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Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics4 Marks
Question
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
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Answer

Let $ A  =\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right] $
$ \therefore|A|  =\left|\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right| $
$ =1(-3-0)-0+0$
$ =-3 \neq 0$
$\therefore \mathrm{A}^{-1}$ exists.
First we have to find the co $-$ factor matrix
$=\left[\mathrm{A}_{i j}\right]_{3 \times 3}$, where $\mathrm{A}_{i j}=(-1)^{i+i} \mathrm{M}_{i j}$
Now $ \mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{lr}3 & 0 \\ 2 & -1\end{array}\right|=-3-0=-3$
$ \mathrm{~A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=-\left|\begin{array}{rr}3 & 0 \\ 5 & -1\end{array}\right|=-(-3-0)=3 $
$ \mathrm{~A}_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{ll}3 & 3 \\ 5 & 2\end{array}\right|=6-15=-9 $
$ \mathrm{~A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{lr}0 & 0 \\ 2 & -1\end{array}\right|=-(0-0)=0 $
$ \mathrm{~A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}1 & 0 \\ 5 & -1\end{array}\right|=-1-0=-1 $
$ \mathrm{~A}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 0 \\ 5 & 2\end{array}\right|=-(2-0)=-2 $
$ \mathrm{~A}_{31}=(-1)^{3+1} \mathrm{M}_{31}=\left|\begin{array}{ll}0 & 0 \\ 3 & 0\end{array}\right|=0-0=0 $
$ \mathrm{~A}_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{ll}1 & 0 \\ 3 & 0\end{array}\right|=-(0-0)=0 $
$ \mathrm{~A}_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{ll}1 & 0 \\ 3 & 3\end{array}\right|=3-0=3$
$\therefore$ the co $-$ factor matrix
$=\left(\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right)=\left[\begin{array}{rrr}-3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3\end{array}\right] $
$ \therefore \operatorname{adj} A=\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{|A|}(\text { adj } A) \\ =\frac{1}{-3}\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{3}\left[\begin{array}{rrr}3 & 0 & 0 \\ -3 & 1 & 0 \\ 9 & 2 & -3\end{array}\right]$

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