EXERCISE 3.2 (MATRIX)

Class 12 ncert solution math exercise 3.2 matrix

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

Find each of the following

(a) $A+B$

(b) $A-B$

(d) $AB$

(e) $BA$

Solution:(a) $A+B$

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

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

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

(b) $A-B$

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

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

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

(c) $3A-C$

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

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

$=\begin{bmatrix} 8 &7\\ 6&2 \end{bmatrix}$

(d) $AB$

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

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

$=\begin{bmatrix} 2-8 &6+20\\ 3-4&9+10 \end{bmatrix}$

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

(e) $BA$

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

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

$=\begin{bmatrix} 2+9&4+6\\ -4+15&-8+10 \end{bmatrix}$

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

Question 2: Compute the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 3: Compute the indicated products:

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

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

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

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

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

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

Solution: (a) $\begin{bmatrix} a & b\\ -b &a \end{bmatrix}\begin{bmatrix} a & -b\\ b &a \end{bmatrix}$

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

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

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

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

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

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

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

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

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

$=\begin{bmatrix} -3&-4&1\\ 8 &13&9 \end{bmatrix}$

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

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

$=\begin{bmatrix} 2+0+12 & -6+6+0&10+12+20\\ 3+0+15&-9+8+0 &15+16+25\\4+0+18&-12+10+0&20+20+30\end{bmatrix}$

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

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

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

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

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

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

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

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

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

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

Then compute A+B and B-C. Also, verify that A+(B+C)=(A+B)-C

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

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

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

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

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

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

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

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

Hence, A+(B-C)=(A+B)-C

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

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

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

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

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

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

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

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

Question 7: Find X and Y,if

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

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

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

Solution: (a) $X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}--(1)$

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

Adding equation (1) and (2)

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

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

$X=\frac{1}{2}\begin{bmatrix}10&0\\2&8\end{bmatrix}$

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

Since

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

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

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

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

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

Multiplying equation (1) by 2and multiplying (2) by 3

$4X+6Y=\begin{bmatrix}4&6\\8&0\end{bmatrix}--(3)$

$9X+6Y=\begin{bmatrix}6&-6\\-3&15\end{bmatrix}--(4)$

subtracting (3) to (4)

$(4X+6Y)-(9X+6Y)=\begin{bmatrix}4&6\\8&0\end{bmatrix}-\begin{bmatrix}6&-6\\-3&15\end{bmatrix}$

$-5X=\begin{bmatrix}4-6&6+6\\8+3&0-15\end{bmatrix}$

$X=-\frac{1}{5}\begin{bmatrix}-2&12\\11&-15\end{bmatrix}$

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

Now,

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

$\Rightarrow3\begin{bmatrix}\frac{2}{5}&-\frac{12}{5}\\-\frac{11}{5}&3\end{bmatrix}+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}$

$2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}-\begin{bmatrix}\frac{6}{5}&-\frac{36}{5}\\-\frac{33}{5}&9\end{bmatrix}$

$\Rightarrow2Y=\begin{bmatrix}\frac{4}{5}&-\frac{26}{5}\\-\frac{28}{5}&-4\end{bmatrix}$

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

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

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

$\Rightarrow2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-Y$

$\Rightarrow2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}$

$\Rightarrow2X=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}$

$\Rightarrow X=\frac{1}{2}\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}$

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

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

Solution: Since

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

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

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

Comparing the corresponding elements two matrices:

$2+y=5$

$\Rightarrow y=3$

$2x+2=8$

$x=3$

Therefore x=3 and y=3

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

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

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

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

Comparing the corresponding elements two matrices:

$2x+3=9$

$\Rightarrow 2x=6$

$\Rightarrow x=3$

$2y=12$

$\Rightarrow y=6$

$2z-3=15$

$\Rightarrow 2z=18$

$\Rightarrow z=9$

$2t+6=18$

$\Rightarrow 2t=12$

$\Rightarrow t=6$

Hence the values of x=3,y=6,z=9 and t=6

Question 11: If $x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$

Findthe value of x and y

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

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

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

Comparing the corresponding elements two matrices:

$2x-y=10--(1)$

$3x+y=5--(2)$

Adding (1)and (2) equation

$5x=15$

$\Rightarrow x=3$

put the value of x in (1) Equation

$2(3)-y=10$

$\Rightarrow 6-y=10$

$\Rightarrow -y=4$

$\Rightarrow y=-4$

Hence the value of x=3 and 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 value of x,y,z and w.

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

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

Comparing the corresponding elements two matrices:

$3x=x+4$

$\Rightarrow 2x=4$

$\Rightarrow x=2$

Since

$3y=6+x+y$

$\Rightarrow 2y=6+x$

$\Rightarrow 2y=6+2$

$\Rightarrow 2y=8$

$\Rightarrow y=4$

Since

$3w=2w+3$

$\Rightarrow w=3$

Since

$3z=-1+z+w$

$\Rightarrow 2z=-1+w$

$\Rightarrow 2z=-1+3$

$\Rightarrow 2z=2$

$\Rightarrow z=1$

Therefore x=2,y=4,z=1 and w=3

Question 13: If $F(x)=\begin{bmatrix}cosx & -sinx&0\\ sinx&cosx &0\\0&0&1\end{bmatrix}$ Show that $F(x)F(y)=F(x+y)$ .

Solution: $F(x)=\begin{bmatrix}cosx & -sinx&0\\ sinx&cosx &0\\0&0&1\end{bmatrix}$ and $F(y)=\begin{bmatrix}cosy & -siny&0\\ siny&cosy&0\\0&0&1\end{bmatrix}$

Now

$F(x)F(x)=\begin{bmatrix}cosx & -sinx&0\\ sinx&cosx &0\\0&0&1\end{bmatrix}\begin{bmatrix}cosy & -siny&0\\ siny&cosy&0\\0&0&1\end{bmatrix}$

$=\begin{bmatrix}cosxcosy+sinxsiny+0 & -cosxsiny-sinxcosy+0&0+0+1\\ sinxcosy+cosxsiny+0&-sinxsiny+cosxcosy +0&0+0+0\\0+0+0&0+0+0&0+0+1\end{bmatrix}$

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

$=F(x+y)$

Hence F(x)F(y)=F(x+y)

Question 14: Show that (a) $\begin{bmatrix} 5 &-1\\ 6&7 \end{bmatrix}\begin{bmatrix} 2 &1\\ 3&4 \end{bmatrix}\neq\begin{bmatrix} 2 &1\\ 3&4 \end{bmatrix}\begin{bmatrix} 5 &-1\\ 6&7 \end{bmatrix}$

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

Solution: (a) $\begin{bmatrix} 5 &-1\\ 6&7 \end{bmatrix}\begin{bmatrix} 2 &1\\ 3&4 \end{bmatrix}\neq\begin{bmatrix} 2 &1\\ 3&4 \end{bmatrix}\begin{bmatrix} 5 &-1\\ 6&7 \end{bmatrix}$

LHS.

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

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

$\Rightarrow\begin{bmatrix} 7 &1\\ 28&34 \end{bmatrix}$

RHS.

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

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

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

Therefore , $LHS\neq RHS$

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

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

$\Rightarrow\begin{bmatrix} -1+0+6 & 1-2+9&0+2+12\\ 0+0+0&0-1+0 &0+1+0\\-1+0+0&1-1+0&0+1+0\end{bmatrix}$

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

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

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

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

Therefore, $LHS\neq RHS$

Question 15. Find $A^2-5A+6I$ , if $A=\begin{bmatrix} 2 & 0&1\\ 2&1 &3\\1&-1&0\end{bmatrix}$

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

$A^2=AA$

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

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

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

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

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

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

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

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

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

$A^3=A^2A$

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

$=\begin{bmatrix} 5+0+16& 0+0+0&10+0+24\\ 2+0+10&0+8+0 &4+4+15\\8+0+26&0+0+0&1+0+39\end{bmatrix}$

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

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

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

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

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

Hence $A^3-6A^2+7A+2I=0$ Proved

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 $A^2=kA-2I$ .

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

$A^2=AA$

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

$\Rightarrow A^2=\begin{bmatrix} 3(3)-2(4) &3(-2)-2(-2)\\ 4(3)-2(4)&4(-2)-2(-2) \end{bmatrix}$

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

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

Now $A^2=ka-2I$

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

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

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

Ccmparing the corresponding element of two matrices

$4k=4$

$\Rightarrow k=1$

Question18: If $A=\begin{bmatrix} 0 &-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2} &0\end{bmatrix}$ and I is the indentity matrix of order 2 ,Show that

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

Solution :

$LHS=I+A$

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

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

$RHS=(I-A)\begin{bmatrix} cos\alpha &-\sin\alpha\\ \sin\alpha&\cos\alpha \end{bmatrix}$

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

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

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

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

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

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

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

$=LHS$

Question 19: A trust fund has ₹ 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 ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a)₹ 1800 (b)₹ 2000

Solution: (a) $\begin{bmatrix}x&(30000-x)\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=1800$

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

$5x+210000-7x=180000$

$-2x=-30000$

$x=15000$

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

(b) Let ₹ x be inested in

the first bond. Then the sum of money invested in the second bond will be ₹(30000-x)

Therefore, in order to obtain an annual total interest of ₹1800, we have

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

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

$5x+210000-7x=200000$

$-2x=-10000$

$x=5000$

Thus, in order to obtain an annual total interest of ₹2000, the trust fund should invest ₹5000 in the first bond and the remaining ₹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 ₹80 , ₹ 60and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra

Solution :books matrix = $\begin{bmatrix} 120 &96&120\end{bmatrix}$