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

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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 :
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2\end{array}\right]$