle 3\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]-\left[\begin{array}{cc}0& -1\\ 0& -1\end{array}\right]}$

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

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

∴ 8X = ${\displaystyle \left[\begin{array}{cc}3& -2\\ -3& 4\end{array}\right]}$

∴ X = ${\displaystyle \frac{1}{8}\left[\begin{array}{cc}3& -2\\ -3& 4\end{array}\right]}$

∴ X = ${\displaystyle \left[\begin{array}{cc}\frac{3}{8}& \frac{-2}{8}\\ \frac{-3}{8}& \frac{4}{8}\end{array}\right]}$

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

By (i) – (ii) x 3, we get

8Y = ${\displaystyle \left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]-3\left[\begin{array}{cc}0& -1\\ 0& -1\end{array}\right]}$

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

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

∴ 8Y = ${\displaystyle \left[\begin{array}{cc}1& 2\\ -11& 4\end{array}\right]}$

∴ Y = ${\displaystyle \frac{1}{8}\left[\begin{array}{cc}1& 2\\ -11& 4\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}\frac{1}{8}& \frac{2}{8}\\ \frac{-1}{8}& \frac{4}{8}\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}\frac{1}{8}& \frac{1}{4}\\ \frac{-1}{8}& \frac{1}{2}\end{array}\right]}$.

Exercise 2.2 | Q 12 | Page 47

QUESTION

Find matrices A and B, if 2A – B = ${\displaystyle \left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]}$ and A – 2B = ${\displaystyle \left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]}$.

SOLUTION

Given equations are

2A – B = ${\displaystyle \left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]}$           ...(i)

and A – 2B = ${\displaystyle \left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]}$    ...(ii)

By (i) – (ii) x 2, we get

3B = ${\displaystyle \left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]-2\left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]}$ 

= ${\displaystyle \left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]-\left[\begin{array}{ccc}6& 4& 16\\ -4& 2& -14\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{ccc}6-6& -6-4& 0-16\\ -4+4& 2-2& 1+14\end{array}\right]}$

∴ 3B = ${\displaystyle \left[\begin{array}{ccc}0& -10& -16\\ 0& 0& 15\end{array}\right]}$

∴ B = ${\displaystyle \frac{1}{3}\left[\begin{array}{ccc}0& -10& -16\\ 0& 0& 15\end{array}\right]}$

∴ B = ${\displaystyle \left[\begin{array}{ccc}0& \frac{-10}{3}& \frac{-16}{3}\\ 0& 0& 5\end{array}\right]}$

By (i) x 2 –  (ii), we get

3A = ${\displaystyle 2\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]-\left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{ccc}12& -12& 0\\ -8& 4& 2\end{array}\right]-\left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{ccc}12-3& -12-2& 0-8\\ -8+2& 4 -1& 2+7\end{array}\right]}$

∴ 3A = ${\displaystyle \left[\begin{array}{ccc}9& -14& -8\\ -6& 3& 9\end{array}\right]}$

∴ A = ${\displaystyle \frac{1}{3}\left[\begin{array}{ccc}9& 14& -8\\ -6& 3& 9\end{array}\right]}$

∴ A = ${\displaystyle \left[\begin{array}{ccc}3& \frac{-14}{3}& \frac{-8}{3}\\ -2& 1& 3\end{array}\right]}$.

Exercise 2.2 | Q 13 | Page 47

QUESTION

Find x and y, if ${\displaystyle \left[\begin{array}{ccc}2x+y& -1& 1\\ 3& 4y& 4\end{array}\right]\left[\begin{array}{ccc}-1& 6& 4\\ 3& 0& 3\end{array}\right]=\left[\begin{array}{ccc}3& 5& 5\\ 6& 18& 7\end{array}\right]}$.

SOLUTION

${\displaystyle \left[\begin{array}{ccc}2x+y& -1& 1\\ 3& 4y& 4\end{array}\right]\left[\begin{array}{ccc}-1& 6& 4\\ 3& 0& 3\end{array}\right]=\left[\begin{array}{ccc}3& 5& 5\\ 6& 18& 7\end{array}\right]}$

∴ ${\displaystyle \left[\begin{array}{ccc}2x+y-1& -1+6& 1+4\\ 3+3& 4y+0& 4+3\end{array}\right]=\left[\begin{array}{ccc}3& 5& 5\\ 6& 18& 7\end{array}\right]}$

∴ ${\displaystyle \left[\begin{array}{ccc}x+y-1& 5& 5\\ 6& 4y& 7\end{array}\right]=\left[\begin{array}{ccc}3& 5& 5\\ 6& 18& 7\end{array}\right]}$

∴ By equality of matrices, we get
2x + y – 1 = 3 and 4y = 18

∴ 2x + y = 4 and y = ${\displaystyle \frac{18}{4}=\frac{9}{2}}$

∴ ${\displaystyle 2+\frac{9}{2}}$ = 4

∴ 2x = ${\displaystyle 4-\frac{9}{2}}$

∴ 2x = ${\displaystyle -\frac{1}{2}}$

∴ x = ${\displaystyle -\frac{1}{4}}$ and y = ${\displaystyle \frac{9}{2}}$.

Exercise 2.2 | Q 14 | Page 47

QUESTION

If ${\displaystyle \left[\begin{array}{cc}2\text{a}+\text{b}& 3\text{a}-\text{b}\\ \text{c}+2\text{d}& 2\text{c}-\text{d}\end{array}\right]=\left[\begin{array}{cc}2& 3\\ 4& -1\end{array}\right]}$, find a, b, c and d.

SOLUTION

${\displaystyle \left[\begin{array}{cc}2\text{a}+\text{b}& 3\text{a}-\text{b}\\ \text{c}+2\text{d}& 2\text{c}-\text{d}\end{array}\right]=\left[\begin{array}{cc}2& 3\\ 4& -1\end{array}\right]}$

∴ By  equality of matrices, we get
2a + b = 2        ....(i)
3a – b = 3        ....(ii)
c + 2d = 4       ....(iii)
2c –d = – 1      ....(iv)
Adding (i) and (ii), we get
5a = 5
∴ a = 1
Substituting a = 1 in (i), we get
2(1) + b = 2
∴ b = 0
By (iii) + (iv) x 2, we get
5c = 2

∴ c = ${\displaystyle \frac{2}{5}}$
Substituting c = ${\displaystyle \frac{2}{5}}$ i (iii), we get

${\displaystyle \frac{2}{5}+2d}$ = 4

∴ 2d = ${\displaystyle 4-\frac{2}{5}}$

∴ 2d = ${\displaystyle \frac{18}{5}}$

∴ d = ${\displaystyle \frac{9}{5}}$.

Exercise 2.2 | Q 15.1 | Page 47

QUESTION

There are two book shops own by Suresh and Ganesh. Their sales ( in Rupees) for books in three subject - Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B. July sales ( in Rupees) :
Physics Chemistry Mathematics
A = ${\displaystyle \left[\begin{array}{ccc}5600& 6750& 8500\\ 6650& 7055& 8905\end{array}\right]\left[\begin{array}{c}\text{Suresh}\\ \text{Ganesh}\end{array}\right]}$
August Sales (in Rupees :
B = ${\displaystyle \left[\begin{array}{ccc}6650& 7055& 8905\\ 7000& 7500& 10200\end{array}\right]\left[\begin{array}{c}\text{Suresh}\\ \text{Ganesh}\end{array}\right]}$
Find the increase in sales in Rupees from July to August 2017.

SOLUTION

Increase  sales rrupees from july to August 2017.
For Suresh :
Increase in sales for Physics books
= 6650 – 5600 = ₹ 1050
Increase in sales for  Chemistry books
= 7055 – 6750 = ₹ 305
Increase in sales for Mathematics books
= 8905 – 8500 = ₹ 405
For Ganesh :
Increase in sales for Physics books
= 7000 – 6650 = ₹ 350
Increase in sales for  Chemistry books
= 7500 – 7055 = ₹ 455
Increase in sales for Mathematics books
= 10200 – 8905 = ₹ 1295.

Exercise 2.2 | Q 15.2 | Page 47

QUESTION

There are two book shops own by Suresh and Ganesh. Their sales ( in Rupees) for books in three subject - Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B. July sales ( in Rupees) :
Physics Chemistry Mathematics
A = ${\displaystyle \left[\begin{array}{ccc}5600& 6750& 8500\\ 6650& 7055& 8905\end{array}\right]\left[\begin{array}{c}\text{Suresh}\\ \text{Ganesh}\end{array}\right]}$
August Sales (in Rupees :
B = ${\displaystyle \left[\begin{array}{ccc}6650& 7055& 8905\\ 7000& 7500& 10200\end{array}\right]\left[\begin{array}{c}\text{Suresh}\\ \text{Ganesh}\end{array}\right]}$
If both book shops get 10% profit in the month of August 2017, find the profit for each book seller in each subject in that month.

SOLUTION

Both book shops got 10% profit in the month of August 2017.
For Suresh :
Profit for Physics books = ${\displaystyle \frac{6650\times 10}{100}}$ = ₹ 665

Profit for Chemistry books = ${\displaystyle \frac{7055\times 10}{100}}$ = ₹ 705.50

Profit for Mathematics books = ${\displaystyle \frac{8905\times 10}{100}}$ = ₹ 890.50
For Ganesh :
Profit for Physics books = ${\displaystyle \frac{7000\times 10}{100}}$ = ₹ 700

Profit for Chemistry books = ${\displaystyle \frac{7500\times 10}{100}}$ = ₹ 750

Profit for Mathematics books = ${\displaystyle \frac{10200\times 10}{100}}$ = ₹ 1020

Concept: Algebra of Matrices

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board

Chapter 2 Matrices

Exercise 2.2 [PAGES 46 - 47