KC Sinha Mathematics Solution Class 12 Chapter 7 वर्ग आव्यूह का सहखंडज और प्रतिलोम (adjoint and inverse of matrix) Exercise 7.1 (Q4-Q6)

Exercise 7.1




Question 4
Find the inverse of the following matrices
(i) $\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$
Sol :
$A=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$

$|A|=\left|\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right|$
$\left(\begin{array}{l}R_{1} \rightarrow R_{1}-R_{2} \\ R_{2} \rightarrow R_{2}-R_{3}\end{array}\right)$

$=\left|\begin{array}{rrr}0 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 3 & 4\end{array}\right|$

$=1\left|\begin{array}{cc}-1 & 0 \\ 1 & -1\end{array}\right|$

=1-0

=1≠0

<to be added>
$A_{11}=\left|\begin{array}{rr}4 & 3 \\ 3 & 4\end{array}\right|$
=16-9
=7

$A_{12}=-\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array}\right|-(4-3)$
=-1

$A_{13}=\left[\begin{array}{ll}1 & 4 \\ 1 & 3\end{array} \right]$
=3-4
=-1

$A_{21}=-\left|\begin{array}{ll}3 & 3 \\ 3 & 4\end{array}\right|$
=-(12-9)
=-3

$A_{22}=\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array} \right|$
=4-3
=1

$A_{23}=-\left|\begin{array}{ll}1 & 3 \\ 1 & 3\end{array}\right|$
=-(3-3)
=0

$A_{31}=\left|\begin{array}{ll}3 & 3 \\ 4 & 3\end{array}\right|$
=9-12
=3

$A_{32}=-\left|\begin{array}{ll}1 & 3 \\ 1 & 3\end{array}\right|$
=-(3-3)
=0

$A_{33}=\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array}\right|$
=4-3
=1

$a d j A=\left[\begin{array}{ccc}7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1\end{array}\right]^{\prime}$

$=\left[\begin{array}{rrr}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$

$\therefore A^{-1}=\frac{1}{| A|} \cdot a d j A$

$=\frac{1}{1}\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$

$A^{-1}=\left[\begin{array}{ccc}7^{0} & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$

(ii) $\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Sol :

(iii) $\left[\begin{array}{rrr}-1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4\end{array}\right]$
Sol :

(iv) $\left[\begin{array}{rrr}1 & 2 & 3 \\ -3 & 5 & 0 \\ 0 & 1 & 1\end{array}\right]$
Sol :

(v) $\left[\begin{array}{rrr}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$
Sol :

(vi) $\left[\begin{array}{rrr}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{array}\right]$
Sol :

(vii) $\left[\begin{array}{rrr}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right]$
Sol :

(viii) $\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Sol :

(ix) diag[1 3 2]
Sol :
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2\end{array}\right]$

(x) $\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$
Sol :

(xi) $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
Sol :

(xii) $\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$
Sol :

(xiii) $\left[\begin{array}{rrr}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
Sol :

(xiv) $\left[\begin{array}{rrr}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$
Sol :



Question 5
आव्यूह A निकाले ताकि 
[Find the matrix A such that]:

(i)$\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right] \mathbf{A}=\left[\begin{array}{rr}-16 & -6 \\ 7 & 2\end{array}\right]$
Sol :
$\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right] A=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$

माना $X=\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right]$ , $Y=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$

XA=Y

A=X-1Y

$|x|=\left|\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right|$

=15-14=1≠0

C11=3 , C12=-(-2)=2

C21=-(-7)=7 ,C22=5

adj X$=\left[\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right]^{\prime}$

$=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$

$x^{-1}=\frac{1}{|x|} \cdot$ adj X

$=\frac{1}{1}\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$

$=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$

$\therefore A=x^{-1} y=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$

$A=\left[\begin{array}{lr}-48+49 & -18+14 \\ -32+35 & -12+10\end{array}\right]$

$A=\left[\begin{array}{ll}1 & -4 \\ 3 & -2\end{array}\right]$

(ii)$A\left[\begin{array}{rr}3 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$
Sol :

Question 6

(i) यदि(If) $A=\left[\begin{array}{rr}-1 & -1 \\ 2 & -2\end{array}\right]$ (दिखाएँ कि) show that
$A^{2}+3 A+4 I_{2}=0$ hence find $A^{-1}$
Sol :
L.H.S
$A^{2}+3 A+4 I_{2}$

$=\left[\begin{array}{cc}-1 & -1 \\ 2 & -2\end{array}\right]\left[\begin{array}{cc}-1 & -1 \\ 2 & -2\end{array}\right]+3\left[\begin{array}{cc}-1 & -1 \\ 2 & -2\end{array}\right]+4\left[\begin{array}{c}1&0 \\ 0&1\end{array}\right]$

$=\left[\begin{array}{ccc}1-2 & 1+2 \\ -2-4 & -2+4\end{array}\right]+\left[\begin{array}{cc}-3 & -3 \\ 6 & -6\end{array}\right]+\left[\begin{array}{cc}4 & 0 \\ 0 & 4\end{array}\right]$

$=\left[\begin{array}{cc}-1 & 3 \\ -6 & 2\end{array}\right]+\left[\begin{array}{cc}1 & -3 \\ 6 & -2\end{array}\right]$

$=\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]=0$

$\because A^{2}+3 A+4 I_{2}=0$

<to be added>

$A^{2} \cdot A^{-1}+3 A \cdot A^{-1}+4 I_{2} \cdot A^{-1}=0 \cdot A^{-1}$

$A+3 I+4 A^{-1}=0$

$\left[\begin{array}{rr}-1 & -1 \\ 2 & -2\end{array}\right]+3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+4 A^{-1}=0$

$\left[\begin{array}{cc}-1 & -1 \\ 2 & -2\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]+4 A^{-1}=0$

$\left[\begin{array}{cc}2 & -1 \\ 2 & 1\end{array}\right]+4 A^{-1}=0$

$4 A^{-1}=\left[\begin{array}{cc}-2 & 1 \\ -2 & -1\end{array}\right]$

$A^{-1}=\frac{1}{4}\left[\begin{array}{cc}-2 & 1 \\ -2 & -1\end{array}\right]$

$A^{-1}=\left[\begin{array}{cc}-\frac{1}{2} & \frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{4}\end{array}\right]$

(ii) यदि (If) $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ , दिखाएँ कि (show that) $\mathrm{A}^{2}-4 \mathrm{~A}+\mathrm{I}=\mathrm{O}$ तथा इससे (and hence find) $\mathrm{A}^{-1}$ निकालें।
Sol :


(iii) यदि (If) $\mathrm{A}=\left[\begin{array}{rrr}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$ तो सत्यापित करें कि (verify that) $\mathrm{A}^{3}-6 \mathrm{~A}^{2}+9 \mathrm{~A}-4 \mathrm{I}=\mathrm{O}$ तथा इससे (and hence find) $\mathrm{A}^{-1}$ निकालें।
Sol :



(iv) If $A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$ show that
$A^{3}-6 A^{2}+5 A+11 I=O$
Hence find $\mathrm{A}^{-1}$
Sol :
$A^{2}=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$

$=\left[\begin{array}{cccc}1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9\end{array}\right]$

$A^{3}=A^{2} \cdot A=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$

$=\left[\begin{array}{ccc}4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42\end{array}\right]$

$A^{3}=\left[\begin{array}{ccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]$

(iv) If $A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$ show that
$A^{3}-6 A^{2}+5 A+11 I=O$
Hence find $\mathrm{A}^{-1}$
Sol :
$A^{2}=\left[\begin{array}{rrr}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]$ , $A^{3}=\left[\begin{array}{ccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]$

L.H.S
$A^{3}-6 A^{2}+5 A+11 I$

$=\left[\begin{array}{ccccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]-6\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]+5\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]+11\left[\begin{array}{c}1&0&0 \\ 0&1&0 \\ 0&0&1\end{array}\right]$

$=\left[\begin{array}{ccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]-\left[\begin{array}{ccc}24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{array}\right]+\left[\begin{array}{ccc}5 & 5 & 5 \\ 5 & 10 & -15 \\ 20 & -5 & 15\end{array}\right]+\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0\\0&0&11 \end{array}\right]$

$=\left[\begin{array}{ccc}24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{array}\right]-\left[\begin{array}{ccc}24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{array}\right]$

$=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

=0


(iv) If $A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$ show that
$A^{3}-6 A^{2}+5 A+11 I=O$
Hence find $\mathrm{A}^{-1}$
Sol :
$A^{2}=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]$

$A^{3}-6 A^{2}+5 A+11 I=0$

<to be added>
$A^{3} A^{-1}-6 A^{2} \cdot A^{-1}+5 \cdot A \cdot A^{-1}+11 I \cdot A^{-1}=0$

$A^{2}-6 A+5 I+11 A^{-1}=0$

$\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]-6\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]+5\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]+11 A^{-1}=0$

$\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]-\left[\begin{array}{ccc}6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18\end{array}\right]+\left[\begin{array}{ccc}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]+11 A^{-1}=0$

$\left[\begin{array}{ccc}9 & 2 & 1 \\ -3 & 13 & -14 \\ 7 & -3 & 19\end{array}\right]-\left[\begin{array}{ccc}6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18\end{array}\right]+11 A^{-1}=0$

$\left[\begin{array}{ccc}3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1\end{array}\right]+11 A^{-1}=0$

$11 A^{-1}=\left[\begin{array}{rrr}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right]$

$A^{-1}=\frac{1}{11}\left[\begin{array}{ccc}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right]$

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