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

Exercise 2.2 | Q 1.1 | Page 46

#### QUESTION

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

#### SOLUTION

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

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

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

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

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

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

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

Exercise 2.2 | Q 1.2 | Page 46

#### QUESTION

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

#### SOLUTION

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

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

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

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

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

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

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

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

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

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

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

Exercise 2.2 | Q 2 | Page 46

#### QUESTION

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

#### SOLUTION

A – 2B + 6I =

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

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

Exercise 2.2 | Q 3 | Page 46

#### QUESTION

If A = $\left[\begin{array}{ccc}1& 2& -3\\ -3& 7& -8\\ 0& -6& 1\end{array}\right],\text{B}=\left[\begin{array}{ccc}9& -1& 2\\ -4& 2& 5\\ 4& 0& -3\end{array}\right]$ 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)

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

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

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

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

Exercise 2.2 | Q 4 | Page 46

#### QUESTION

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

#### SOLUTION

3A – 4B + 5X = C
∴ 5X =C + 4B – 3A

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

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

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

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

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

Exercise 2.2 | Q 5 | Page 46

#### QUESTION

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

#### SOLUTION

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

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

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

Exercise 2.2 | Q 6 | Page 46

#### QUESTION

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

#### SOLUTION

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

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

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

Exercise 2.2 | Q 7 | Page 47

#### QUESTION

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

#### SOLUTION

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

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

Since A is a symmetric matrix,
A = AT

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

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

∴ By equality of matrices, we get

a = – 4, b = $\frac{3}{5}$, c = – 7.

Exercise 2.2 | Q 8 | Page 47

#### QUESTION

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

#### SOLUTION

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

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

Since A is a skew-symmetric matrix,
A = AT

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

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

∴ By equality of matrices, we get

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

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.

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



#### SOLUTION

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

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

∴ AT = A i.e., A = AT
∴ 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.

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



#### SOLUTION

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

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

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

∴ A ≠ AT and A ≠ – AT
∴ 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.

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



#### SOLUTION

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

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

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

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

Exercise 2.2 | Q 10 | Page 47

#### QUESTION

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

#### SOLUTION

A = [aij]3x3

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

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

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

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

∴ AT = – A i.e., A = – AT
∴ 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 = $\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$  and X – 3Y = $\left[\begin{array}{cc}0& -1\\ 0& -1\end{array}\right]$.

#### SOLUTION

Given equations are

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

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

By (i) x 3 – (ii), we get

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

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

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

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

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

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

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

By (i) – (ii) x 3, we get

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

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

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

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

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

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

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

Exercise 2.2 | Q 12 | Page 47

#### QUESTION

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

#### SOLUTION

Given equations are

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

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

By (i) – (ii) x 2, we get

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

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

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

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

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

By (i) x 2 –  (ii), we get

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

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

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

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

Exercise 2.2 | Q 13 | Page 47

#### QUESTION

Find x and y, if .

#### SOLUTION

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

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

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

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

∴ $2+\frac{9}{2}$ = 4

∴ 2x = $4-\frac{9}{2}$

∴ 2x = $-\frac{1}{2}$

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

Exercise 2.2 | Q 14 | Page 47

#### QUESTION

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

#### SOLUTION

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

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

$\frac{2}{5}+2d$ = 4

∴ 2d = $4-\frac{2}{5}$

∴ 2d = $\frac{18}{5}$

∴ d = $\frac{9}{5}$.

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 = $\left[\begin{array}{ccc}5600& 6750& 8500\\ 6650& 7055& 8905\end{array}\right]\left[\begin{array}{c}\text{Suresh}\\ \text{Ganesh}\end{array}\right]$
August Sales (in Rupees :
B = $\left[\begin{array}{ccc}6650& 7055& 8905\\ 7000& 7500& 10200\end{array}\right]\left[\begin{array}{c}\text{Suresh}\\ \text{Ganesh}\end{array}\right]$
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 = $\left[\begin{array}{ccc}5600& 6750& 8500\\ 6650& 7055& 8905\end{array}\right]\left[\begin{array}{c}\text{Suresh}\\ \text{Ganesh}\end{array}\right]$
August Sales (in Rupees :
B = $\left[\begin{array}{ccc}6650& 7055& 8905\\ 7000& 7500& 10200\end{array}\right]\left[\begin{array}{c}\text{Suresh}\\ \text{Ganesh}\end{array}\right]$
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 = $\frac{6650×10}{100}$ = ₹ 665

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

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

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

Profit for Mathematics books = $\frac{10200×10}{100}$ = ₹ 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

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HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Important-formula

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