Class 12 ncert solution math exercise 3.2 matrix

 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

(c)3A-c

(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

 

(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}

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}

HenceA^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}

price matrix=\begin{bmatrix}80\\60\\40\end{bmatrix}

Total cost =\begin{bmatrix} 120 &96&120\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}

=120(80)+96(60)+120(40)

=9600+5760+4800

=20160

Thus, the bookshop will receive ₹

20160 from the sale of all these books.

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.

Question 21: The restriction on n, k and p so that PY + WY will be defined are:

(A)k=3,p=n                 (B)k is arbitrary p=2

(C)p is arbitrary k=3   (D)k=2,p=3

Solution: Order of P=p×k

Order of Y=3×k

Order of PY= (p×k)(3×k)=p×k

Hence k=3

Order of W=n × 3

Order of WY=(n × 3)(3×k)=n×k

order of PY + WY =p×k+n×k

It is possible if p=n

and k=3

  hence option (A) is true

Queestion 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

Solution: Order of matrix X=  2× n

Order  of  matrix = 2× p

Order of marix  7X-5Z=2× n or 2× p

Hence the  correct option is (B)

Team Gmath

Leave a Reply

Your email address will not be published. Required fields are marked *