Matrices Class 12 Ncert Solutions

                                                                  Chapter 3 Matrices

                                                                    Exercise 3.2

Looking for Ex 3.1 

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

Find each of the following:

(i) A+B     (ii) A-B   (iii) 3A-C   (iv) AB     (v) BA

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

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

                   = \(\begin{bmatrix}3&7\\1&7\\ \end{bmatrix}\)

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

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

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

(iii)3A - C  = \(3\begin{bmatrix}2&4\\3&2\\ \end{bmatrix}\) - \(\begin{bmatrix}-2&5\\3&4\\ \end{bmatrix}\)

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

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

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

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


             =  \(\begin{bmatrix}2\times1+4\times(-2)&2\times3+4\times5\\3\times1+2\times(-2)&3\times3+2\times5\\ \end{bmatrix}\)

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

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

             =  \(\begin{bmatrix}1\times2+3\times3&1\times4+3\times2\\(-2)\times2+5\times3&(-2)\times4+5\times2\\ \end{bmatrix}\)

            =  \(\begin{bmatrix}11&10\\11&12\\ \end{bmatrix}\)

2. Compute the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

= \(\begin{bmatrix}1&1\\1&1\\ \end{bmatrix}\)

 3.Compute the indicated products. 

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

Sol. \(\begin{bmatrix}a\times a+b\times b&a\times(-b)+b\times a\\(-b)\times a+a\times b&(-b)\times(-b)&a\times a\\ \end{bmatrix}\)

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

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

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

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

Sol.   = \(\begin{bmatrix}1\times2&1\times3&1\times4\\2\times2&2\times3&2\times4\\3\times2&3\times3&3\times4\\ \end{bmatrix}\)

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

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

Sol.  \(\begin{bmatrix}1\times1+(-2)\times2&1\times2+(-2)\times3&1\times3+(-2)\times1\\2\times1+3\times2&2\times2+3\times3&2\times3+3\times1\\ \end{bmatrix}\)  

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

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

Sol. \(\begin{bmatrix}2\times1+3\times0+4\times3&2\times1+3\times2+4\times0&2\times5+3\times4+4\times5\\ 3\times1+4\times0+5\times3&3\times(-3)+4\times2+5\times0&3\times5+4\times4+5\times5\\4\times1+5\times0+6\times3&4\times(-3)+5\times2+6\times0&4\times5+5\times4+6\times5\\ \end{bmatrix}\)

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

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

Sol.  \(\begin{bmatrix}2\times1+1\times(-1)&2\times0+1\times2&2\times1+1\times1\\3\times1+2\times(-1)&3\times0+2\times2&3\times1+2\times1\\-1\times1+1\times(-1)&-1\times0+1\times2&-1\times1+1\times1\\ \end{bmatrix}\)

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

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

Sol. \(\begin{bmatrix}3\times2+(-1)\times1+3\times3&3\times(-3)+(-1)\times0+3\times1\\-1\times2+0\times1+2\times3&(-1)\times(-3)+0\times0+2\times1\\ \end{bmatrix}\)

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

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.

Sol. 

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.

Sol. 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\(\begin{bmatrix}\frac{2}{5}&\frac{3}{5}&1\\ \frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\ \frac{7}{5}&\frac{6}{5}&\frac{2}{5}\\ \end{bmatrix}\) 

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

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

=\(\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\\ \end{bmatrix}\)

6. Simplify  \(cos\theta\begin{bmatrix}cos\theta&sin\theta\\-sin\theta&cos\theta\\ \end{bmatrix} +  sin\theta\begin{bmatrix}sin\theta&-cos\theta\\cos\theta&sin\theta\\ \end{bmatrix}\)

 Sol.  \(\begin{bmatrix}cos\theta\times cos\theta&cos\theta\times sin\theta\\cos\theta\times(-sin\theta)&cos\theta\times cos\theta\\ \end{bmatrix}\) + \(\begin{bmatrix} sin\theta\times sin\theta &sin\theta\times(-cos\theta)\\sin\theta\times cos\theta&sin\theta\times sin\theta\\\end{bmatrix}\)

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

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

= \(\begin{bmatrix}1&0\\0&1\\ \end{bmatrix}\)

Class 12 Ncert Solutions Maths Matrices

7. Find X and Y, if 

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

Sol.  Adding the given two equations we get
 X+Y +X - Y= \(\begin{bmatrix}7&0\\2&5\\ \end{bmatrix}\) + \(\begin{bmatrix}3&0\\0&3\\ \end{bmatrix}\)
\(\implies\) 2X = \(\begin{bmatrix}7+3&0+0\\2+0&5+3\\ \end{bmatrix}\)

\(\implies\) 2X = \(\begin{bmatrix}10&0\\2&8\\ \end{bmatrix}\)

\(\implies\) X = \(\frac{1}{2}\begin{bmatrix}10&0\\2&8\\ \end{bmatrix}\)

\(\implies\) X = \(\begin{bmatrix}\frac{10}{2}&\frac{0}{2}\\ \frac{2}{2}&\frac{8}{2}\\ \end{bmatrix}\)

\(\implies\) X = \(\begin{bmatrix}5&0\\1&4\\ \end{bmatrix}\)

Similarly, subtracting the given equations we get

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

\(\implies\) 2Y = \(\begin{bmatrix}4&0\\2&2\\ \end{bmatrix}\)

\(\implies\) Y = \(\frac{1}{2}\begin{bmatrix}10&0\\2&8\\ \end{bmatrix}\)

\(\implies\) Y = \(\begin{bmatrix}\frac{4}{2}&\frac{0}{2}\\ \frac{2}{2}&\frac{2}{2}\\ \end{bmatrix}\)

\(\implies\) Y = \(\begin{bmatrix}2&0\\1&1\\ \end{bmatrix}\)

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

Sol. \begin{align}
2X+3Y = \begin{bmatrix}2&3\\4&0\\ \end{bmatrix}\tag1\\
3X+2Y = \begin{bmatrix}2&-2\\-1&5\\ \end{bmatrix}\tag2\\
(1)\times3\implies 6X+9Y = \begin{bmatrix}6&9\\12&0\\ \end{bmatrix}\tag3\\
(2)\times2\implies 6X+4Y = \begin{bmatrix}4&-4\\-2&10\\ \end{bmatrix}\tag4\\
(3)-(4)\implies 9Y-4Y =  \begin{bmatrix}6&9\\12&0\\ \end{bmatrix}-  \begin{bmatrix}4&-4\\-1&20\\ \end{bmatrix}\\
\implies 5Y =  \begin{bmatrix}2&13\\14&-10\\ \end{bmatrix}\\
\implies Y =  \begin{bmatrix}\frac{2}{5}&\frac{13}{5}\\ \frac{14}{5}&-2\\ \end{bmatrix}
\end{align}

\begin{align}
(3)\times2\implies 4X+6Y = \begin{bmatrix}4&6\\8&0\\ \end{bmatrix}\tag5\\
(4)\times3\implies 9X+4Y = \begin{bmatrix}6&-6\\-3&15\\ \end{bmatrix}\tag6\\
(6)-(5)\implies 5X = \begin{bmatrix}2&-12\\-5&15\\ \end{bmatrix}\\
\implies X= \begin{bmatrix}\frac{2}{5}&\frac{-12}{5}\\ \frac{-11}{5}&3\\ \end{bmatrix}
\end{align}

8. Find X, if Y =  \(\begin{bmatrix}3&2\\1&4\\ \end{bmatrix}\)  and 2X + Y = \(\begin{bmatrix}1&0\\-3&2\\ \end{bmatrix}\)

Sol. 2X + Y = \(\begin{bmatrix}1&0\\-3&2\\ \end{bmatrix}\)
\(\implies\) 2X =  \(\begin{bmatrix}1&0\\-3&2\\ \end{bmatrix}\) - Y
\(\implies\) 2X =  \(\begin{bmatrix}1&0\\-3&2\\ \end{bmatrix}\) -   \(\begin{bmatrix}3&2\\1&4\\ \end{bmatrix}\)
\(\implies\) 2X =  \(\begin{bmatrix}-2&-2\\4&2\\ \end{bmatrix}\)
\(\implies\) X =  \(\begin{bmatrix}-1&-1\\2&1\\ \end{bmatrix}\)

9. Find x and y, if   \(2\begin{bmatrix}1&3\\0&x\\ \end{bmatrix} +\begin{bmatrix}y&0\\1&2\\ \end{bmatrix} = \begin{bmatrix}5&6\\1&8\\ \end{bmatrix}\) 

Sol. \(2\begin{bmatrix}1&3\\0&x\\ \end{bmatrix} +\begin{bmatrix}y&0\\1&2\\ \end{bmatrix} = \begin{bmatrix}5&6\\1&8\\ \end{bmatrix}\)

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

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

Equating the corresponding elements by the equality of matrices we get

2+y = 5 \(\implies\) y=3

and 2x+2 = 8 \(\implies\) x= 3

10. Find x and y, if  \(2\begin{bmatrix}x&z\\y&t\\ \end{bmatrix} + 3\begin{bmatrix}1&-1\\0&2\\ \end{bmatrix} = 3\begin{bmatrix}3&5\\4&6\\ \end{bmatrix}\) 

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

= \(\begin{bmatrix}2x-3&2z-3\\2y-0&2t-6\\ \end{bmatrix}  = \begin{bmatrix}9&15\\12&18\\ \end{bmatrix}\)
Equating the corresponding elements by the equality of matrices
we get
2x+3=9\(\implies\)x = 3
2z - 3=15\(\implies\)z = 9
2y =12\(\implies\)y =6
2t +6=18\(\implies\)t=6

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

Sol.   \(x\begin{bmatrix}2\\3\\ \end{bmatrix} + y\begin{bmatrix}-1\\1\\ \end{bmatrix} = 3\begin{bmatrix}10&5\\ \end{bmatrix}\)

=  \(\begin{bmatrix}2x\\3x\\ \end{bmatrix} + \begin{bmatrix}-y\\y\\ \end{bmatrix} = 3\begin{bmatrix}10&5\\ \end{bmatrix}\)

= \(\begin{bmatrix}2x -y\\3x+y\\ \end{bmatrix}  = 3\begin{bmatrix}10&5\\ \end{bmatrix}\)

Equating the corresponding elements by the equality of matrices we get
2x-y =10

3x +y = 5

Solving these equations we get x = 3, y=-4

12. Given  \(3\begin{bmatrix}x&y\\z&w\\ \end{bmatrix} = \begin{bmatrix}x&6\\-1&2w\\ \end{bmatrix} + \begin{bmatrix}4&x+y\\z+w&3\\ \end{bmatrix}\) 

Sol.\(3\begin{bmatrix}x&y\\z&w\\ \end{bmatrix} = \begin{bmatrix}x&6\\-1&2w\\ \end{bmatrix} + \begin{bmatrix}4&x+y\\z+w&3\\ \end{bmatrix}\)

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

Equating the corresponding elements by the equality of matrices
3x=x+4\(\implies\) x=2
3y=6+x+y\(\implies\) 3y-y = 6 +2 \(\implies\) y=4
3w = 2w + 3\(\implies\)w=3
3z = -1+z+w \(\implies\) 3z-z=-1+3\(\implies\) z=1


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)

Sol.  F(x) = \(\begin{bmatrix}cosx&-sinx&0\\sinx&cosx&0\\0&0&1\\ \end{bmatrix}\)

F(y) =\(\begin{bmatrix}cosy&-siny&0\\siny&cosy&0\\0&0&1\\ \end{bmatrix}\)

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

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

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

= F(x+y) = R.H.S

Class 12 Ncert Solutions Maths Matrices

14. Show that

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

Sol. Do it yourself.

 (ii) Do it yourself.

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

Sol. 

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

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

=\(\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2 \\ \end{bmatrix}\)

\(A^2 -5A+6I\) 

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

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

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

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

Sol.  \(A^2\)= A.A=\(\begin{bmatrix}1&0&2\\0&2&1\\2&03\\ \end{bmatrix}\)\(\begin{bmatrix}1&0&2\\0&2&1\\2&03\\ \end{bmatrix}\)

= \(\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\\ \end{bmatrix}\)

=\(\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\\ \end{bmatrix}\)

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

=  \(\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\\ \end{bmatrix}\)\(\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\\ \end{bmatrix}\)

= \(\begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+33\\ \end{bmatrix}\)

= \(\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\\ \end{bmatrix}\)

\(\therefore A^3 - 6A^2 + 7A+2I \) = \(\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\\ \end{bmatrix}\)- 6\(\begin{bmatrix}1&0&2\\0&2&1\\2&03\\ \end{bmatrix}\)\(\begin{bmatrix}1&0&2\\0&2&1\\2&03\\ \end{bmatrix}\)+7\(\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\\ \end{bmatrix}\) + 2\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\ \end{bmatrix}\)

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

=\(\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\\ \end{bmatrix}\)

= 0 =R.H.S  Hence proved

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 \(A^2 = kA-2I\)

Sol. \(A^2\) =AA= \(\begin{bmatrix}3&-2\\4&-2\\ \end{bmatrix}\)\(\begin{bmatrix}3&-2\\4&-2\\ \end{bmatrix}\)

= \(\begin{bmatrix} 9-8&-6+4\\12-8&-8+4\\ \end{bmatrix}\)

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

\(A^2 = kA-2I\)
\(\begin{bmatrix} 1&-2\\4&-4\\ \end{bmatrix}\) = k\(\begin{bmatrix}3&-2\\4&-2\\ \end{bmatrix}\) - 2\(\begin{bmatrix}1&0\\0&1\\ \end{bmatrix}\)

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

Equating the corresponding elements by the equality of matrices we get
3k-1 = 1\(\implies\)k=1

Class 12 Ncert Solutions Maths Matrices

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)\(\begin{bmatrix}cos\alpha&-sin\alpha\\sin\alpha&cos\alpha\\ \end{bmatrix}\)

Sol. L.H.S = I+A = \(\begin{bmatrix} 1&0\\0&1\\ \end{bmatrix}\) + \(\begin{bmatrix} 0&-tan\frac{\alpha}{2}\\tan\frac{\alpha}{2}&0\\ \end{bmatrix}\)

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

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

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

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

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

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

19. 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 matris multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund obtain an annual totalk interest of:  (a) Rs 1800   (b) Rs 2000

Sol. (a) Let the amount to be invested in the first bond be Rs x so 30000-x will be invested in the second bond.
Interest in first bond = 5%.
Interest in second bond =7%.

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


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

\(\frac{5x}{100}+\frac{7}{100}\times3000 - \frac{7x}{100}= 1800\)

\(\frac{5x}{100}+2100 - \frac{7x}{100}= 1800\)

\(-\frac{2x}{100}= -300\)

\(\frac{x}{50}= 300\)

x=15000

Hence , the trust fund must invest Rs 15000 in the first bond and the another Rs 15000 insecond bond to get Rs1800 as interest

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

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

\(\frac{5x}{100}+\frac{7}{100}\times3000 - \frac{7x}{100}= 2000\)

\(\frac{5x}{100}+2100 - \frac{7x}{100}= 2000\)

\(-\frac{2x}{100}= -100\)

\(\frac{x}{50}= 100\)

x=5000

Hence , the trust fund must invest Rs 5000 in the first bond and the another Rs 25000 insecond bond to get Rs2000 as interest 

20. The booksjop 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 recieve from the sellling all the books using  matrix algebra.

Sol. 1 dozen =12. So, 10 dozen = 120, 8dozen =96,

According to matrix algebra
\(\begin{bmatrix}120&96&120\\ \end{bmatrix}\begin{bmatrix}80\\60\\40\\ \end{bmatrix}=120\times80+96\times60+120\times40=20160\)

Hence, the total amount the bookshop will recieve after selling the books is Rs 20160

Assume X,Y, Z, W and P are the matrices of order \(2\times n\), \(3\times k\), \(2\times p\) \(n\times 3\) \(p\times k\) respectively. Choose the correct anser in Exercises 21 and 22.

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

Sol. The order of yhe matrix P is  \(p\times k\) and Y is \(3\times k\).So, the matrix product will be feasible if k=3. Beacuse the product is possible only when the no of columns of the first matrix is equal to the no of rows of the second matrix. The order of the resultant matrix is \(p\times k\) or \(p\times3\)

For ex. the order of the rresultant matrix of the product of two matrices having order \(3times5\) and \(5\times2\) is \(3\times2\)

Now, the order of the matrix W is \(n\times3\) and Y is \(3\times k\). So the order of the resultant matrix is \(n\times k\)

We have to add two matrix of order \(p\times3\) and \(n\times k\). We can add matrix only if the orders are same. So order will be same if p=n and k=3
The correct option is (A)

22. If n = p , then the order of the matrix 7X - 5Z is :

(A) \(p\times2\)   (B) \(2\times n\)  (C) \(n\times3\)  (D) \(p\times n\)

Sol. The order of X is \(2\times n\) and Z is \(2\times p\). The order of the resultant matrix is \(2\times p\). Since n=p, order will be \(2\times p\).
The correct answer is (B)

                              Class 12 Ncert Solutions Maths Matrices

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