NCERT Solutions for Class 12 Maths Chapter 3 Matrices

A + B

Answer:

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

(i) A + B

The addition of a matrix can be done as follows

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

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

$A+B = \begin{bmatrix} 3 &7 \\ 1 & 7 \end{bmatrix}$

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

A - B

Answer:

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

(ii) A - B

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

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

$A-B = \begin{bmatrix} 1 &1 \\ 5 & -3 \end{bmatrix}$

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

3A - C

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

(iii) 3A - C

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

$3A -C = 3\begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

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

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

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

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

Answer:

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

(iv) AB

$AB = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $\times \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$AB = \begin{bmatrix} 2\times 1+4\times -2 & \, \, \, 2\times 3+4\times 5 \\ 3\times 1+2\times -2 & \, \, \, 3\times 3+2 \times 5 \end{bmatrix}$

$AB = \begin{bmatrix} -6 &26 \\ -1 & 19 \end{bmatrix}$

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

$\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}$

Answer:

The multiplication is performed as follows

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

$BA = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ $\times \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$

$BA = \begin{bmatrix} 1\times 2+3\times 3 &1\times 4+3\times 2 \\ -2\times 2+5\times 3& -2\times 4+2\times 5 \end{bmatrix}$

$BA = \begin{bmatrix} 11 &10 \\ 11& 2 \end{bmatrix}$

Question 2(i): Compute the following:

Answer:

(i) $\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}$

$= \begin{bmatrix} a+a &b+b \\ -b+b & a+a \end{bmatrix}$

$= \begin{bmatrix} 2a &2b \\ 0 & 2a \end{bmatrix}$

Question 2(ii): Compute the following:

$\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}$

Answer:

(ii) The addition operation can be performed as follows

$\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}$

$=\begin{bmatrix} a^2 + b^2+2ab& b^2+c^2+2bc\\ a^2 + c^2-2ac& a^2 + b^2-2ab \end{bmatrix}$

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

Question 2(iii): Compute the following:

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

Answer:

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

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

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

$=\begin{bmatrix} 11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix}$

Question 2(iv): Compute the following:

$\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}$

Answer:

(iv) the addition is done as follows

$\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}$

$=\begin{bmatrix} \cos^2+ \sin^2 x &\sin^2 x+\cos^2 x\\ \sin^2 x+\cos^2 x & \cos^2x+ \sin^2 x \end{bmatrix}$ since $sin^2x+cos^2x=1$

$=\begin{bmatrix} 1 &1\\ 1 & 1 \end{bmatrix}$

Question 3(i): Compute the indicated products.

$\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

Answer:

(i) The multiplication is performed as follows

$\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

$=\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \times \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

$=\begin{bmatrix} a\times a+b\times b &a\times -b+b\times a \\ -b\times a+a\times b &-b\times -b+a\times a \end{bmatrix}$

$=\begin{bmatrix} a^{2}+b^{2} & 0 \\ 0 & b^{2}+a^{2} \end{bmatrix}$

Question 3(ii): Compute the indicated products.

$\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}$

Answer:

(ii) the multiplication can be performed as follows

$\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}$

$=\begin{bmatrix} 1\times 2 &1\times 3&1\times 4\\ 2\times 2&2\times 3&2\times 4\\3\times 2&3\times 3&3\times 4 \end{bmatrix}$

$=\begin{bmatrix} 2 &3& 4\\ 4&6&8\\6&9&12 \end{bmatrix}$

Question 3(iii): Compute the indicated products.

$\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}$

Answer:

(iii) The multiplication can be performed as follows

$\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}$

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

Question 3(iv): Compute the indicated products.

$\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

Answer:

(iv) The multiplication is performed as follows

$\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

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

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

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

Question 3(v): Compute the indicated products.

$\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

Answer:

(v) The product can be computed as follows

$\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

$=\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\times \begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

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

$=\begin{bmatrix} 1 &2&3 \\ 1 & 4&5\\ -2 & 2&0 \end{bmatrix}$

Question 3(vi): Compute the indicated products.

$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

Answer:

(vi) The given product can be computed as follows

$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 \times 2+(-1)\times 1+3\times 3\, \, \, & 3 \times (-3)+(-1)\times 0+3\times 1 \\ (-1) \times 2+ 0 \times 1+2\times 3 \, \, \, & (-1) \times -3+0\times 0+2\times 1 \end{bmatrix}$

$=\begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

$3A-5B = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$3A-5B = 0$

Question 6: Simplify $\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$.

Answer:

The simplification is explained in the following step

$\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^{2}\theta & \sin\theta \cos\theta \\ -\sin\theta \cos\theta & \cos^{2}\theta \end{bmatrix} +\begin{bmatrix} \sin^{2}\theta & - \sin\theta \cos\theta\\ \sin\theta\cos\theta & \sin^{2}\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^{2}\theta+\sin^{2}\theta & \sin\theta \cos\theta - \sin\theta \cos\theta \\ -\sin\theta \cos\theta + \sin\theta \cos\theta & \cos^{2}\theta + \sin^{2}\theta\end{bmatrix}$

$= \begin{bmatrix} 1&0 \\ 0 & 1\end{bmatrix} =I$

the final answer is an identity matrix of order 2

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

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

Answer:

(i) The given matrices are

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

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

$X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}.............................2$

Adding equation 1 and 2, we get

$2 X = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ $+ \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

$2 X = \begin{bmatrix} 7+3 &0+0 \\ 2+0 &5+3 \end{bmatrix}$

$2 X = \begin{bmatrix} 10 &0 \\ 2 &8 \end{bmatrix}$

$X = \begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$

Putting the value of X in equation 1, we get

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

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

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

$Y = \begin{bmatrix} 2 &0 \\ 1 &1 \end{bmatrix}$

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

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

Answer:

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

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

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

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

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

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

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

$5Y= \begin{bmatrix} 2 &13 \\ 14 & -10 \end{bmatrix}$

$Y= \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}$

Putting value of Y in equation 1 , we get

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

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

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

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

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

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

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

Question 8: Find X, if $Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$ and $2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

Answer:

$Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$

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

Substituting the value of Y in the above equation

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

$2X = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}- \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$

$2X = \begin{bmatrix} 1-3 &0-2 \\ -3-1 & 2-4 \end{bmatrix}$

$2X = \begin{bmatrix} -2 &-2 \\ -4 & -2 \end{bmatrix}$

$X = \begin{bmatrix} -1 &-1 \\ -2 & -1 \end{bmatrix}$

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

Answer:

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

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

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

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

Now equating LHS and RHS we can write the following equations

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

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

$y=3$ $2x=6$

$x=3$

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

Answer:

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

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

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

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

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

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

Dividing by 2 on both sides

$\begin{bmatrix}x & z \\ y &t \end{bmatrix} = \begin{bmatrix} 3 &9\\ 6 & 6 \end{bmatrix}$

$x=3,y=6,z=9\, \, and\, \, t=6$

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

Answer:

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

$\begin{bmatrix}2x\\3x \end{bmatrix} + \begin{bmatrix} -y\\y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

Adding both the matrices in the LHS and rewriting

$\begin{bmatrix}2x-y\\3x+y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

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

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

Adding equations 1 and 2, we get

$5x=15$

$x=3$

Put the value of x in equation 2, we have

$3x+y=5$

$3\times 3+y=5$

$9+y=5$

$y=5-9$

$y=-4$

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

Answer:

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

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

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

Thus, we have

$3x=x+4$

$3x-x=4$

$2x=4$

$x=2$

$3y=6+x+y$

Put the value of x

$3y-y=6+2$

$2y=8$

$y=4$

$3w=2w+3$

$3w-2w=3$

$w=3$

$3z=-1+z+w$

$3z-z=-1+3$

$2z=2$

$z=1$

Hence, we have $x=2,y=4,z=1\, \, and\, \, w=3.$

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

Answer:

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

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

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

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

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

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

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

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

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

Question 14(i): Show that

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

Answer:

To prove:

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

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

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

$= \begin{bmatrix}7 &1\\33&34 \end{bmatrix}$

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

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

$= \begin{bmatrix} 16 & 5\\ 39 & 25 \end{bmatrix}$

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

Question 14(ii): Show that

$\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

Answer:

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

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

$= \begin{bmatrix}1\times(-1)+2\times 0+3\times 2 \, \, \, & 1\times(1)+2\times (-1)+3\times 3\, \, \, &1\times(0)+2\times 1+3\times 4\\0\times(-1)+1\times 0+0\times 2\, \, \, &0\times(1)+1\times (-1)+0\times 3\, \, \, &0\times(0)+1\times 1+0\times 4\\1\times(-1)+1\times 0+0\times 2\, \, \, &1\times(1)+1\times (-1)+0\times 3\, \, \, &1\times(0)+1\times 1+0\times 4 \end{bmatrix}$

$= \begin{bmatrix}5& 8&14\\0&-1&1\\-1&0&1\end{bmatrix}$

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

$= \begin{bmatrix}-1\times(1)+1\times 0+0\times 1 \, \, \, & -1\times(2)+1\times (1)+0\times 1\, \, \, &-1\times(3)+1\times 0+0\times 0\\0\times(1)+-(1)\times 0+1\times 1\, \, \, &0\times(2)+(-1)\times (1)+1\times 1\, \, \, &0\times(3)+(-1)\times 0+1\times 0\\2\times(1)+3\times 0+4\times 1\, \, \, &2\times(2)+3\times (1)+4\times 1\, \, \, &2\times(3)+3\times 0+4\times 0 \end{bmatrix}$

$= \begin{bmatrix}-1& -1&-3\\1&0&0\\6&11&6\end{bmatrix}$

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

Question 15: Find$A^2 -5A + 6I$, if

$A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

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

$A\times A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}\times \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

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

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

$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$\therefore$ $A^2 -5A + 6I$

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

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

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

$= \begin{bmatrix} 1 & -1 & -3\\ -1 & -1 &-10 \\ -5 & 4 & 4 \end{bmatrix}$

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

Answer:

$A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

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

$A\times A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$$\times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

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

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

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

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

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

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

$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$\therefore$ $A^3 - 6A^2 + 7A + 2I = 0$

L.H.S :

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

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

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

$=\begin{bmatrix}30-30&0&48-48\\12-12&24-24&30-30\\48-48&0&78-78 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 &0 \\ 0 & 0 & 0 \end{bmatrix}=0$

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

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

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

Answer:

$A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$A \times A= \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$$\times \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

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

$A^{2} = \begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}$

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

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$$k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$$2 \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$$k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$$\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$

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

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

$\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ $=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

We have,$3=3k$

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

Hence, the value of k is 1.

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

Answer:

$A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

To prove : $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

L.H.S : $I+A$

$I+A = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$$+ \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$

$I+A = \begin{bmatrix} 1+0&0-\tan\frac{\alpha}{2}\\0+\tan\frac{\alpha}{2}&1+ 0\end{bmatrix}$

$I+A = \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}$

R.H.S : $(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$$= (\begin{bmatrix}1 &0\\0&1 \end{bmatrix}-$ $\begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix})$$\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$ $=\begin{bmatrix} 1-0&0-(-\tan\frac{\alpha}{2})\\0-\tan\frac{\alpha}{2}&1- 0\end{bmatrix}$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$$=\begin{bmatrix} 1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix}$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$=\begin{bmatrix} \cos\alpha + \sin \alpha\tan\frac{\alpha}{2} &- \sin \alpha+ \cos \alpha \tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} \cos\alpha + \sin \alpha &\tan\frac{\alpha}{2} \sin\alpha + \cos \alpha \end{bmatrix}$

$=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}\tan\frac{\alpha}{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ (2\cos^{2} \frac{\alpha }{2} -1)\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} (2\cos^{2} \frac{\alpha }{2} -1) + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} &\tan\frac{\alpha}{2} 2\sin\frac{\alpha } {2} \ cos \frac{\alpha }{2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}$

$=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin^{2}\frac{\alpha }{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} -\tan\frac{\alpha}{2}\\-2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+\tan\frac{\alpha}{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} & 2\sin^{2}\frac{\alpha } {2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}$

$= \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}$

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

i.e. $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$.

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

$\begin{bmatrix}x &(30000-x) \end{bmatrix}$ $\begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix}$ $=1800$ (simple interest for 1 year $=\frac{pricipal\times rate}{100}$ )

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

$5x+210000-7x=180000$

$210000-180000=7x-5x$

$30000=2x$

$x=15000$

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

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

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

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

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

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

$210000 - 2x = 200000$

$210000 - 200000 = 2x$

$10000 = 2x$

$x = 5000$

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{bmatrix}10 &8&10 \end{bmatrix}$ $\begin{bmatrix}80\\60\\40 \end{bmatrix}$

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

$= 12(800+480+ 400)$

$= 12(1680)$

$=20160$

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 restrictions on $n, k$ and $p$ so that $P Y+W Y$ will be defined are:
(A) $k=3, p=n$
(B) k is arbitrary, $p=2$
(C) p is arbitrary, $k=3$
(D) $k=2, p=3$

Answer:
P and Y are of order $p \times k$ and $\mathbf{3} \times k$ 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 $7X - 5Z$ is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

Answer:

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

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

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

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

Matrices $7X$ and $5Z$ 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 $7X - 5Z$ is $2 \times n$.

Option (B) is correct.

Matrices Class 12 Chapter 3 Question Answers
Exercise: 3.3
Page number: 66-68
Total questions: 12

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

$\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}$

The transpose of the given matrix is

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

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

$\begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}$

Interchanging the rows and columns of the matrix A, we get

$A^{T}=\begin{bmatrix} 1 & 2\\ -1 & 3 \end{bmatrix}$

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

$\begin{bmatrix} -1 & 5 & 6\\ \sqrt3& 5 &6 \\ 2 &3 &-1 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} -1 & 5 & 6\\ \sqrt3& 5 &6 \\ 2 &3 &-1 \end{bmatrix}$

Transpose is obtained by interchanging the rows and columns of a matrix

$A^{T} = \begin{bmatrix} -1 & \sqrt3 & 2\\ 5& 5 &3 \\ 6 &6 &-1 \end{bmatrix}$

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

$(A + B)' = A' + B'$

Answer:

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

$(A + B)' = A' + B'$

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

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

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

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

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

R.H.S : $A' + B'$

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

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

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

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

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

$(A - B)' = A' - B'$

Answer:

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

$(A - B)' = A' - B'$

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

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

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

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

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

R.H.S : $A' - B'$

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

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

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

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

$(A - B)' = A' - B'$.

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

$(A + B)' = A' + B'$

Answer:

$A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$

To prove: $(A + B)' = A' + B'$

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

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

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

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

$\therefore \, \, \, (A+B)' = \begin{bmatrix} 2 & 5\\ 1 &4\\1 & 4 \end{bmatrix}$

R.H.S: $A' + B'$

$A'+B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $+ \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}$

$A'+B' = \begin{bmatrix} 2 & 5\\ 1 &4 \\ 1 & 4 \end{bmatrix}$

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

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

$(A - B)' = A' - B'$

Answer:

$A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}$

To prove: $(A - B)' = A' - B'$

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

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

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

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

$\therefore \, \, \, (A-B)' = \begin{bmatrix} 4 & 3\\ -3 &0\\-1 & -2 \end{bmatrix}$

R.H.S: $A' - B'$

$A'-B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}$ $- \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}$

$A'-B' = \begin{bmatrix} 4 & 3\\ -3 &0 \\ -1 & -2 \end{bmatrix}$

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

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

Answer:

$B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$

$A' = \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}$

$A=(A')' = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}$

$(A + 2B)'$ :

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

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

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

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

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

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

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

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

Answer:

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

To prove : $(AB)' = B'A'$

$L.H.S : (AB)'$

$AB = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix}$$\begin{bmatrix} -1& 2 &1 \end{bmatrix}$

$AB = \begin{bmatrix} -1&2&1\\4&-8&-4 \\-3 &6&3\end{bmatrix}$

$(AB)' = \begin{bmatrix} -1&4&-3\\2&-8&6 \\1 &-4&3\end{bmatrix}$

$R.H.S : B'A'$

$B' = \begin{bmatrix} -1\\2 \\1 \end{bmatrix}$

$A' = \begin{bmatrix} 1& -4 &3 \end{bmatrix}$

$B'A' = \begin{bmatrix} -1\\2 \\1 \end{bmatrix}$$\begin{bmatrix} 1& -4 &3 \end{bmatrix}$

$B'A' = \begin{bmatrix} -1&4&-3\\2&-8&6 \\1&-4&3 \end{bmatrix}$

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

so it is verified that $(AB)' = B'A'$.

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

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

Answer:

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

To prove : $(AB)' = B'A'$

$L.H.S : (AB)'$

$AB = \begin{bmatrix} 0\\1 \\2 \end{bmatrix}$$\begin{bmatrix} 1& 5 &7 \end{bmatrix}$

$AB = \begin{bmatrix} 0&0&0\\1&5&7 \\2 &10&14\end{bmatrix}$

$(AB)' = \begin{bmatrix} 0&1&2\\0&5&10 \\0 &7&14\end{bmatrix}$

$R.H.S : B'A'$

$B' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix}$

$A' = \begin{bmatrix} 0& 1 &2 \end{bmatrix}$

$B'A' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix}$$\begin{bmatrix} 0& 1 &2 \end{bmatrix}$

$B'A' = \begin{bmatrix} 0&1&2\\0&5&10 \\0&7&14 \end{bmatrix}$

Hence, L.H.S =R.H.S i.e.$(AB)' = B'A'$.

Question 6(i). If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$, then verify that $A'A =I$

Answer:

$A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

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

$A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$

To prove: $A'A =I$

L.H.S :$A'A$

$A'A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ $\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} \cos^{2}\alpha + \sin ^{2}\alpha & \sin\alpha \cos \alpha - \sin \alpha \ cos \alpha \\ \sin\alpha \cos \alpha - \sin \alpha \cos \alpha & \ sin^{2}\alpha +\cos^{2}\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S$

Question 6(ii). If $A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$, then verify that $A'A = I$

Answer:

$A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$

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

$A' = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$

To prove: $A'A =I$

L.H.S :$A'A$

$A'A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$ $\begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S$

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

Answer:

$A = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$

The transpose of A is

$A' = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}$

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

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

Answer:

$A = \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$

The transpose of A is

$A' = \begin{bmatrix} 0 & -1 & 1\\ 1 & 0 &-1 \\- 1 & 1 &0 \end{bmatrix}$

$A' =- \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}$

$A' =- A$

Since $ A'=-A$ so given matrix is a skew-symmetric matrix.

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

$(A + A')$ is a symmetric matrix.

Answer:

$A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$

$A' = \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A + A'= \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$ $+ \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A + A'= \begin{bmatrix} 1+1 & 5+6\\ 6+5 & 7+7 \end{bmatrix}$

$A + A'= \begin{bmatrix}2 & 11\\ 11& 14 \end{bmatrix}$

$(A + A')'= \begin{bmatrix}2 & 11\\ 11& 14 \end{bmatrix}$

We have $A+A'=(A + A')'$

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

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

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

Answer:

$A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$

$A' = \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A - A'= \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}$ $- \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}$

$A - A'= \begin{bmatrix} 1-1 & 5-6\\ 6-5 & 7-7 \end{bmatrix}$

$A - A'= \begin{bmatrix}0 & -1\\ 1& 0 \end{bmatrix}$

$(A - A')'= \begin{bmatrix}0 & 1\\ -1& 0 \end{bmatrix}=-(A-A')$

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 $A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$

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

$A' = \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix}$

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

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

$\frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2}(A+A') = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

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

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

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

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

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

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

$\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$

$A'=\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A+A'=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$$+\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & 6\\ 6 & -2 \end{bmatrix}$

Let

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

$B'=\begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}=B$

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

$A-A'=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}$$-\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix}$$= \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix}$

$C'= \begin{bmatrix} 0 & -2\\ 2 & 0 \end{bmatrix}$

$C=-C'$

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

Represent A as the sum of B and C.

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

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

$\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$$+ \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A+A'=\begin{bmatrix} 12 & -4 & 4\\ -4 & 6 & -2\\ 4 & -2 & 6 \end{bmatrix}$

Let

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

$B'= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}=B$

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

$A-A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$$- \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 0&0\\ 0 & 0&0 \\0&0&0\end{bmatrix}$

Let

$C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$$=\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$

$C'=\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}$

$C=-C'$

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

Represent A as the sum of B and C.

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

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

$\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$

Answer:

$A=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$

$A'=\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$$+\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 6 & 1 & -5\\ 1& -4 & -4\\ -5 & -4 & 4 \end{bmatrix}$

Let

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

$B'= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}=B$

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

$A-A'=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}$$-\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 5&3\\ -5 & 0&6 \\-3&-6&0\end{bmatrix}$

Let

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

$C'=\begin{bmatrix} 0 &- \frac{5}{2}&-\frac{3}{2}\\ \frac{5}{2}&0 &- 3\\\frac{3}{2}&3&0 \end{bmatrix}$

$C=-C'$

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

Represent A as the sum of B and C.

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

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

$\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$

Answer:

$A =\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$

$A'=\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$$+\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A+A'=\begin{bmatrix} 2 & 4\\ 4 & 4 \end{bmatrix}$

Let

$B=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 2 & 4\\ 4 & 4 \end{bmatrix}$$=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}$

$B'=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}=B$

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

$A-A'=\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}$$-\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}$

$A-A'=\begin{bmatrix} 0 & 6\\ -6 & 0 \end{bmatrix}$

Let

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

$C'= \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix}$

$C=-C'$

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

Represent A as the sum of B and C.

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

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

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

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

Answer:

If A and B are symmetric matrices, then

$A'=A$ and $B' = B$

we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$

$=BA-AB$

$= -(AB-BA)$

Hence, we have $(AB-BA) = -(AB-BA)'$

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

Option A is correct.

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

If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$ and $A+A' =I$, then the value of $\alpha$ is

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{3}$

(D) $\frac{3\pi}{2}$

Answer:

$A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$

$A' = \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$

$A+A' = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$$+ \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$$= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$

$A+A' = \begin{bmatrix} 2\cos\alpha & 0\\ 0 & 2\cos\alpha \end{bmatrix}$$= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$

$2 cos \alpha=1$

$cos \alpha=\frac{1}{2}$

$\alpha=\frac{\pi}{3}$

Option B is correct.

Matrices Class 12 Chapter 3 Question Answers
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) AB=BA

(B) AB=BA=0

(D) AB=BA=I

Answer:

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

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

Hence, matrices A and B will be inverse of each other only if $AB=BA=I$.

Option D is correct.

Matrices Class 12 Chapter 3 Question Answers
Miscellaneous Exercise
Page number: 72-73
Total questions: 11

Question:1 Let $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$, show that $(aI + bA)^n = a^n I + na^{n-1} bA$, where I is the identity matrix of order 2 and $n \in N$.

Answer:

Given :

$A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

To prove : $(aI + bA)^n = a^n I + na^{n-1} bA$

For n=1, $aI + bA = a I + a^{0} bA =a I + bA$

The result is true for n=1.

Let the result be true for n=k,

$(aI + bA)^k = a^k I + ka^{k-1} bA$

Now, we prove that the result is true for n=k+1,

$(aI + bA)^{k+1} = (aI + bA)^k (aI + bA)$

$= (a^k I + ka^{k-1} bA)$$(aI + bA)$

$=a^{k+1}I+Ka^{k}bAI+a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$A^{2} = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}=0$

Put the value of $A^{2}$ in above equation,

$(aI + bA)^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$(aI + bA)^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+0$

$=a^{k+1}I+(k+1)a^{k}bAI$

Hence, the result is true for n=k+1.

Thus, we have $(aI + bA)^n = a^n I + na^{n-1} bA$ where $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$,$n \in N$.

Question 2. If $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$ then show that $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$, $n\in N$.

Answer:

Given :

$A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$

To prove:

$A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$

For n=1, we have

$A^1 =\begin{bmatrix} 3^{1-1} & 3^{1-1} &3^{1-1} \\ 3^{1-1}& 3^{1-1} & 3^{1-1}\\ 3^{1-1} & 3^{1-1}& 3^{1-1} \end{bmatrix}$$=\begin{bmatrix} 3^{0} & 3^{0} &3^{0} \\ 3^{0}& 3^{0} & 3^{0}\\ 3^{0} & 3^{0}& 3^{0} \end{bmatrix}$$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}=A$

Thus, the result is true for n=1.

Now, take n=k,

$A^k =\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

For n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$$\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix}3. 3^{k-1} & 3.3^{k-1} &3.3^{k-1} \\3. 3^{k-1}& 3.3^{k-1} & 3.3^{k-1}\\3. 3^{k-1} & 3.3^{k-1}&3. 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix} 3^{(K+1)-1} &3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1}&3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1} & 3^{(K+1)-1}& 3^{(K+1)-1}\end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$, $n\in N$ where $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$.

Question 3. If $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$, then prove that $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$, where n is any positive integer.

Answer:

Given :

$A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$

To prove:

$A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$

For n=1, we have

$A^1 = \begin{bmatrix} 1+2\times 1 & -4\times 1\\ 1 & 1-2\times 1 \end{bmatrix}$$= \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}=A$

Thus, the result is true for n=1.

Now, the result is true for n=k,

$A^k = \begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

For n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$$\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

$=\begin{bmatrix} 3(1+2k)-4k & -12k-4(1-2k)\\ (1+2k)-k &-4k-(1-2k) \end{bmatrix}$

$=\begin{bmatrix} 3+6k-4k & -12k-4k+8k\\ 1+k &-4k-1+2k \end{bmatrix}$

$=\begin{bmatrix} 3+2k & -4k-4k\\ 1+k &-2k-1 \end{bmatrix}$

$=\begin{bmatrix} 1+2(k+1)& -4(k+1)\\ 1+k &1-2(k+1) \end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$, where $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$.

Question 4. If A and B are symmetric matrices, prove that $AB - BA$ is a skew-symmetric matrix.

Answer:

If A and B are symmetric matrices, then

$A'=A$ and $B' = B$

we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$

$=BA-AB$

$= -(AB-BA)$

Hence, we have $(AB-BA) = -(AB-BA)'$

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

Question 5. Show that the matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.

Answer:

Let be a symmetric matrix, then $A'=A$

Consider, $(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by $A$

$=B'(AB)$

i.e. $(B'AB)'$ $=B'(AB)$

Thus, if A is a symmetric matrix than $B'(AB)$ is a symmetric matrix.

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

$(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by -$A$,

$=B'(-AB)$

$= - B'AB$

i.e. $(B'AB)'$ $= - B'AB$.

Thus, if A is a skew-symmetric matrix, then $- B'AB$ is a skew-symmetric matrix.

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

Question 6. Find the values of x, y, z if the matrix $A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$ satisfy the equation $A'A = I$

Answer:

$A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$

$A' = \begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$

$A'A = I$

$\begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$$\begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

$\begin{bmatrix} x^{2}+x^{2} & xy-xy& -xz+xz\\ xy-xy& 4y^{2}+y^{2}+y^{2} & 2yz-yz-yz\\ -zx+zx & 2yz-yz-yz &z^{2}+z^{2}+z^{2}\end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

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

Thus equating the terms element-wise

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

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

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

Question 7. For what values of x: $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$?

Answer:

$\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 6& 2& 4 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 0+4+4x \end{bmatrix} = O$

$4+4x=0$

$4x=-4$

$x=-1$

Thus, the value of x is -1.

Question 8. If $A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 -5A + 7I= 0$.

Answer:

$A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$$\begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

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

$A^{2} = \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$

$I= \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

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

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

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

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

$=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix} =0=R.H.S$

Hence, we proved that

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

Question 9. Find x, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$.

Answer:

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

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

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

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

$\begin{bmatrix}x ^{2}-2x-40+2x-8\end{bmatrix} = 0$

$\therefore \, \, x ^{2}-48= 0$

$x ^{2}=48$

thus the value of x is

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

Question 10(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 :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

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

$= 25000+3000+18000$

$= 46000$

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

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

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

$= 15000+30000+8000$

$= 53000$

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

Question 10(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 :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

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

$= 20000+2000+9000$

$= 31000$

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

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

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

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

$= 12000+20000+4000$

$= 36000$

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

Question 11. Find the matrix X so that $X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

Answer:

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

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 $\begin{bmatrix} a & c\\ b & d \end{bmatrix}$

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

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

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

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

Taking, $a+4c=-7$

$a=-4c-7$

$2a+5c=-8$

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

$-3c=6$

$c=-2$

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

$a=8-7=1$

$b+4d=2$

$b=-4d+2$

$2b+5d=4$

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

$\Rightarrow -3d=0$

$\Rightarrow d=0$

$b=-4d+2$

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

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

Matrix X is $\begin{bmatrix} 1 & -2\\ 2 & 0 \end{bmatrix}$.

Question 12. If A and B are square matrices of the same order such that $AB = BA$, then prove by induction that $AB^n = B^n A$. Further, prove that $(AB)^n = A^n B^n$for all $n \in N$.

Answer:

A and B are square matrices of the same order such that $AB = BA$,

To prove : $AB^n = B^n A$, $n \in N$

For n=1, we have $AB^1 = B^1 A$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $AB^k = B^k A$

Now, taking n=k+1 , we have $AB^{k+1} = AB^k .B$

$AB^{k+1} = (B^kA) .B$

$AB^{k+1} = (B^k) .AB$

$AB^{k+1} = (B^k) .BA$

$AB^{k+1} = (B^k.B) .A$

$AB^{k+1} = (B^{k+1}) .A$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$, $n \in N$.

To prove: $(AB)^n = A^n B^n$

For n=1, we have $(AB)^1 = A^1 B^1$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $(AB)^k = A^k B^k$

Now, taking n=k+1 , we have $(AB)^{k+1} = (A B)^k.(AB)$

$(AB)^{k+1} = A^k B^k.(AB)$

$(AB)^{k+1} = A^{K}( B^kA)B$

$(AB)^{k+1} = A^{K}( AB^k)B$

$(AB)^{k+1} = (A^{K}A)(B^kB)$

$(AB)^{k+1} = (A^{k+1})(B^{k+1})$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$ and $(AB)^n = A^n B^n$for all $n \in N$.

Question 13 Choose the correct answer in the following questions:

If $A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ is such that $A^2 = I$

(A) $1 + \alpha^2 + \beta \gamma = 0$

(B) $1 - \alpha^2 + \beta \gamma = 0$

(D) $1 + \alpha^2 - \beta \gamma = 0$

Answer:

$A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$

$A^2 = I$

$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&\alpha \beta-\alpha \beta\\\alpha \gamma-\alpha \gamma&\beta \gamma+\alpha^{2} \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&0\\0&\beta \gamma+\alpha^{2} \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

Thus, we obtained that

$\alpha^{2} +\beta \gamma=1$

$\Rightarrow 1-\alpha^{2} -\beta \gamma=0$

Option C is correct.

Question 14. 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 $A'=-A$

$A'=A'$

$\Rightarrow \, \, \, \, \, \, \, A=-A$

$\Rightarrow \, \, \, \, \, \, \, A+A=0$

$\Rightarrow \, \, \, \, \, \, \, 2A=0$

$\Rightarrow \, \, \, \, \, \, \, A=0$

Hence, A is a zero matrix.

Option B is correct.

Question 15. 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 $A^{2}=A$

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

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

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

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

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

$=I+A-A$

$=I$

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

Option C is correct.

Matrices Class 12 NCERT Solutions: Exercise-wise

Exercise-wise NCERT Solutions of Matrices Class 12 Maths Chapter 3 are provided in the links below.

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Matrices Class 12 NCERT Solutions: Topics

Topics you will learn in NCERT Class 12 Maths Chapter 3 Matrices include:

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Class 12 Maths Chapter 3 Solutions - Important Formulae

Matrix Definition and Properties:

A matrix is an ordered rectangular array of numbers or functions.

A matrix of order m × n consists of m rows and n columns.

The order of a matrix is written as m × n, where m is the number of rows and n is the number of columns.

A matrix is called a square matrix when m = n.

A diagonal matrix A = [a_ij]_m×m has a_ij = 0 when i ≠ j.

A scalar matrix A = [a_ij]_n×n has a_ij = 0 when i ≠ j, aij = k (where k is a constant)

when i = j.

An identity matrix A = [a_ij]n×n has a_ij = 1 when i = j and a_ij = 0 when i ≠ j.

A zero matrix contains all its elements as zero.

A column matrix is of the form [A]_{n × 1}.

A row matrix is of the form [A]_{1 × n}.

Equality of Matrices:

Two matrices A and B are equal (A = B) if they have the same order and a_ij = b_ij for all the corresponding values of i and j.

Operations on Matrices:

Matrix Addition:

If A = [a_ij]_{m × n} and B = [b_ij]_{m × n}, then A + B = [a_ij + b_ij]m × n.

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NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Matrix Subtraction:

If A = [a_ij]_{m × n} and B = [b_ij]_{m × n}, then A - B = [a_ij - b_ij]_{m × n}.

Multiplication of a Matrix by a Scalar:

Let A = [a_ij]_{m × n} be a matrix and k is a scalar, then kA is obtained by multiplying each element of A by the scalar k, i.e., kA = [ka_ij]_{m × n}.

Multiplication of Matrices:

Let A be an m × p matrix, and B be a p × n matrix. Their product AB is defined if the number of columns in A is equal to the number of rows in B. The resulting matrix is an m × n matrix, and the elements are calculated as follows: (AB)_ij = Σ(a_i * b_j), where the sum is taken over all values of p.

Transpose of a Matrix:

The transpose of a matrix A, denoted as AT, is obtained by interchanging its rows and columns.

Symmetric and Skew-Symmetric Matrices:

A matrix A is symmetric if A =AT (i.e., it is equal to its transpose).

A matrix A is skew-symmetric if AT = -A (i.e., the transpose of A is equal to the negative of A).

Elementary Operation or Transformation of a Matrix:

Elementary row operations include:

Interchanging any two rows.
Multiplying a row by a non-zero scalar.
Adding or subtracting a multiple of one row from another row.

The inverse of a Matrix by Elementary Operations:

You can find the inverse of a matrix using elementary row operations. If the matrix A is invertible, you can transform it into the identity matrix I through row operations on an augmented matrix [A | I], where I is the identity matrix of the same order as A. If this process is successful, the resulting matrix on the left will be I, and the matrix on the right will be the inverse of A.

Approach to Solve Questions of Matrices Class 12

Matrices play a significant role in Class 12 mathematics, and using these approaches, students can tackle the Matrices Class 12 Chapter 3 Question Answers with greater confidence.

Recognise the problem type: Before starting to solve a matrix-related problem, always try to identify what type of question you are dealing with in the first step. Some basic categories are Matrix operations like addition or multiplication, Transpose of a Matrix, Symmetric/Skew-Symmetric Matrix, Inverse of a Matrix, Solving Linear Equations using Matrix, etc.
Conceptual clarity: There are many concepts related to matrices. Before trying to solve any matrix questions, you should always learn the key concepts and formulas of matrices. A clear understanding of these concepts will help you solve the questions easily.
Simplify the problems: After learning the properties, you should apply them in the solutions to simplify the problem and reduce calculation time. Try to break the large and complex problems into simple parts and then solve them.
Some common errors to avoid: There are some basic common mistakes students make while solving, like incorrectly multiplying matrices as they don't follow the row-by-column rule properly, adding matrices of different orders, applying the wrong inverse formula, etc. Always remember to avoid these types of mistakes.
Tips and tricks to Improve Speed & Accuracy: To improve your speed and accuracy, you have to practice many different types of questions from the ncert book, the exemplar book, and the previous year papers. Also, you can revise the key concepts and formulas periodically to boost your memory.

Chapter Summary of NCERT Solutions for Class 12 Maths Chapter 3 Matrices

All the basics associated with matrices and its different kinds, and matrix operations have been explained comprehensively in this chapter. Matrix addition, multiplication, transpose, symmetric matrix, skew symmetric matrix, inverse of matrix and properties of matrices have all been explained in this chapter. NCERT solutions help students understand these concepts and solve problems based on matrices easily. Chapter 3 consists of a total of 55 questions across a total of 4 exercises, which is enough for students to get practice for understanding the concepts. Regular practice of these questions makes them perfect in solving matrices as well as other questions related to Higher Maths.

Expert Review of NCERT Solutions for Class 12 Maths Chapter 3 Matrices

As per the expert opinion of Mathematics experts at Careers360, Matrices is one of the highest-scoring chapters of Class 12th Mathematics as several questions are based on basic formulae and concepts. Students who are aware of matrix operations and their properties will be able to answer the questions effectively. NCERT Solutions help students in understanding concepts clearly, enhance calculation abilities and analytical skills via in-depth explanations. It is also recommended by experts that students should practice all the exercises comprehensively and revise all the important properties regularly so as to avoid any silly mistake. Consistent practice helps students perform well in CBSE Board examinations as well as competitive exams like JEE Main, JEE Advanced.

What Extra Should Students Study Beyond the NCERT for JEE?

For JEE aspirants, it is important to go beyond NCERT. Below are some extra topics that can help you build a deeper understanding and handle challenging problems with confidence.

Concept Name	JEE	NCERT
Determine The Order of the Matrix	✅	✅
Types of Matrices	✅	✅
Triangular matrix (Upper and Lower triangular matrix)	✅	✅
Matrix operations	✅	✅
Matrix Multiplication	✅	✅
Transpose of a Matrix	✅	✅
Symmetric Matrix & Skew Symmetric Matrix	✅	✅
Conjugate of a Matrix	✅	✅
Hermitian matrix	✅	❌
Skew Hermitian Matrix	✅	❌
Trace of a matrix and properties	✅	❌
Orthogonal matrix and Unitary matrix and Idempotent matrix	✅	❌
Periodic matrix and Nilpotent matrix and Involutory matrix	✅	❌
Elementary row operations	✅	❌
Singular Matrix	✅	❌
Minors And Cofactors	✅	❌
Adjoint and Inverse of a Matrix	✅	✅
Inverse Matrix	✅	✅
Solving Linear Equations Using Matrix	✅	✅

NCERT Solutions for Class 12 Maths: Chapter-Wise

We at Careers360 compiled all the NCERT class 12 Maths solutions in one place for easy student reference. Access them by using the following links.

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

NCERT Solutions for Class 12 Maths Chapter 3 Matrices

NCERT Solutions for Class 12 Maths Chapter 4 Determinants

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

NCERT Solutions for Class 12 Maths Chapter 7 Integrals

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 11 Three-Dimensional Geometry

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

NCERT Solutions for Class 12 Maths Chapter 13 Probability

Also Read,

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 Books and NCERT Syllabus

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

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Frequently Asked Questions (FAQs)

Q: What is a matrix?

Matrix is an array of numbers, variables, or expressions having two dimensions i.e., rows and columns.

Q: Why is the Matrices chapter important for Class 12 Maths?

This chapter forms the basis for the topic Determinants and Linear Programming and comes regularly in board examinations as well as competitive examinations.

Q: What are the various types of matrices?

Some common types of matrices are Row matrix, Column matrix, Square matrix, Diagonal matrix, Identity matrix, Zero matrix, Symmetric matrix and Skew-symmetric matrix.

Q: What is the transpose of a matrix?

Interchanging the row and columns of a matrix is called its transpose.

Q: What is the inverse of a matrix?

The matrix which upon multiplication by the original matrix yields the identity matrix is known as its inverse.

Q: How NCERT Solutions are helpful in this chapter?

These solutions are comprehensive and explain each problem from scratch that helps enhance the conceptual knowledge of the students and practice with more and more problems.

Q: Is this chapter useful in the JEE Main and JEE Advanced exam?

Definitely, questions related to determinants and matrices are generally asked in theJEE Main andJEE Advanced examination. Additionally, the questions based on this chapter are very popular in the NDA and CUET exam as well.

Q: What topics should students focus on the most?

Students have to concentrate on matrix operations including transpose, inverse, and properties of various matrices.

Q: How to get good marks in this chapter?

Regular practice with NCERT exercises of NCERT, revisiting the formulas and working on the questions based on applications could help the students.

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Questions related to CBSE Class 12th

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Have a question related to CBSE Class 12th ?

2025-2026 cbse haferily question pepar

Hello Ananya,

Please specify the class for which you need the question papers. I am providing Class 10 and 12 papers.

Here are the links to the CBSE Half-yearly Question Papers (2025-2026).

Cbsc2025-2026 haile question papear

Kushal Chakraborty

26 Jun'26

Hello Ananya,

Please specify the class for which you need the question papers. I am providing Class 10 and 12 papers.

Here are the links to the CBSE Half-yearly Question Papers (2025-2026).

Question paper previous year 2026math

Kushal Chakraborty

26 Jun'26

Hello Pawan,

CBSE Class 10 Mathematics 2026 and previous year question paper:

https://school.careers360.com/boards/cbse/cbse-class-10-question-paper-2026

CBSE Class 12 Mathematics 2026 and previous year question paper:

https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers-class-12-maths

Ncrt 12 th prevois question paper

Saniya Khatri

16 Jun'26

Hello Dharani,

Check the link below to download NCERT Class 12 previous year question papers in PDF format for all subjects.

https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers-class-12