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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
If $A=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]$, find $\operatorname{adj}(A)$
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Answer

$\begin{aligned} & A_{11}=(-1)^{1+1} M_{11}=1\left|\begin{array}{ll}0 & 1 \\ 4 & 3\end{array}\right|=1(0-4)=-4 \end{aligned} $
$ A_{12}=(-1)^{1+2} M_{12}=(-1)\left|\begin{array}{ll}1 & 1 \\ 4 & 3\end{array}\right|=(-1)(3-4)=(-1)(-1)=1  $
$ A_{13}=(-1)^{1+3} M_{13}=1\left|\begin{array}{ll}1 & 0 \\ 4 & 4\end{array}\right|=1(4-0)=(1)(4)=4  $
$ A_{21}=(-1)^{2+1} M_{21}=(-1)\left|\begin{array}{cc}-3 & -3 \\ 4 & 3\end{array}\right|=(-1)(-9+12)=(-1)$
$ (3)=-3  $
$\begin{aligned} & A_{22}=(-1)^{2+2} M_{22}=1\left|\begin{array}{cc}-4 & -3 \\ 4 & 3\end{array}\right|=1(-12+12)=1(0)=0 \end{aligned} $
$ A_{23}=(-1)^{2+3} M_{23}=(-1)\left|\begin{array}{cc}-4 & -3 \\ 4 & 4\end{array}\right|=(-1)(-16+12)=(-1)  $
$ (-4)=4$
$\begin{aligned} & A_{31}=(-1)^{3+1} M_{31}=1\left|\begin{array}{cc} -3 & -3 \\ 0 & 1 \end{array}\right|=1(-3-0)=-3 \end{aligned} $
$ A_{32}=(-1)^{3+2} M_{32}=(-1)\left|\begin{array}{cc} -4 & -3 \\ 1 & 1 \end{array}\right|=(-1)(-4+3)=(-1)(-1) =1 $
$A_{33}=(-1)^{3+3} M_{33}=1\left|\begin{array}{cc}-4 & -3 \\ 1 & 0\end{array}\right|=1(0+3)=(1)(3)=3$
$\begin{aligned} & \operatorname{adj}(A)=\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{21} & A_{32} & A_{33}\end{array}\right] \end{aligned} $
$ =\left[\begin{array}{ccc}-4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & -3\end{array}\right] $
$ =\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]$

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