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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS1 Mark
Question
If $\text{A}=\begin{bmatrix}2&0&0\\-1&2&3\\3&3&5\end{bmatrix},$ then find A (adj A).
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Answer

Given,
$\text{A}=\begin{bmatrix}2&0&0\\-1&2&3\\3&3&5\end{bmatrix}$
$\text{M}_{11}=(-1)^2\begin{bmatrix}2&3\\3&5 \end{bmatrix}=10-9=1$
$\text{M}_{12}=-\begin{bmatrix}1&3\\3&5 \end{bmatrix}=-[-5-9]=14$
$\text{M}_{13}=(-1)^4.\begin{bmatrix}-1&2\\3&3 \end{bmatrix}=[-3-6]=-9$
$\text{M}_{21}=-\begin{bmatrix}0&0\\3&5\end{bmatrix}=0$
$\text{M}_{22}=\begin{bmatrix}2&0\\3&5\end{bmatrix}=10$
$\text{M}_{23}=-\begin{bmatrix}2&0\\3&3\end{bmatrix}=-6$
$\text{M}_{31}=\begin{bmatrix}0&0\\2&3\end{bmatrix}=0$
$\text{M}_{32}=-\begin{bmatrix}2&0\\-1&3\end{bmatrix}=-6$
$\text{M}_{33}=\begin{bmatrix}2&0\\-1&2\end{bmatrix}=4$
$\text{Adj.(A)}=\begin{bmatrix}1&0&0\\14&10&-6\\-9&-6&4\end{bmatrix}$
Then,
$\text{A.(adj A)}=\begin{bmatrix}2&0&0\\-1&2&3\\3&3&5\end{bmatrix}.\begin{bmatrix}1&0&0\\14&10&-6\\-9&-6&4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\times1+(0)\times14+0\times9&2\times0+0+0&2\times0+0\times0\\-1\times1+2\times14+3\times-9&-1\times0+2\times10+3\times-6&-1\times0+2\times-6+14\\3\times1+3\times14+5\times-9&3\times0+3\times10+5\times-6&3\times0+3\times-6+5\times4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}=2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A.(adj A)}=2\text{I}$

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