NCERT Solutions for Exercise 3.2 Class 12 Maths Chapter 3 - Matrices

Question 1(ii) Let $A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ , $B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$ , $C = [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$

A + B

Answer:

$A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ $B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$

(i) A + B

The addition of matrix can be done as follows

$A + B = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ $+ [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$

$A + B = [\begin{matrix} 2 + 1 & 4 + 3 \\ 3 + (- 2) & 2 + 5 \end{matrix}]$

$A + B = [\begin{matrix} 3 & 7 \\ 1 & 7 \end{matrix}]$

Question 1(iii) Let $A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ , $B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$ , $C = [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$

A - B

Answer:

$A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ $B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$

(ii) A - B

$A - B = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ $- [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$

$A - B = [\begin{matrix} 2 - 1 & 4 - 3 \\ 3 - (- 2) & 2 - 5 \end{matrix}]$

$A - B = [\begin{matrix} 1 & 1 \\ 5 & - 3 \end{matrix}]$

Question 1(iv)Let $A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ , $B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$ , $C = [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$

3A - C

Answer:

$A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ $C = [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$

(iii) 3A - C

First multiply each element of A with 3 and then subtract C

$3 A - C = 3 [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ $- [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$

$3 A - C = [\begin{matrix} 6 & 12 \\ 9 & 6 \end{matrix}]$ $- [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$

$3 A - C = [\begin{matrix} 6 - (- 2) & 12 - 5 \\ 9 - 3 & 6 - 4 \end{matrix}]$

$3 A - C = [\begin{matrix} 8 & 7 \\ 6 & 2 \end{matrix}]$

Question 1(v) Let $A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ , $B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$ , $C = [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$

Answer:

$A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ $B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$

(iv) AB

$A B = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ $\times [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$

$A B = [\begin{matrix} 2 \times 1 + 4 \times - 2 & 2 \times 3 + 4 \times 5 \\ 3 \times 1 + 2 \times - 2 & 3 \times 3 + 2 \times 5 \end{matrix}]$

$A B = [\begin{matrix} - 6 & 26 \\ - 1 & 19 \end{matrix}]$

$[\begin{matrix} a & b \\ - b & a \end{matrix}] + [\begin{matrix} a & b \\ b & a \end{matrix}]$

Answer:

The multiplication is performed as follows

$A = [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ , $B = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$

$B A = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$ $\times [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$

$B A = [\begin{matrix} 1 \times 2 + 3 \times 3 & 1 \times 4 + 3 \times 2 \\ - 2 \times 2 + 5 \times 3 & - 2 \times 4 + 2 \times 5 \end{matrix}]$

$B A = [\begin{matrix} 11 & 10 \\ 11 & 2 \end{matrix}]$

Question 2(i). Compute the following:

Answer:

(i) $[\begin{matrix} a & b \\ - b & a \end{matrix}] + [\begin{matrix} a & b \\ b & a \end{matrix}]$

$= [\begin{matrix} a + a & b + b \\ - b + b & a + a \end{matrix}]$

$= [\begin{matrix} 2 a & 2 b \\ 0 & 2 a \end{matrix}]$

Question 2(ii). Compute the following:

$[\begin{matrix} a^{2} + b^{2} & b^{2} + c^{2} \\ a^{2} + c^{2} & a^{2} + b^{2} \end{matrix}] + [\begin{matrix} 2 a b & 2 b c \\ - 2 a c & - 2 a b \end{matrix}]$

Answer:

(ii) The addition operation can be performed as follows

$[\begin{matrix} a^{2} + b^{2} & b^{2} + c^{2} \\ a^{2} + c^{2} & a^{2} + b^{2} \end{matrix}] + [\begin{matrix} 2 a b & 2 b c \\ - 2 a c & - 2 a b \end{matrix}]$

$= [\begin{matrix} a^{2} + b^{2} + 2 a b & b^{2} + c^{2} + 2 b c \\ a^{2} + c^{2} - 2 a c & a^{2} + b^{2} - 2 a b \end{matrix}]$

$= [\begin{matrix} (a + b)^{2} & (b + c)^{2} \\ (a - c)^{2} & (a - b)^{2} \end{matrix}]$

Question 2(iii). Compute the following:

$[\begin{matrix} - 1 & 4 & - 6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{matrix}] + [\begin{matrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{matrix}]$

Answer:

(iii) The addition of given three by three matrix is performed as follows

$[\begin{matrix} - 1 & 4 & - 6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{matrix}] + [\begin{matrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{matrix}]$

$= [\begin{matrix} - 1 + 12 & 4 + 7 & - 6 + 6 \\ 8 + 8 & 5 + 0 & 16 + 5 \\ 2 + 3 & 8 + 2 & 5 + 4 \end{matrix}]$

$= [\begin{matrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{matrix}]$

Question 2(iv). Compute the following:

$[\begin{matrix} \cos^{2} x & \sin^{2} x \\ \sin^{2} x & \cos^{2} x \end{matrix}] + [\begin{matrix} \sin^{2} x & \cos^{2} x \\ \cos^{2} x & \sin^{2} x \end{matrix}]$

Answer:

(iv) the addition is done as follows

$[\begin{matrix} \cos^{2} x & \sin^{2} x \\ \sin^{2} x & \cos^{2} x \end{matrix}] + [\begin{matrix} \sin^{2} x & \cos^{2} x \\ \cos^{2} x & \sin^{2} x \end{matrix}]$

$= [\begin{matrix} \cos^{2} + \sin^{2} x & \sin^{2} x + \cos^{2} x \\ \sin^{2} x + \cos^{2} x & \cos^{2} x + \sin^{2} x \end{matrix}]$ since $s i n^{2} x + c o s^{2} x = 1$

$= [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]$

Question 3(i). Compute the indicated products.

$[\begin{matrix} a & b \\ - b & a \end{matrix}] [\begin{matrix} a & - b \\ b & a \end{matrix}]$

Answer:

(i) The multiplication is performed as follows

$[\begin{matrix} a & b \\ - b & a \end{matrix}] [\begin{matrix} a & - b \\ b & a \end{matrix}]$

$= [\begin{matrix} a & b \\ - b & a \end{matrix}] \times [\begin{matrix} a & - b \\ b & a \end{matrix}]$

$= [\begin{matrix} a \times a + b \times b & a \times - b + b \times a \\ - b \times a + a \times b & - b \times - b + a \times a \end{matrix}]$

$= [\begin{matrix} a^{2} + b^{2} & 0 \\ 0 & b^{2} + a^{2} \end{matrix}]$

Question 3(ii). Compute the indicated products.

$[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}] [\begin{matrix} 2 & 3 & 4 \end{matrix}]$

Answer:

(ii) the multiplication can be performed as follows

$[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}] [\begin{matrix} 2 & 3 & 4 \end{matrix}]$

$= [\begin{matrix} 1 \times 2 & 1 \times 3 & 1 \times 4 \\ 2 \times 2 & 2 \times 3 & 2 \times 4 \\ 3 \times 2 & 3 \times 3 & 3 \times 4 \end{matrix}]$

$= [\begin{matrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{matrix}]$

Question 3(iii). Compute the indicated products.

$[\begin{matrix} 1 & - 2 \\ 2 & 3 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix}]$

Answer:

(iii) The multiplication can be performed as follows

$[\begin{matrix} 1 & - 2 \\ 2 & 3 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix}]$

$= [\begin{matrix} 1 \times 1 + (- 2) \times 2 & 1 \times 2 + (- 2) \times 3 & 1 \times 3 + (- 2) \times 1 \\ 2 \times 1 + 3 \times 2 & 2 \times 2 + 3 \times 3 & 2 \times 3 + 3 \times 1 \end{matrix}]$

Question 3(iv). Compute the indicated products.

$[\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{matrix}] [\begin{matrix} 1 & - 3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{matrix}]$

Answer:

(iv) The multiplication is performed as follows

$[\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{matrix}] [\begin{matrix} 1 & - 3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{matrix}]$

$= [\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{matrix}] \times [\begin{matrix} 1 & - 3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{matrix}]$

$= [\begin{matrix} 2 \times 1 + 3 \times 0 + 4 \times 3 & 2 \times (- 3) + 3 \times 2 + 4 \times 0 & 2 \times 5 + 3 \times 4 + 4 \times 5 \\ 3 \times 1 + 4 \times 0 + 5 \times 3 & 3 \times (- 3) + 4 \times 2 + 5 \times 0 & 3 \times 5 + 4 \times 4 + 5 \times 5 \\ 4 \times 1 + 5 \times 0 + 6 \times 3 & 4 \times (- 3) + 5 \times 2 + 6 \times 0 & 4 \times 5 + 5 \times 4 + 6 \times 5 \end{matrix}]$

$= [\begin{matrix} 14 & 0 & 42 \\ 18 & - 1 & 56 \\ 22 & - 2 & 70 \end{matrix}]$

Question 3(v). Compute the indicated products.

$[\begin{matrix} 2 & 1 \\ 3 & 2 \\ - 1 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 1 \\ - 1 & 2 & 1 \end{matrix}]$

Answer:

(v) The product can be computed as follows

$[\begin{matrix} 2 & 1 \\ 3 & 2 \\ - 1 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 1 \\ - 1 & 2 & 1 \end{matrix}]$

$= [\begin{matrix} 2 & 1 \\ 3 & 2 \\ - 1 & 1 \end{matrix}] \times [\begin{matrix} 1 & 0 & 1 \\ - 1 & 2 & 1 \end{matrix}]$

$= [\begin{matrix} 2 \times 1 + 1 \times (- 1) & 2 \times 0 + 1 \times (2) & 2 \times 1 + 1 \times (1) \\ 3 \times 1 + 2 \times (- 1) & 3 \times 0 + 2 \times (2) & 3 \times 1 + 2 \times (1) \\ (- 1) \times 1 + 1 \times (- 1) & (- 1) \times 0 + 1 \times (2) & (- 1) \times 1 + 1 \times (1) \end{matrix}]$

$= [\begin{matrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ - 2 & 2 & 0 \end{matrix}]$

Question 3(vi). Compute the indicated products.

$[\begin{matrix} 3 & - 1 & 3 \\ - 1 & 0 & 2 \end{matrix}] [\begin{matrix} 2 & - 3 \\ 1 & 0 \\ 3 & 1 \end{matrix}]$

Answer:

(vi) The given product can be computed as follows

$[\begin{matrix} 3 & - 1 & 3 \\ - 1 & 0 & 2 \end{matrix}] [\begin{matrix} 2 & - 3 \\ 1 & 0 \\ 3 & 1 \end{matrix}]$

$= [\begin{matrix} 3 & - 1 & 3 \\ - 1 & 0 & 2 \end{matrix}] \times [\begin{matrix} 2 & - 3 \\ 1 & 0 \\ 3 & 1 \end{matrix}]$

$= [\begin{matrix} 3 \times 2 + (- 1) \times 1 + 3 \times 3 & 3 \times (- 3) + (- 1) \times 0 + 3 \times 1 \\ (- 1) \times 2 + 0 \times 1 + 2 \times 3 & (- 1) \times - 3 + 0 \times 0 + 2 \times 1 \end{matrix}]$

$= [\begin{matrix} 14 & - 6 \\ 4 & 5 \end{matrix}]$

Question 4. If $A = [\begin{matrix} 1 & 2 & - 3 \\ 5 & 0 & 2 \\ 1 & - 1 & 1 \end{matrix}]$ , $B = [\begin{matrix} 3 & - 1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix}]$ and $C = [\begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & - 2 & 3 \end{matrix}]$ , then compute (A+B) and (B-C). Also verify that A + (B - C) = (A + B) - C

Answer:

$A = [\begin{matrix} 1 & 2 & - 3 \\ 5 & 0 & 2 \\ 1 & - 1 & 1 \end{matrix}]$ , $B = [\begin{matrix} 3 & - 1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix}]$ and $C = [\begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & - 2 & 3 \end{matrix}]$

$A + B = [\begin{matrix} 1 & 2 & - 3 \\ 5 & 0 & 2 \\ 1 & - 1 & 1 \end{matrix}]$ $+ [\begin{matrix} 3 & - 1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix}]$

$A + B = [\begin{matrix} 1 + 3 & 2 + (- 1) & - 3 + 2 \\ 5 + 4 & 0 + 2 & 2 + 5 \\ 1 + 2 & - 1 + 0 & 1 + 3 \end{matrix}]$

$A + B = [\begin{matrix} 4 & 1 & - 1 \\ 9 & 2 & 7 \\ 3 & - 1 & 4 \end{matrix}]$

$B - C = [\begin{matrix} 3 & - 1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix}]$ $- [\begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & - 2 & 3 \end{matrix}]$

$B - C = [\begin{matrix} 3 - 4 & - 1 - 1 & 2 - 2 \\ 4 - 0 & 2 - 3 & 5 - 2 \\ 2 - 1 & 0 - (- 2) & 3 - 3 \end{matrix}]$

$B - C = [\begin{matrix} - 1 & - 2 & 0 \\ 4 & - 1 & 3 \\ 1 & 2 & 0 \end{matrix}]$

Now, to prove A + (B - C) = (A + B) - C

$L . H . S : A + (B - C)$

$A + (B - C) = [\begin{matrix} 1 & 2 & - 3 \\ 5 & 0 & 2 \\ 1 & - 1 & 1 \end{matrix}]$ $+ [\begin{matrix} - 1 & - 2 & 0 \\ 4 & - 1 & 3 \\ 1 & 2 & 0 \end{matrix}]$ (Puting value of $B - C$ from above)

$A + (B - C) = [\begin{matrix} 1 - 1 & 2 - 2 & - 3 + 0 \\ 5 + 4 & 0 + (- 1) & 2 + 3 \\ 1 + 1 & - 1 + 2 & 1 + 0 \end{matrix}]$

$A + (B - C) = [\begin{matrix} 0 & 0 & - 3 \\ 9 & - 1 & 5 \\ 2 & 1 & 1 \end{matrix}]$

$R . H . S : (A + B) - C$

$(A + B) - C = [\begin{matrix} 4 & 1 & - 1 \\ 9 & 2 & 7 \\ 3 & - 1 & 4 \end{matrix}]$ $- [\begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & - 2 & 3 \end{matrix}]$

$(A + B) - C = [\begin{matrix} 4 - 4 & 1 - 1 & - 1 - 2 \\ 9 - 0 & 2 - 3 & 7 - 2 \\ 3 - 1 & - 1 - (- 2) & 4 - 3 \end{matrix}]$

$(A + B) - C = [\begin{matrix} 0 & 0 & - 3 \\ 9 & - 1 & 5 \\ 2 & 1 & 1 \end{matrix}]$

Hence, we can see L.H.S = R.H.S = $[\begin{matrix} 0 & 0 & - 3 \\ 9 & - 1 & 5 \\ 2 & 1 & 1 \end{matrix}]$

Question 5. If $A = [\begin{matrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{matrix}]$ and $B = [\begin{matrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{matrix}]$ , then compute 3A - 5B

Answer:

$A = [\begin{matrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{matrix}]$ and $B = [\begin{matrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{matrix}]$

$3 A - 5 B = 3 \times [\begin{matrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{matrix}]$ $- 5 \times [\begin{matrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{matrix}]$

$3 A - 5 B = [\begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix}]$ $- [\begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix}]$

$3 A - 5 B = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

$3 A - 5 B = 0$

Question 6. Simplify $\cos θ [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] + \sin θ [\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}]$ .

Answer:

The simplification is explained in the following step

$\cos θ [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] + \sin θ [\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}]$

$= [\begin{matrix} \cos^{2} θ & \sin θ \cos θ \\ - \sin θ \cos θ & \cos^{2} θ \end{matrix}] + [\begin{matrix} \sin^{2} θ & - \sin θ \cos θ \\ \sin θ \cos θ & \sin^{2} θ \end{matrix}]$

$= [\begin{matrix} \cos^{2} θ + \sin^{2} θ & \sin θ \cos θ - \sin θ \cos θ \\ - \sin θ \cos θ + \sin θ \cos θ & \cos^{2} θ + \sin^{2} θ \end{matrix}]$

$= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$

the final answer is an identity matrix of order 2

Question 7(i). Find X and Y, if

$X + Y = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}]$ and $X - Y = [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}]$

Answer:

(i) The given matrices are

$X + Y = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}]$ and $X - Y = [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}]$

$X + Y = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1$

$X - Y = [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2$

Adding equation 1 and 2, we get

$2 X = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}]$ $+ [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}]$

$2 X = [\begin{matrix} 7 + 3 & 0 + 0 \\ 2 + 0 & 5 + 3 \end{matrix}]$

$2 X = [\begin{matrix} 10 & 0 \\ 2 & 8 \end{matrix}]$

$X = [\begin{matrix} 5 & 0 \\ 1 & 4 \end{matrix}]$

Putting the value of X in equation 1, we get

$[\begin{matrix} 5 & 0 \\ 1 & 4 \end{matrix}]$ $+ Y = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}]$

$Y = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}] -$ $[\begin{matrix} 5 & 0 \\ 1 & 4 \end{matrix}]$

$Y = [\begin{matrix} 7 - 5 & 0 - 0 \\ 2 - 1 & 5 - 4 \end{matrix}]$

$Y = [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}]$

Question 7(ii). Find X and Y, if

$2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$ and $3 X + 2 Y = [\begin{matrix} 2 & - 2 \\ - 1 & 5 \end{matrix}]$

Answer:

(ii) $2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$ and $3 X + 2 Y = [\begin{matrix} 2 & - 2 \\ - 1 & 5 \end{matrix}]$

$2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] . . . . . . . . . . . . . . . . . . . . . . . . . .1$

$3 X + 2 Y = [\begin{matrix} 2 & - 2 \\ - 1 & 5 \end{matrix}] . . . . . . . . . . . . . . . . . . . . . .2$

Multiply equation 1 by 3 and equation 2 by 2 and subtract them,

$3 (2 X + 3 Y) - 2 (3 X + 2 Y) = 3 \times [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$ $- 2 \times [\begin{matrix} 2 & - 2 \\ - 1 & 5 \end{matrix}]$

$6 X + 9 Y - 6 X - 4 Y = [\begin{matrix} 6 & 9 \\ 12 & 0 \end{matrix}]$ $- [\begin{matrix} 4 & - 4 \\ - 2 & 10 \end{matrix}]$

$9 Y - 4 Y = [\begin{matrix} 6 - 4 & 9 - (- 4) \\ 12 - (- 2) & 0 - 10 \end{matrix}]$

$5 Y = [\begin{matrix} 2 & 13 \\ 14 & - 10 \end{matrix}]$

$Y = [\begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & - 2 \end{matrix}]$

Putting value of Y in equation 1 , we get

$2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$

$2 X + 3 [\begin{matrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & - 2 \end{matrix}] = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$

$2 X + [\begin{matrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & - 6 \end{matrix}] = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$

$2 X = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] - [\begin{matrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & - 6 \end{matrix}]$

$2 X = [\begin{matrix} 2 - \frac{6}{5} & 3 - \frac{39}{5} \\ 4 - \frac{42}{5} & 0 - (- 6) \end{matrix}]$

$2 X = [\begin{matrix} \frac{4}{5} & - \frac{24}{5} \\ - \frac{22}{5} & 6 \end{matrix}]$

$X = [\begin{matrix} \frac{2}{5} & - \frac{12}{5} \\ - \frac{11}{5} & 3 \end{matrix}]$

Question 8. Find X, if $Y = [\begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix}]$ and $2 X + Y = [\begin{matrix} 1 & 0 \\ - 3 & 2 \end{matrix}]$

Answer:

$Y = [\begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix}]$

$2 X + Y = [\begin{matrix} 1 & 0 \\ - 3 & 2 \end{matrix}]$

Substituting the value of Y in the above equation

$2 X + [\begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 \\ - 3 & 2 \end{matrix}]$

$2 X = [\begin{matrix} 1 & 0 \\ - 3 & 2 \end{matrix}] - [\begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix}]$

$2 X = [\begin{matrix} 1 - 3 & 0 - 2 \\ - 3 - 1 & 2 - 4 \end{matrix}]$

$2 X = [\begin{matrix} - 2 & - 2 \\ - 4 & - 2 \end{matrix}]$

$X = [\begin{matrix} - 1 & - 1 \\ - 2 & - 1 \end{matrix}]$

Question 9. Find x and y, if $2 [\begin{matrix} 1 & 3 \\ 0 & x \end{matrix}] + [\begin{matrix} y & 0 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 5 & 6 \\ 1 & 8 \end{matrix}]$

Answer:

$2 [\begin{matrix} 1 & 3 \\ 0 & x \end{matrix}] + [\begin{matrix} y & 0 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 5 & 6 \\ 1 & 8 \end{matrix}]$

$[\begin{matrix} 2 & 6 \\ 0 & 2 x \end{matrix}] + [\begin{matrix} y & 0 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 5 & 6 \\ 1 & 8 \end{matrix}]$

$[\begin{matrix} 2 + y & 6 + 0 \\ 0 + 1 & 2 x + 2 \end{matrix}] = [\begin{matrix} 5 & 6 \\ 1 & 8 \end{matrix}]$

$[\begin{matrix} 2 + y & 6 \\ 1 & 2 x + 2 \end{matrix}] = [\begin{matrix} 5 & 6 \\ 1 & 8 \end{matrix}]$

Now equating LHS and RHS we can write the following equations

$2 + y = 5$ $2 x + 2 = 8$

$y = 5 - 2$ $2 x = 8 - 2$

$y = 3$ $2 x = 6$

$x = 3$

Question 10. Solve the equation for x, y, z and t, if $2 [\begin{matrix} x & z \\ y & t \end{matrix}] + 3 [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] = 3 [\begin{matrix} 3 & 5 \\ 4 & 6 \end{matrix}]$

Answer:

$2 [\begin{matrix} x & z \\ y & t \end{matrix}] + 3 [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] = 3 [\begin{matrix} 3 & 5 \\ 4 & 6 \end{matrix}]$

Multiplying with constant terms and rearranging we can rewrite the matrix as

$[\begin{matrix} 2 x & 2 z \\ 2 y & 2 t \end{matrix}] = [\begin{matrix} 9 & 15 \\ 12 & 18 \end{matrix}] - 3 [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}]$

$[\begin{matrix} 2 x & 2 z \\ 2 y & 2 t \end{matrix}] = [\begin{matrix} 9 & 15 \\ 12 & 18 \end{matrix}] - [\begin{matrix} 3 & - 3 \\ 0 & 6 \end{matrix}]$

$[\begin{matrix} 2 x & 2 z \\ 2 y & 2 t \end{matrix}] = [\begin{matrix} 9 - 3 & 15 - (- 3) \\ 12 - 0 & 18 - 6 \end{matrix}]$

$[\begin{matrix} 2 x & 2 z \\ 2 y & 2 t \end{matrix}] = [\begin{matrix} 6 & 18 \\ 12 & 12 \end{matrix}]$

Dividing by 2 on both sides

$[\begin{matrix} x & z \\ y & t \end{matrix}] = [\begin{matrix} 3 & 9 \\ 6 & 6 \end{matrix}]$

$x = 3, y = 6, z = 9 a n d t = 6$

Question 11. If $x [\begin{matrix} 2 \\ 3 \end{matrix}] + y [\begin{matrix} - 1 \\ 1 \end{matrix}] = [\begin{matrix} 10 \\ 5 \end{matrix}]$ , find the values of x and y.

Answer:

$x [\begin{matrix} 2 \\ 3 \end{matrix}] + y [\begin{matrix} - 1 \\ 1 \end{matrix}] = [\begin{matrix} 10 \\ 5 \end{matrix}]$

$[\begin{matrix} 2 x \\ 3 x \end{matrix}] + [\begin{matrix} - y \\ y \end{matrix}] = [\begin{matrix} 10 \\ 5 \end{matrix}]$

Adding both the matrix in LHS and rewriting

$[\begin{matrix} 2 x - y \\ 3 x + y \end{matrix}] = [\begin{matrix} 10 \\ 5 \end{matrix}]$

$2 x - y = 10. . . . . . . . . . . . . . . . . . . . . . . .1$

$3 x + y = 5. . . . . . . . . . . . . . . . . . . . . . . .2$

Adding equation 1 and 2, we get

$5 x = 15$

$x = 3$

Put the value of x in equation 2, we have

$3 x + y = 5$

$3 \times 3 + y = 5$

$9 + y = 5$

$y = 5 - 9$

$y = - 4$

Question 12. Given $3 [\begin{matrix} x & y \\ z & w \end{matrix}] = [\begin{matrix} x & 6 \\ - 1 & 2 w \end{matrix}] + [\begin{matrix} 4 & x + y \\ z + w & 3 \end{matrix}]$ , find the values of x, y, z and w.

Answer:

$3 [\begin{matrix} x & y \\ z & w \end{matrix}] = [\begin{matrix} x & 6 \\ - 1 & 2 w \end{matrix}] + [\begin{matrix} 4 & x + y \\ z + w & 3 \end{matrix}]$

$[\begin{matrix} 3 x & 3 y \\ 3 z & 3 w \end{matrix}] = [\begin{matrix} x + 4 & 6 + x + y \\ - 1 + z + w & 2 w + 3 \end{matrix}]$

If two matrices are equal than corresponding elements are also equal.

Thus, we have

$3 x = x + 4$

$3 x - x = 4$

$2 x = 4$

$x = 2$

$3 y = 6 + x + y$

Put the value of x

$3 y - y = 6 + 2$

$2 y = 8$

$y = 4$

$3 w = 2 w + 3$

$3 w - 2 w = 3$

$w = 3$

$3 z = - 1 + z + w$

$3 z - z = - 1 + 3$

$2 z = 2$

$z = 1$

Hence, we have $x = 2, y = 4, z = 1 a n d w = 3.$

Question 13. If $F (x) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}]$ , show that $F (x) F (y) = F (x + y)$ .

Answer:

$F (x) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}]$

To prove : $F (x) F (y) = F (x + y)$

$R . H . S : F (x + y)$

$F (x + y) = [\begin{matrix} \cos (x + y) & - \sin (x + y) & 0 \\ \sin (x + y) & \cos (x + y) & 0 \\ 0 & 0 & 1 \end{matrix}]$

$L . H . S : F (x) F (y)$

$F (x) F (y) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] \times [\begin{matrix} \cos y & - \sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{matrix}]$

$F (x) F (y) = [\begin{matrix} \cos x \cos y - \sin x \sin y + 0 & - \cos x \sin y - \sin x \cos y + 0 & 0 + 0 + 0 \\ s i n x \cos y + \cos x \sin y + 0 & - \sin x \sin y + \cos x \cos y + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 1 \end{matrix}]$

$F (x) F (y) = [\begin{matrix} \cos (x + y) & - \sin (x + y) & 0 \\ \sin (x + y) & \cos (x + y) & 0 \\ 0 & 0 & 1 \end{matrix}]$

Hence, we have L.H.S. = R.H.S i.e. $F (x) F (y) = F (x + y)$ .

Question 14(i). Show that

$[\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}] [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}] \neq [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}] [\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}]$

Answer:

To prove:

$[\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}] [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}] \neq [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}] [\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}]$

$L . H . S : [\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}] [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}]$

$= [\begin{matrix} 5 \times 2 + (- 1) \times 3 & 5 \times 1 + (- 1) \times 4 \\ 6 \times 2 + 7 \times 3 & 6 \times 1 + 7 \times 4 \end{matrix}]$

$= [\begin{matrix} 7 & 1 \\ 33 & 34 \end{matrix}]$

$R . H . S : [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}] [\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}]$

$= [\begin{matrix} 2 \times 5 + 1 \times 6 & 2 \times (- 1) + 1 \times 7 \\ 3 \times 5 + 4 \times 6 & 3 \times (- 1) + 4 \times 7 \end{matrix}]$

$= [\begin{matrix} 16 & 5 \\ 39 & 25 \end{matrix}]$

Hence, the right-hand side not equal to the left-hand side, that is

Question 14(ii). Show that

$[\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}] \neq [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}]$

Answer:

To prove the following multiplication of three by three matrices are not equal

$L . H . S : [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}]$

$= [\begin{matrix} 1 \times (- 1) + 2 \times 0 + 3 \times 2 & 1 \times (1) + 2 \times (- 1) + 3 \times 3 & 1 \times (0) + 2 \times 1 + 3 \times 4 \\ 0 \times (- 1) + 1 \times 0 + 0 \times 2 & 0 \times (1) + 1 \times (- 1) + 0 \times 3 & 0 \times (0) + 1 \times 1 + 0 \times 4 \\ 1 \times (- 1) + 1 \times 0 + 0 \times 2 & 1 \times (1) + 1 \times (- 1) + 0 \times 3 & 1 \times (0) + 1 \times 1 + 0 \times 4 \end{matrix}]$

$= [\begin{matrix} 5 & 8 & 14 \\ 0 & - 1 & 1 \\ - 1 & 0 & 1 \end{matrix}]$

$R . H . S : [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}]$

$= [\begin{matrix} - 1 \times (1) + 1 \times 0 + 0 \times 1 & - 1 \times (2) + 1 \times (1) + 0 \times 1 & - 1 \times (3) + 1 \times 0 + 0 \times 0 \\ 0 \times (1) + - (1) \times 0 + 1 \times 1 & 0 \times (2) + (- 1) \times (1) + 1 \times 1 & 0 \times (3) + (- 1) \times 0 + 1 \times 0 \\ 2 \times (1) + 3 \times 0 + 4 \times 1 & 2 \times (2) + 3 \times (1) + 4 \times 1 & 2 \times (3) + 3 \times 0 + 4 \times 0 \end{matrix}]$

$= [\begin{matrix} - 1 & - 1 & - 3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{matrix}]$

Hence, $L . H . S \neq R . H . S$ i.e. $[\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}] \neq [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}]$ .

Question 15. Find $A^{2} - 5 A + 6 I$ , if

$A = [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}]$

Answer:

$A = [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}]$

First, we will find ou the value of the square of matrix A

$A \times A = [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}] \times [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}]$

$A^{2} = [\begin{matrix} 2 \times 2 + 0 \times 2 + 1 \times 1 & 2 \times 0 + 0 \times 1 + 1 \times - 1 & 2 \times 1 + 0 \times 3 + 1 \times 0 \\ 2 \times 2 + 1 \times 2 + 3 \times 1 & 2 \times 0 + 1 \times 1 + 3 \times - 1 & 2 \times 1 + 1 \times 3 + 3 \times 0 \\ 1 \times 2 + (- 1) \times 2 + 0 \times 1 & 1 \times 0 + (- 1) \times 1 + 0 \times - 1 & 1 \times 1 + (- 1) \times 3 + 0 \times 0 \end{matrix}]$

$A^{2} = [\begin{matrix} 5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2 \end{matrix}]$

$I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$∴$ $A^{2} - 5 A + 6 I$

$= [\begin{matrix} 5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2 \end{matrix}]$ $- 5 [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}]$ $+ 6 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$= [\begin{matrix} 5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2 \end{matrix}]$ $- [\begin{matrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & - 5 & 0 \end{matrix}]$ $+ [\begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix}]$

$= [\begin{matrix} 5 - 10 + 6 & - 1 - 0 + 0 & 2 - 5 + 0 \\ 9 - 10 + 0 & - 2 - 5 + 6 & 5 - 15 + 0 \\ 0 - 5 + 0 & - 1 - (- 5) + 0 & - 2 - 0 + 6 \end{matrix}]$

$= [\begin{matrix} 1 & - 1 & - 3 \\ - 1 & - 1 & - 10 \\ - 5 & 4 & 4 \end{matrix}]$

Question 16. If $A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$ prove that $A^{3} - 6 A^{2} + 7 A + 2 I = 0$ .

Answer:

$A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$

First, find the square of matrix A and then multiply it with A to get the cube of matrix A

$A \times A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$ $\times [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$

$A^{2} = [\begin{matrix} 1 + 0 + 4 & 0 + 0 + 0 & 2 + 0 + 6 \\ 0 + 0 + 2 & 0 + 4 + 0 & 0 + 2 + 3 \\ 2 + 0 + 6 & 0 + 0 + 0 & 4 + 0 + 9 \end{matrix}]$

$A^{2} = [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}]$

$A^{3} = A^{2} \times A$

$A^{2} \times A = [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}]$ $\times [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$

$A^{3} = [\begin{matrix} 5 + 0 + 16 & 0 + 0 + 0 & 10 + 0 + 24 \\ 2 + 0 + 10 & 0 + 8 + 0 & 4 + 4 + 15 \\ 8 + 0 + 26 & 0 + 0 + 0 & 16 + 0 + 39 \end{matrix}]$

$A^{3} = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}]$

$I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$∴$ $A^{3} - 6 A^{2} + 7 A + 2 I = 0$

L.H.S :

$[\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}]$ $- 6 [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}]$ $+ 7 [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$ $+ 2 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$= [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}]$ $- [\begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix}]$ $+ [\begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix}]$ $+ [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}]$

$= [\begin{matrix} 21 - 30 + 7 + 2 & 0 - 0 + 0 + 0 & 34 - 48 + 14 + 0 \\ 12 - 12 + 0 + 0 & 8 - 24 + 14 + 2 & 23 - 30 + 7 + 0 \\ 34 - 48 + 14 + 0 & 0 - 0 + 0 + 0 & 55 - 78 + 21 + 2 \end{matrix}]$

$= [\begin{matrix} 30 - 30 & 0 & 48 - 48 \\ 12 - 12 & 24 - 24 & 30 - 30 \\ 48 - 48 & 0 & 78 - 78 \end{matrix}]$

$= [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = 0$

Hence, L.H.S = R.H.S

i.e. $A^{3} - 6 A^{2} + 7 A + 2 I = 0$ .

Question 17. If $A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$ and $I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , find k so that $A^{2} = k A - 2 I$ .

Answer:

$A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$

$I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$A \times A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$ $\times [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$

$A^{2} = [\begin{matrix} 9 - 8 & - 6 + 4 \\ 12 - 8 & - 8 + 4 \end{matrix}]$

$A^{2} = [\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}]$

$A^{2} = k A - 2 I$

$[\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] =$ $k [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] -$ $2 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] =$ $k [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] -$ $[\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}]$

$[\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] +$ $[\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}]$ $= k [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$

$[\begin{matrix} 1 + 2 & - 2 + 0 \\ 4 + 0 & - 4 + 2 \end{matrix}]$ $= [\begin{matrix} 3 k & - 2 k \\ 4 k & - 2 k \end{matrix}]$

$[\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$ $= [\begin{matrix} 3 k & - 2 k \\ 4 k & - 2 k \end{matrix}]$

We have, $3 = 3 k$

$k = \frac{3}{3} = 1$

Hence, the value of k is 1.

Question 18. If $A = [\begin{matrix} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{matrix}]$ and I is the identity matrix of order 2, show that $I + A = (I - A) [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

Answer:

$A = [\begin{matrix} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{matrix}]$

$I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

To prove : $I + A = (I - A) [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

L.H.S : $I + A$

$I + A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ $+ [\begin{matrix} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{matrix}]$

$I + A = [\begin{matrix} 1 + 0 & 0 - \tan \frac{α}{2} \\ 0 + \tan \frac{α}{2} & 1 + 0 \end{matrix}]$

$I + A = [\begin{matrix} 1 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 1 \end{matrix}]$

R.H.S : $(I - A) [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

$(I - A) [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$ $= ([\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] -$ $[\begin{matrix} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{matrix}])$ $\times [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

$(I - A) [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$ $= [\begin{matrix} 1 - 0 & 0 - (- \tan \frac{α}{2}) \\ 0 - \tan \frac{α}{2} & 1 - 0 \end{matrix}]$ $\times [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

$(I - A) [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$ $= [\begin{matrix} 1 & \tan \frac{α}{2} \\ - \tan \frac{α}{2} & 1 \end{matrix}]$ $\times [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

$= [\begin{matrix} \cos α + \sin α \tan \frac{α}{2} & - \sin α + \cos α \tan \frac{α}{2} \\ - \tan \frac{α}{2} \cos α + \sin α & \tan \frac{α}{2} \sin α + \cos α \end{matrix}]$

$= [\begin{matrix} 1 - 2 \sin^{2} \frac{α}{2} + 2 \sin \frac{α}{2} c o s \frac{α}{2} \tan \frac{α}{2} & - 2 \sin \frac{α}{2} c o s \frac{α}{2} + (2 \cos^{2} \frac{α}{2} - 1) \tan \frac{α}{2} \\ - \tan \frac{α}{2} (2 \cos^{2} \frac{α}{2} - 1) + 2 \sin \frac{α}{2} c o s \frac{α}{2} & \tan \frac{α}{2} 2 \sin \frac{α}{2} c o s \frac{α}{2} + 1 - 2 \sin^{2} \frac{α}{2} \end{matrix}]$

$= [\begin{matrix} 1 - 2 \sin^{2} \frac{α}{2} + 2 \sin^{2} \frac{α}{2} & - 2 \sin \frac{α}{2} c o s \frac{α}{2} + 2 \sin \frac{α}{2} c o s \frac{α}{2} - \tan \frac{α}{2} \\ - 2 \sin \frac{α}{2} c o s \frac{α}{2} + \tan \frac{α}{2} + 2 \sin \frac{α}{2} c o s \frac{α}{2} & 2 \sin^{2} \frac{α}{2} + 1 - 2 \sin^{2} \frac{α}{2} \end{matrix}]$

$= [\begin{matrix} 1 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 1 \end{matrix}]$

Hence, we can see L.H.S = R.H.S

i.e. $I + A = (I - A) [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$ .

Question 19(i). A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 1800

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

$[\begin{matrix} x & (30000 - x) \end{matrix}]$ $[\begin{matrix} \frac{5}{100} \\ \frac{7}{100} \end{matrix}]$ $= 1800$ (simple interest for 1 year $= \frac{p r i c i p a l \times r a t e}{100}$ )

$\frac{5}{100} x + \frac{7}{100} (30000 - x) = 1800$

$5 x + 210000 - 7 x = 180000$

$210000 - 180000 = 7 x - 5 x$

$30000 = 2 x$

$x = 15000$

Thus, to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs 15000 in the first bond and Rs 15000 in the second bond.

Question 19(ii). A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 2000

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

$[\begin{matrix} x & (30000 - x) \end{matrix}] [\begin{matrix} \frac{5}{100} \\ \frac{7}{100} \end{matrix}] = 2000$

(Simple interest for 1 year = $\frac{Principal \times Rate}{100}$ )

$\frac{5}{100} x + \frac{7}{100} (30000 - x) = 2000$

$\frac{5 x + 210000 - 7 x}{100} = 2000$

$\frac{210000 - 2 x}{100} = 2000$

$210000 - 2 x = 200000$

$210000 - 200000 = 2 x$

$10000 = 2 x$

$x = 5000$

Thus, to obtain an annual total interest of Rs. 2000, the trust fund should invest Rs 5000 in the first bond and Rs 25000 in the second bond.

Question 20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.

Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively.

The total amount the bookshop will receive from selling all the books:

$12$ $[\begin{matrix} 10 & 8 & 10 \end{matrix}]$ $[\begin{matrix} 80 \\ 60 \\ 40 \end{matrix}]$

$= 12 (10 \times 80 + 8 \times 60 + 10 \times 40)$

$= 12 (800 + 480 + 400)$

$= 12 (1680)$

$= 20160$

The total amount the bookshop will receive from selling all the books is 20160.

Question 21 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

Q21. The restriction on n, k and p so that PY + WY will be defined are:
(A) $k = 3, p = n$

(B) k is arbitrary, $p = 2$

(D) $k = 2, p = 3$

Answer:

P and Y are of order $p \times k$ and $3 \times k$ respectively.

Therefore, $P Y$ will be defined only if $k = 3$ , i.e., the order of $P Y$ is $p \times k$ .

W and Y are of order $n \times 3$ and $3 \times k$ respectively.

Therefore, $W Y$ is defined because the number of columns of $W$ is equal to the number of rows of $Y$ , which is 3, i.e., the order of $W Y$ is $n \times k$ .

Matrices $P Y$ and $W Y$ can only be added if they both have the same order, i.e., $p \times k = n \times k \Rightarrow p = n$ .

Therefore, $k = 3$ , $p = n$ are restrictions on $n$ , $k$ , and $p$ so that $P Y + W Y$ will be defined.

Option (A) is correct.

Question 22 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively. Choose the correct answer in Exercises 21 and 22. If n = p, then the order of the matrix $7 X - 5 Z$ is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

Answer:

$X$ has order $2 \times n$ .

Therefore, $7 X$ also has order $2 \times n$ .

$Z$ has order $2 \times p$ .

Therefore, $5 Z$ also has order $2 \times p$ .

Matrices $7 X$ and $5 Z$ can only be subtracted if they both have the same order, i.e., $2 \times n = 2 \times p$ , and it is given that $p = n$ .

We can say that both matrices have order $2 \times n$ .

Therefore, the order of $7 X - 5 Z$ is $2 \times n$ .

Option (B) is correct.

Topics covered in Chapter 3: Matrices: Exercise 3.2

Introduction
Topics covered
Addition of matrices: If $A = [a_{i j}]$ and $B = [b_{i j}]$ are two matrices of the same order, say $m \times n$ . Then, the sum of the two matrices A and B is defined as a matrix $C = {[c_{i j}]}_{m \times n}$ , where $c_{i j} = a_{i j} + b_{i j}$ , for all possible values of $i$ and $j$ .
Multiplication of a matrix by a scalar: Multiplication of a matrix by a scalar as follows: if $A = {[a_{i j}]}_{m \times n}$ is a matrix and $k$ is a scalar, then $k A$ is another matrix which is obtained by multiplying each element of A by the scalar $k$ .
Difference of matrices: If $A = [a_{i j}], B = [b_{i j}]$ are two matrices of the same order, say $m \times n$ , then difference $A - B$ is defined as a matrix $D = [d_{i j}]$ , where $d_{i j} = a_{i j} - b_{i j}$ , for all value of $i$ and $j$ . In other words, $D = A - B = A + (- 1) B$ , that is, the sum of the matrix A and the matrix - B.
Properties of matrix addition

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Commutative Law: If $A = [a_{i j}], B = [b_{i j}]$ are matrices of the same order, say $m \times n$ , then $A + B = B + A$ .
Associative Law: For any three matrices $A = [a_{i j}], B = [b_{i j}], C = [c_{i j}]$ of the same order, say $m \times n, (A + B) + C = A + (B + C)$ .
Existence of additive identity: Let $A = [a_{i j}]$ be an $m \times n$ matrix and O be an $m \times n$ zero matrix, then $A + O = O + A = A$ . In other words, O is the additive identity for matrix addition.
The existence of additive inverse: Let $A = {[a_{i j}]}_{m \times n}$ be any matrix, then we have another matrix as $- A = {[- a_{i j}]}_{m \times n}$ such that $A + (- A) = (- A) + A = O$ . So, - A is the additive inverse of A or the negative of A.

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Properties of scalar multiplication of a matrix

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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$k (A + B) = k A + k B$
$k (A + B) = k A + k B$
$(k + l) A = k A + l A$

Multiplication of matrices: If $A = {[a_{i j}]}_{m \times n}, B = {[b_{j k}]}_{n \times p}$ , then the $i^{th}$ row of A is $[a_{i 1} a_{i 2} \dots a_{i n}]$ and the $k^{th}$ column of B is $[\begin{matrix} b_{1 k} \\ b_{2 k} \\ ⋮ \\ b_{n k} \end{matrix}]$ , then $c_{i k} = a_{i 1} b_{1 k} + a_{i 2} b_{2 k} + a_{i 3} b_{3 k} + \dots + a_{i n} b_{n k} = \sum_{j = 1}^{n} a_{i j} b_{j k}$ .

The matrix $C = {[c_{i k}]}_{m m \times p}$ is the product of A and B.

Properties of Multiplication

The associative law: For any three matrices A, B, and C. We have $(A B) C = A (B C)$ whenever both sides of the equality are defined.
The distributive law: For three matrices $A, B$ and C.

(i) $A (B + C) = AB + AC$

(ii) $(A + B) C = AC + BC$ , whenever both sides of equality are defined.

The existence of multiplicative identity: For every square matrix $A$ , there exists an identity matrix of the same order such that $IA = AI = A$ .

Also, read,

NCERT solutions for class 12 maths chapter 3

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Students can check these NCERT exemplar links for additional practice.

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Frequently Asked Questions (FAQs)

1. Can I get free NCERT solutions for Class 12 maths?

Click here to get NCERT solutions for class 12 maths. Links for solutions to each chapters of NCERT syllabus Class 12 Mathematics is available here. All the exercise questions of NCERT textbook are covered. For more questions solve use NCERT exemplar.

2. What is the use of matrix ?

Matrix is an important tool useful in science, statistics, research, representation of data, mechanics, optics, electromagnetism, quantum mechanics, and quantum electrodynamics, etc.

3. What is zero matrix ?

A matrix all of whose entries are zero is called a zero matrix.

4. What is equal matrix ?

Two matrices are equal matrices if the order and correspondence entities of both matrices are the same.

5. Does scalar matrix is square matrix ?

Yes, the scalar matrix is a square matrix.

6. How do you identify a scalar matrix ?

A scalar matrix is a square matrix whose all diagonal elements are equal and all other elements are zero.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

quartely question paper for mathematics cbse

Manasvini Gupta 30 January,2025

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

Jefferson 25 September,2024

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Kanishka kaushikii 17 September,2024

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

Yuvan 30 August,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.