NCERT Solutions for Class 12 Maths Chapter 3 Matrices

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(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:

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

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

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

Answer:

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

(iv) AB

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

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

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

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

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

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 the 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 $\sin^{2} x + \cos^{2} x = 1$

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

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 then 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 is 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 is 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 out the value of the square of matrix A

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

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

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

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

$∴$ $A^{2} - 5 A + 6 I$

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

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

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

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

Question16. 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 $= \frac{p r i c i p a l \times r a t e}{100}$ )

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

$5 x + 210000 - 7 x = 200000$

$210000 - 200000 = 7 x - 5 x$

$10000 = 2 x$

$x = 5000$

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.

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 $p * k$ and $3 * k$ respectively.

$∴$ PY will be defined only if k=3, i.e. order of PY is $p * k$ .

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

$∴$ WY 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 WY is $n * k$

Matrices PY and WY can only be added if they both have same order i.e = $n * k$ implies p=n.

Thus, $k = 3, p = n$ are restrictions on n, k, and p so that PY + WY 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 of order $2 * n$ .

$∴$ 7X also has of order $2 * n$ .

Z has of order $2 * p$ .

$∴$ 5Z also has of order $2 * p$ .

Mtarices 7X and 5Z can only be subtracted if they both have same order i.e $2 * n$ = $2 * p$ and it is given that p=n.

We can say that both matrices have order of $2 * n$ .

Thus, order of $7 X - 5 Z$ is $2 * n$ .

Option (B) is correct.

Class 12 Maths chapter 3 solutions Exercise: 3.3
Page number: 66-68
Total questions: 12

Question 1(i). Find the transpose of each of the following matrices:

$[\begin{matrix} 5 \\ \frac{1}{2} \\ - 1 \end{matrix}]$

Answer:

$A = [\begin{matrix} 5 \\ \frac{1}{2} \\ - 1 \end{matrix}]$

The transpose of the given matrix is

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

Question 1(ii). Find the transpose of each of the following matrices:

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

Answer:

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

interchanging the rows and columns of the matrix A we get

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

Question 1(iii). Find the transpose of each of the following matrices:

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

Answer:

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

Transpose is obtained by interchanging the rows and columns of matrix

$A^{T} = [\begin{matrix} - 1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & - 1 \end{matrix}]$

Question 2(i). If $A = [\begin{matrix} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{matrix}]$ and $B = [\begin{matrix} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix}]$ , then verify

$(A + B)^{'} = A^{'} + B^{'}$

Answer:

$A = [\begin{matrix} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{matrix}]$ and $B = [\begin{matrix} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix}]$

$(A + B)^{'} = A^{'} + B^{'}$

L.H.S : $(A + B)^{'}$

$A + B = [\begin{matrix} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{matrix}]$ $+ [\begin{matrix} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix}]$

$A + B = [\begin{matrix} - 1 + (- 4) & 2 + 1 & 3 + (- 5) \\ 5 + 1 & 7 + 2 & 9 + 0 \\ - 2 + 1 & 1 + 3 & 1 + 1 \end{matrix}]$

$A + B = [\begin{matrix} - 5 & 3 & - 2 \\ 6 & 9 & 9 \\ - 1 & 4 & 2 \end{matrix}]$

$(A + B)^{'} = [\begin{matrix} - 5 & 6 & - 1 \\ 3 & 9 & 4 \\ - 2 & 9 & 2 \end{matrix}]$

R.H.S : $A^{'} + B^{'}$

$A^{'} + B^{'} = [\begin{matrix} - 1 & 5 & - 2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{matrix}]$ $+ [\begin{matrix} - 4 & 1 & 1 \\ 1 & 2 & 3 \\ - 5 & 0 & 1 \end{matrix}]$

$A^{'} + B^{'} = [\begin{matrix} - 1 + (- 4) & 5 + 1 & - 2 + 1 \\ 2 + 1 & 7 + 2 & 1 + 3 \\ 3 + (- 5) & 9 + 0 & 1 + 1 \end{matrix}]$

$A^{'} + B^{'} = [\begin{matrix} - 5 & 6 & - 1 \\ 3 & 9 & 4 \\ - 2 & 9 & 2 \end{matrix}]$

Thus we find that the LHS is equal to RHS and hence verified.

Question 2(ii). If $A = [\begin{matrix} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{matrix}]$ and $B = [\begin{matrix} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix}]$ , then verify

$(A - B)^{'} = A^{'} - B^{'}$

Answer:

$A = [\begin{matrix} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{matrix}]$ and $B = [\begin{matrix} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix}]$

$(A - B)^{'} = A^{'} - B^{'}$

L.H.S : $(A - B)^{'}$

$A - B = [\begin{matrix} - 1 & 2 & 3 \\ 5 & 7 & 9 \\ - 2 & 1 & 1 \end{matrix}]$ $- [\begin{matrix} - 4 & 1 & - 5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix}]$

$A - B = [\begin{matrix} - 1 - (- 4) & 2 - 1 & 3 - (- 5) \\ 5 - 1 & 7 - 2 & 9 - 0 \\ - 2 - 1 & 1 - 3 & 1 - 1 \end{matrix}]$

$A - B = [\begin{matrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ - 3 & - 2 & 0 \end{matrix}]$

$(A - B)^{'} = [\begin{matrix} 3 & 4 & - 3 \\ 1 & 5 & - 2 \\ 8 & 9 & 0 \end{matrix}]$

R.H.S : $A^{'} - B^{'}$

$A^{'} - B^{'} = [\begin{matrix} - 1 & 5 & - 2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{matrix}]$ $- [\begin{matrix} - 4 & 1 & 1 \\ 1 & 2 & 3 \\ - 5 & 0 & 1 \end{matrix}]$

$A^{'} - B^{'} = [\begin{matrix} - 1 - (- 4) & 5 - 1 & - 2 - 1 \\ 2 - 1 & 7 - 2 & 1 - 3 \\ 3 - (- 5) & 9 - 0 & 1 - 1 \end{matrix}]$

$A^{'} - B^{'} = [\begin{matrix} 3 & 4 & - 3 \\ 1 & 5 & - 2 \\ 8 & 9 & 0 \end{matrix}]$

Hence, L.H.S = R.H.S. so verified that

$(A - B)^{'} = A^{'} - B^{'}$ .

Question 3(i). If $A^{'} = [\begin{matrix} 3 & 4 \\ - 1 & 2 \\ 0 & 1 \end{matrix}]$ and $B = [\begin{matrix} - 1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix}]$ , then verify

$(A + B)^{'} = A^{'} + B^{'}$

Answer:

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

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

To prove: $(A + B)^{'} = A^{'} + B^{'}$

$L . H . S : (A + B)^{'} =$

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

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

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

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

R.H.S: $A^{'} + B^{'}$

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

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

Hence, L.H.S = R.H.S i.e. $(A + B)^{'} = A^{'} + B^{'}$ .

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

$(A - B)^{'} = A^{'} - B^{'}$

Answer:

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

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

To prove: $(A - B)^{'} = A^{'} - B^{'}$

$L . H . S : (A - B)^{'} =$

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

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

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

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

R.H.S: $A^{'} - B^{'}$

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

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

Hence, L.H.S = R.H.S i.e. $(A - B)^{'} = A^{'} - B^{'}$ .

Question 4. If $A^{'} = [\begin{matrix} - 2 & 3 \\ 1 & 2 \end{matrix}]$ and $B = [\begin{matrix} - 1 & 0 \\ 1 & 2 \end{matrix}]$ , then find $(A + 2 B)^{'}$

Answer:

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

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

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

$(A + 2 B)^{'}$ :

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

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

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

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

Transpose is obtained by interchanging rows and columns and the transpose of A+2B is

$(A + 2 B)^{'} = [\begin{matrix} - 4 & 5 \\ 1 & 6 \end{matrix}]$

Question 5(i) For the matrices A and B, verify that $(A B)^{'} = B^{'} A^{'}$ , where

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

Answer:

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

To prove : $(A B)^{'} = B^{'} A^{'}$

$L . H . S : (A B)^{'}$

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

$A B = [\begin{matrix} - 1 & 2 & 1 \\ 4 & - 8 & - 4 \\ - 3 & 6 & 3 \end{matrix}]$

$(A B)^{'} = [\begin{matrix} - 1 & 4 & - 3 \\ 2 & - 8 & 6 \\ 1 & - 4 & 3 \end{matrix}]$

$R . H . S : B^{'} A^{'}$

$B^{'} = [\begin{matrix} - 1 \\ 2 \\ 1 \end{matrix}]$

$A^{'} = [\begin{matrix} 1 & - 4 & 3 \end{matrix}]$

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

$B^{'} A^{'} = [\begin{matrix} - 1 & 4 & - 3 \\ 2 & - 8 & 6 \\ 1 & - 4 & 3 \end{matrix}]$

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

so it is verified that $(A B)^{'} = B^{'} A^{'}$ .

Question 5(ii) For the matrices A and B, verify that $(A B)^{'} = B^{'} A^{'}$ , where

$A = [\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}]$ , $B = [\begin{matrix} 1 & 5 & 7 \end{matrix}]$

Answer:

$A = [\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}]$ , $B = [\begin{matrix} 1 & 5 & 7 \end{matrix}]$

To prove : $(A B)^{'} = B^{'} A^{'}$

$L . H . S : (A B)^{'}$

$A B = [\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}]$ $[\begin{matrix} 1 & 5 & 7 \end{matrix}]$

$A B = [\begin{matrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{matrix}]$

$(A B)^{'} = [\begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{matrix}]$

$R . H . S : B^{'} A^{'}$

$B^{'} = [\begin{matrix} 1 \\ 5 \\ 7 \end{matrix}]$

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

$B^{'} A^{'} = [\begin{matrix} 1 \\ 5 \\ 7 \end{matrix}]$ $[\begin{matrix} 0 & 1 & 2 \end{matrix}]$

$B^{'} A^{'} = [\begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{matrix}]$

Heence, L.H.S =R.H.S i.e. $(A B)^{'} = B^{'} A^{'}$ .

Question 6(i). If $A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$ , then verify that $A^{'} A = I$

Answer:

$A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$

By interchanging rows and columns we get transpose of A

$A^{'} = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

To prove: $A^{'} A = I$

L.H.S : $A^{'} A$

$A^{'} A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$ $[\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$

$A^{'} A = [\begin{matrix} \cos^{2} α + \sin^{2} α & \sin α \cos α - \sin α c o s α \\ \sin α \cos α - \sin α \cos α & s i n^{2} α + \cos^{2} α \end{matrix}]$

$A^{'} A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I = R . H . S$

Question 6(ii). If $A = [\begin{matrix} \sin α & \cos α \\ - \cos α & \sin α \end{matrix}]$ , then verify that $A^{'} A = I$

Answer:

$A = [\begin{matrix} \sin α & \cos α \\ - \cos α & \sin α \end{matrix}]$

By interchanging columns and rows of the matrix A we get the transpose of A

$A^{'} = [\begin{matrix} \sin α & - \cos α \\ \cos α & \sin α \end{matrix}]$

To prove: $A^{'} A = I$

L.H.S : $A^{'} A$

$A^{'} A = [\begin{matrix} \sin α & - \cos α \\ \cos α & \sin α \end{matrix}]$ $[\begin{matrix} \sin α & \cos α \\ - \cos α & \sin α \end{matrix}]$

$A^{'} A = [\begin{matrix} \cos^{2} α + \sin^{2} α & \sin α \cos α - \sin α c o s α \\ \sin α \cos α - \sin α \cos α & s i n^{2} α + \cos^{2} α \end{matrix}]$

$A^{'} A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I = R . H . S$

Question 7(i). Show that the matrix $A = [\begin{matrix} 1 & - 1 & 5 \\ - 1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix}]$ is a symmetric matrix.

Answer:

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

the transpose of A is

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

Since, $A^{'} = A$ so given matrix is a symmetric matrix.

Question 7(ii) Show that the matrix $A = [\begin{matrix} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ 1 & - 1 & 0 \end{matrix}]$ is a skew-symmetric matrix.

Answer:

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

The transpose of A is

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

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

$A^{'} = - A$

Since, $A^{'} = - A$ so given matrix is a skew-symmetric matrix.

Question 8(i). For the matrix $A = [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]$ , verify that

$(A + A^{'})$ is a symmetric matrix.

Answer:

$A = [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]$

$A^{'} = [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}]$

$A + A^{'} = [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]$ $+ [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}]$

$A + A^{'} = [\begin{matrix} 1 + 1 & 5 + 6 \\ 6 + 5 & 7 + 7 \end{matrix}]$

$A + A^{'} = [\begin{matrix} 2 & 11 \\ 11 & 14 \end{matrix}]$

$(A + A^{'})^{'} = [\begin{matrix} 2 & 11 \\ 11 & 14 \end{matrix}]$

We have $A + A^{'} = (A + A^{'})^{'}$

Hence, $(A + A^{'})$ is a symmetric matrix.

Question 8(ii) For the matrix $A = [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]$ , verify that

$(A - A^{'})$ is a skew symmetric matrix.

Answer:

$A = [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]$

$A^{'} = [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}]$

$A - A^{'} = [\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]$ $- [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}]$

$A - A^{'} = [\begin{matrix} 1 - 1 & 5 - 6 \\ 6 - 5 & 7 - 7 \end{matrix}]$

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

$(A - A^{'})^{'} = [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] = - (A - A^{'})$

We have $A - A^{'} = - (A - A^{'})^{'}$

Hence, $(A - A^{'})$ is a skew-symmetric matrix.

Question 9. Find $\frac{1}{2} (A + A^{'})$ and $\frac{1}{2} (A - A^{'})$ , when $A = [\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}]$

Answer:

$A = [\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}]$

the transpose of the matrix is obtained by interchanging rows and columns

$A^{'} = [\begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix}]$

$\frac{1}{2} (A + A^{'}) = \frac{1}{2} ([\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}]$ $+ [\begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix}])$

$\frac{1}{2} (A + A^{'}) = \frac{1}{2} ([\begin{matrix} 0 + 0 & a + (- a) & b + (- b) \\ - a + a & 0 + 0 & c + (- c) \\ - b + b & - c + c & 0 + 0 \end{matrix}])$

$\frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

$\frac{1}{2} (A + A^{'}) = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

$\frac{1}{2} (A + A^{'}) = 0$

$\frac{1}{2} (A - A^{'}) = \frac{1}{2} ([\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}]$ $- [\begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix}])$

$\frac{1}{2} (A - A^{'}) = \frac{1}{2} ([\begin{matrix} 0 - 0 & a - (- a) & b - (- b) \\ - a - a & 0 - 0 & c - (- c) \\ - b - b & - c - c & 0 - 0 \end{matrix}])$

$\frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{matrix} 0 & 2 a & 2 b \\ - 2 a & 0 & 2 c \\ - 2 b & - 2 c & 0 \end{matrix}]$

$\frac{1}{2} (A - A^{'}) = [\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}]$

Question 10(i). Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

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

Answer:

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

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

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

$A + A^{'} = [\begin{matrix} 6 & 6 \\ 6 & - 2 \end{matrix}]$

Let

$B = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{matrix} 6 & 6 \\ 6 & - 2 \end{matrix}]$ $= [\begin{matrix} 3 & 3 \\ 3 & - 1 \end{matrix}]$

$B^{'} = [\begin{matrix} 3 & 3 \\ 3 & - 1 \end{matrix}] = B$

Thus, $\frac{1}{2} (A + A^{'})$ is a symmetric matrix.

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

$A - A^{'} = [\begin{matrix} 0 & 4 \\ - 4 & 0 \end{matrix}]$

Let

$C = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{matrix} 0 & 4 \\ - 4 & 0 \end{matrix}]$ $= [\begin{matrix} 0 & 2 \\ - 2 & 0 \end{matrix}]$

$C^{'} = [\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}]$

$C = - C^{'}$

Thus, $\frac{1}{2} (A - A^{'})$ is a skew symmetric matrix.

Represent A as sum of B and C.

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

Question:10(ii). Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

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

Answer:

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

$A^{'} = [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

$A + A^{'} = [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$ $+ [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

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

Let

$B = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{matrix} 12 & - 4 & 4 \\ - 4 & 6 & - 2 \\ 4 & - 2 & 6 \end{matrix}]$ $= [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

$B^{'} = [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] = B$

Thus, $\frac{1}{2} (A + A^{'})$ is a symmetric matrix.

$A - A^{'} = [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$ $- [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

$A - A^{'} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

Let

$C = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$ $= [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

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

$C = - C^{'}$

Thus, $\frac{1}{2} (A - A^{'})$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

$B + C = [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$ $+ [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$ $= [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] = A$

Question 10(iii). Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

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

Answer:

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

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

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

$A + A^{'} = [\begin{matrix} 6 & 1 & - 5 \\ 1 & - 4 & - 4 \\ - 5 & - 4 & 4 \end{matrix}]$

Let

$B = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{matrix} 6 & 1 & - 5 \\ 1 & - 4 & - 4 \\ - 5 & - 4 & 4 \end{matrix}]$ $= [\begin{matrix} 3 & \frac{1}{2} & - \frac{5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}]$

$B^{'} = [\begin{matrix} 3 & \frac{1}{2} & - \frac{5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}] = B$

Thus, $\frac{1}{2} (A + A^{'})$ is a symmetric matrix.

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

$A - A^{'} = [\begin{matrix} 0 & 5 & 3 \\ - 5 & 0 & 6 \\ - 3 & - 6 & 0 \end{matrix}]$

Let

$C = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{matrix} 0 & 5 & 3 \\ - 5 & 0 & 6 \\ - 3 & - 6 & 0 \end{matrix}]$ $= [\begin{matrix} 0 & \frac{5}{2} & \frac{3}{2} \\ - \frac{5}{2} & 0 & 3 \\ \frac{- 3}{2} & - 3 & 0 \end{matrix}]$

$C^{'} = [\begin{matrix} 0 & - \frac{5}{2} & - \frac{3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}]$

$C = - C^{'}$

Thus, $\frac{1}{2} (A - A^{'})$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

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

Question 10(iv). Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

$[\begin{matrix} 1 & 5 \\ - 1 & 2 \end{matrix}]$

Answer:

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

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

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

$A + A^{'} = [\begin{matrix} 2 & 4 \\ 4 & 4 \end{matrix}]$

Let

$B = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [\begin{matrix} 2 & 4 \\ 4 & 4 \end{matrix}]$ $= [\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}]$

$B^{'} = [\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}] = B$

Thus, $\frac{1}{2} (A + A^{'})$ is a symmetric matrix.

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

$A - A^{'} = [\begin{matrix} 0 & 6 \\ - 6 & 0 \end{matrix}]$

Let

$C = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [\begin{matrix} 0 & 6 \\ - 6 & 0 \end{matrix}]$ $= [\begin{matrix} 0 & 3 \\ - 3 & 0 \end{matrix}]$

$C^{'} = [\begin{matrix} 0 & - 3 \\ 3 & 0 \end{matrix}]$

$C = - C^{'}$

Thus, $\frac{1}{2} (A - A^{'})$ is a skew-symmetric matrix.

Represent A as the sum of B and C.

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

Question 11 Choose the correct answer in the Exercises 11 and 12.

If A, B are symmetric matrices of same order, then AB – BA is a

(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix

Answer:

If A, B are symmetric matrices then

$A^{'} = A$ and $B^{'} = B$

we have, ${(A B - B A)}^{'} = {(A B)}^{'} - {(B A)}^{'} = B^{'} A^{'} - A^{'} B^{'}$

$= B A - A B$

$= - (A B - B A)$

Hence, we have $(A B - B A) = - (A B - B A)^{'}$

Thus,( AB-BA)' is skew symmetric.

Option A is correct.

Question 12. Choose the correct answer in the Exercises 11 and 12.

If $A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$ and $A + A^{'} = I$ , then the value of $α$ is

(A) $\frac{π}{6}$

(B) $\frac{π}{3}$

(D) $\frac{3 π}{2}$

Answer:

$A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

$A^{'} = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$

$A + A^{'} = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$ $+ [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$ $= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$A + A^{'} = [\begin{matrix} 2 \cos α & 0 \\ 0 & 2 \cos α \end{matrix}]$ $= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$2 c o s α = 1$

$c o s α = \frac{1}{2}$

$α = \frac{π}{3}$

Option B is correct.

Class 12 Maths chapter 3 solutions Exercise: 3.4
Page number: 69-69
Total questions: 1

Question:1 Matrices A and B will be inverse of each other only if

(A) $A B = B A$

(B) $A B = B A = 0$

(D) $A B = B A = I$

Answer:

We know that if A is a square matrix of order n and there is another matrix B of same order n, such that $A B = B A = I$ , then B is inverse of matrix A.

In this case, it is clear that A is inverse of B.

Hence, m atrices A and B will be inverse of each other only if $A B = B A = I$ .

Option D is correct.

Class 12 Maths chapter 3 solutions Miscellaneous Exercise:
Page number: 72-73
Total questions: 11

Question 1. If A and B are symmetric matrices, prove that $A B - B A$ is a skew symmetric matrix.

Answer:

If A, B are symmetric matrices then

$A^{'} = A$ and $B^{'} = B$

we have, ${(A B - B A)}^{'} = {(A B)}^{'} - {(B A)}^{'} = B^{'} A^{'} - A^{'} B^{'}$

$= B A - A B$

$= - (A B - B A)$

Hence, we have $(A B - B A) = - (A B - B A)^{'}$

Thus,( AB-BA)' is skew symmetric.

Question 2. Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

Let be a A is symmetric matrix , then $A^{'} = A$

Consider, $(B^{'} A B)^{'} = {B^{'} (A B)}^{'}$

$= {(A B)}^{'} (B^{'})^{'}$

$= B^{'} A^{'} (B)$

$= B^{'} (A^{'} B)$

Replace $A^{'}$ by $A$

$= B^{'} (A B)$

i.e. $(B^{'} A B)^{'}$ $= B^{'} (A B)$

Thus, if A is a symmetric matrix than $B^{'} (A B)$ is a symmetric matrix.

Now, let A be a skew-symmetric matrix, then $A^{'} = - A$ .

$(B^{'} A B)^{'} = {B^{'} (A B)}^{'}$

$= {(A B)}^{'} (B^{'})^{'}$

$= B^{'} A^{'} (B)$

$= B^{'} (A^{'} B)$

Replace $A^{'}$ by - $A$ ,

$= B^{'} (- A B)$

$= - B^{'} A B$

i.e. $(B^{'} A B)^{'}$ $= - B^{'} A B$ .

Thus, if A is a skew-symmetric matrix then $- B^{'} A B$ is a skew-symmetric matrix.

Hence, the matrix B′AB is symmetric or skew-symmetric according to as A is symmetric or skew-symmetric.

Question 3. Find the values of x , y , z if the matrix $A = [\begin{matrix} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{matrix}]$ satisfy the equation $A^{'} A = I$

Answer:

$A = [\begin{matrix} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{matrix}]$

$A^{'} = [\begin{matrix} 0 & x & x \\ 2 y & y & - y \\ z & - z & z \end{matrix}]$

$A^{'} A = I$

$[\begin{matrix} 0 & x & x \\ 2 y & y & - y \\ z & - z & z \end{matrix}]$ $[\begin{matrix} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{matrix}]$ $= [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

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

$[\begin{matrix} 2 x^{2} & 0 & 0 \\ 0 & 6 y^{2} & 0 \\ 0 & 0 & 3 z^{2} \end{matrix}]$ $= [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Thus equating the terms elementwise

$2 x^{2} = 1$ $6 y^{2} = 1$ $3 z^{2} = 1$

$x^{2} = \frac{1}{2}$ $y^{2} = \frac{1}{6}$ $z^{2} = \frac{1}{3}$

$x = \pm \frac{1}{\sqrt{2}}$ $y = \pm \frac{1}{\sqrt{6}}$ $z = \pm \frac{1}{\sqrt{3}}$

Question 4. For what values of x: $[\begin{matrix} 1 & 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = O$ ?

Answer:

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

$[\begin{matrix} 1 + 4 + 1 & 2 + 0 + 0 & 0 + 2 + 2 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = O$

$[\begin{matrix} 6 & 2 & 4 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = O$

$[\begin{matrix} 0 + 4 + 4 x \end{matrix}] = O$

$4 + 4 x = 0$

$4 x = - 4$

$x = - 1$

Thus, value of x is -1.

Question 5. If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$ , show that $A^{2} - 5 A + 7 I = 0$ .

Answer:

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

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

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

$A^{2} = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}]$

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

To prove: $A^{2} - 5 A + 7 I = 0$

L.H.S : $A^{2} - 5 A + 7 I$

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

$= [\begin{matrix} 8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7 \end{matrix}]$

$= [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 = R . H . S$

Hence, we proved that

$A^{2} - 5 A + 7 I = 0$ .

Question 6. Find x, if $[\begin{matrix} x & - 5 & - 1 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] [\begin{matrix} x \\ 4 \\ 1 \end{matrix}] = 0$ .

Answer:

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

$[\begin{matrix} x + 0 - 2 & 0 - 10 + 0 & 2 x - 5 - 3 \end{matrix}] [\begin{matrix} x \\ 4 \\ 1 \end{matrix}] = 0$

$[\begin{matrix} x - 2 & - 10 & 2 x - 8 \end{matrix}] [\begin{matrix} x \\ 4 \\ 1 \end{matrix}] = 0$

$[\begin{matrix} x (x - 2) - 40 + (2 x - 8) \end{matrix}] = 0$

$[\begin{matrix} x^{2} - 2 x - 40 + 2 x - 8 \end{matrix}] = 0$

$∴ x^{2} - 48 = 0$

$x^{2} = 48$

thus the value of x is

$x = \pm 4 \sqrt{3}$

Question 7(a) A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

Answer:

The unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively.

The total revenue in the market I with the help of matrix algebra can be represented as :

$[\begin{matrix} 10000 & 2000 & 18000 \end{matrix}] [\begin{matrix} 2.50 \\ 1.50 \\ 1.00 \end{matrix}]$

$= 10000 \times 2.50 + 2000 \times 1.50 + 18000 \times 1.00$

$= 25000 + 3000 + 18000$

$= 46000$

The total revenue in market II with the help of matrix algebra can be represented as :

$[\begin{matrix} 6000 & 20000 & 8000 \end{matrix}] [\begin{matrix} 2.50 \\ 1.50 \\ 1.00 \end{matrix}]$

$= 6000 \times 2.50 + 20000 \times 1.50 + 8000 \times 1.00$

$= 15000 + 30000 + 8000$

$= 53000$

Hence, total revenue in the market I is 46000 and total revenue in market II is 53000.

Question 7(b). A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If the unit costs of the above three commodities are ` 2.00, ` 1.00, and 50 paise respectively. Find the gross profit.

Answer:

The unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively.

The total cost price in market I with the help of matrix algebra can be represented as :

$[\begin{matrix} 10000 & 2000 & 18000 \end{matrix}] [\begin{matrix} 2.00 \\ 1.00 \\ 0.50 \end{matrix}]$

$= 10000 \times 2.00 + 2000 \times 1.00 + 18000 \times 0.50$

$= 20000 + 2000 + 9000$

$= 31000$

Total revenue in the market I is 46000 , gross profit in the market is $= 46000 - 31000$ $= R s . 15000$

The total cost price in market II with the help of matrix algebra can be represented as :

$[\begin{matrix} 6000 & 20000 & 8000 \end{matrix}] [\begin{matrix} 2.00 \\ 1.00 \\ 0.50 \end{matrix}]$

$= 6000 \times 2.0 + 20000 \times 1.0 + 8000 \times 0.50$

$= 12000 + 20000 + 4000$

$= 36000$

Total revenue in market II is 53000, gross profit in the market is $= 53000 - 36000 = R s . 17000$

Question 8. Find the matrix X so that $X [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{matrix}]$

Answer:

$X [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{matrix}]$

The matrix given on R.H.S is $2 \times 3$ matrix and on LH.S is $2 \times 3$ matrix.Therefore, X has to be $2 \times 2$ matrix.

Let X be $[\begin{matrix} a & c \\ b & d \end{matrix}]$

$[\begin{matrix} a & c \\ b & d \end{matrix}]$ $[\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{matrix}]$

$[\begin{matrix} a + 4 c & 2 a + 5 c & 3 a + 6 c \\ b + 4 d & 2 b + 5 d & 3 b + 6 d \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{matrix}]$

$a + 4 c = - 7$ $2 a + 5 c = - 8$ $3 a + 6 c = - 9$

$b + 4 d = 2$ $2 b + 5 d = 4$ $3 b + 6 d = 6$

Taking, $a + 4 c = - 7$

$a = - 4 c - 7$

$2 a + 5 c = - 8$

$- 8 c - 14 + 5 c = - 8$

$- 3 c = 6$

$c = - 2$

$a = - 4 \times - 2 - 7$

$a = 8 - 7 = 1$

$b + 4 d = 2$

$b = - 4 d + 2$

$2 b + 5 d = 4$

$\Rightarrow$ $- 8 d + 4 + 5 d = 4$

$\Rightarrow - 3 d = 0$

$\Rightarrow d = 0$

$b = - 4 d + 2$

$\Rightarrow b = - 4 \times 0 + 2 = 2$

Hence, we have $a = 1, b = 2, c = - 2, d = 0$

Matrix X is $[\begin{matrix} 1 & - 2 \\ 2 & 0 \end{matrix}]$ .

Question 9. Choose the correct answer in the following questions:

If $A = [\begin{matrix} α & β \\ γ & - α \end{matrix}]$ is such that $A^{2} = I$

(A) $1 + α^{2} + β γ = 0$

(B) $1 - α^{2} + β γ = 0$

(D) $1 + α^{2} - β γ = 0$

Answer:

$A = [\begin{matrix} α & β \\ γ & - α \end{matrix}]$

$A^{2} = I$

$[\begin{matrix} α & β \\ γ & - α \end{matrix}]$ $[\begin{matrix} α & β \\ γ & - α \end{matrix}]$ $= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} α^{2} + β γ & α β - α β \\ α γ - α γ & β γ + α^{2} \end{matrix}]$ $= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} α^{2} + β γ & 0 \\ 0 & β γ + α^{2} \end{matrix}]$ $= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

Thus we obtained that

$α^{2} + β γ = 1$

$\Rightarrow 1 - α^{2} - β γ = 0$

Option C is correct.

Question 10. If the matrix A is both symmetric and skew-symmetric, then

(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these

Answer:

If the matrix A is both symmetric and skew-symmetric, then

$A^{'} = A$ and $A^{'} = - A$

$A^{'} = A^{'}$

$\Rightarrow A = - A$

$\Rightarrow A + A = 0$

$\Rightarrow 2 A = 0$

$\Rightarrow A = 0$

Hence, A is a zero matrix.

Option B is correct.

Question 11. If A is square matrix such that $A^{2} = A$ , then $(I + A)^{3} - 7 A$ is equal to

(A) A
(B) I – A
(C) I
(D) 3A

Answer:

A is a square matrix such that $A^{2} = A$

$(I + A)^{3} - 7 A$

$= I^{3} + A^{3} + 3 I^{2} A + 3 I A^{2} - 7 A$

$= I + A^{2} . A + 3 A + 3 A^{2} - 7 A$

$= I + A . A + 3 A + 3 A - 7 A$ (Replace $A^{2}$ by $A$ )

$= I + A^{2} + 6 A - 7 A$

$= I + A - A$

$= I$

Hence, we have $(I + A)^{3} - 7 A = I$

Option C is correct.

If you are interested in Matrices Class 12 NCERT Solutions exercises then these are listed below.

Matrices Class 12 NCERT Solutions Exercise 3.1

Matrices Class 12 NCERT Solutions Exercise 3.2

Matrices Class 12 NCERT Solutions Exercise 3.3

Matrices Class 12 NCERT Solutions Exercise 3.4

Matrices Class 12 NCERT Solutions Miscellaneous Exercise

NCERT solutions for class 12 Maths: Chapter-wise

NCERT Solutions for Class 12 Mathematics Chapter 1: Relations and Functions

NCERT Solutions for Class 12 Mathematics Chapter 2: Inverse Trigonometric Functions

NCERT Solutions for Class 12 Mathematics Chapter 3: Matrices

NCERT Solutions for Class 12 Mathematics Chapter 4: Determinants

NCERT Solutions for Class 12 Mathematics Chapter 5: Differentiability

NCERT Solutions for Class 12 Mathematics Chapter 6: Application of Derivatives

NCERT Solutions for Class 12 Mathematics Chapter 7: Integrals

NCERT Solutions for Class 12 Mathematics Chapter 8: Application of Integrals

NCERT Solutions for Class 12 Mathematics Chapter 9: Differential Equations

NCERT Solutions for Class 12 Mathematics Chapter 10: Vector Algebra

NCERT Solutions for Class 12 Mathematics Chapter 11: Three Dimensional Geometry

NCERT Solutions for Class 12 Mathematics Chapter 12: Linear Programming

NCERT Solutions for Class 12 Mathematics Chapter 13: Probability

Importance of solving NCERT questions for class 12 Math Chapter 3

Learning the formulae of Matrices is not enough if students do not know where and how to use them. For that, solving the NCERT questions is very important. Here are some other crucial points for solving these questions.

Solving various types of problems will give students conceptual clarity of matrix operations such as addition, subtraction, multiplication and inverse.
It will also build a strong foundation for higher mathematics topics like determinants, linear algebra and vector spaces.
The difficulty level of NCERT exercises are from easy to higher level. So students can gradually increase their learning.
Regular practice of these problems will also help students increase their accuracy as well as save crucial time during the exam.

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and NCERT syllabus for class 12

NCERT solutions for class 12 - subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Solutions - Class Wise

Given below are the class-wise solutions of class 12 NCERT:

NCERT Exemplar solutions for class 12 - subject-wise

Given below are the subject-wise exemplar solutions of class 12 NCERT:

Happy learning !!!

Frequently Asked Questions (FAQs)

1. What are the important topics in Class 12 Maths Chapter 3 - Matrices?

The topics covered in matrices for class 12 include the following topics:

Introduction
Matrix
Types of Matrices
Operations on Matrices
Transpose of a Matrix

2. What is the adjoint of a matrix, and how is it used to find the inverse?

The adjoint of a matrix is the transpose of its cofactor matrix, and it's used to find the inverse of a matrix by dividing the adjoint by the determinant of the original matrix. The inverse matrix is also found using the following equation:
$A^{- 1} = adj (A) / \det (A),$
where $adj (A)$ refers to the adjoint of a matrix $A, \det (A)$ refers to the determinant of a matrix $A$ .

3. What is the rank of a matrix, and how is it calculated?

The rank of a matrix is equal to the number of linearly independent rows or columns in it. It cannot be more than its number of rows and columns. To find the rank of a matrix, we can transform the matrix to its row echelon form and count the number of non-zero rows.

4. How can we find the inverse of a matrix using elementary transformations?

To find the inverse of a matrix A using elementary transformations, we can use elementary row operations on A = IA, in a sequence, until we get I = BA. We can also use elementary column operations on A = AI, in a sequence, till we get I = AB. If the inverse of matrix A exists, we can write A = IA and apply a sequence of row operations till we get an identity matrix on the LHS and use the same elementary operations on the RHS to get I = BA

5. What is the difference between symmetric and skew-symmetric matrices?

A square matrix A is said to be symmetric if a_ij = a_j_i for all i and j, where a_ij is an element present at (i,j)^th position (i^th row and j^t^h column in matrix A) and a_ji is an element present at (j,i)^th position (j^th row and i^thcolumn in matrix A) whereas square matrix A is said to be skew-symmetric if a_ij =−a_ji for all i and j. In other words, we can say that matrix A is said to be skew-symmetric if the transpose of matrix A is equal to the negative of matrix A i.e (A^T =−A)

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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.