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If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$ then $A^{-1}=$
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The correct answer is:
$A^3$
Here, $C_{11}=1, C_{12}=-2, C_{13}=-2$
$C_{21}=-1, C_{22}=3, C_{23}=3$
$C_{31}=0, C_{32}=-4, C_{33}=-3$
$\left.\Rightarrow \operatorname{det} A \neq A|=| \begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array} \right\rvert\,=1$
$\Rightarrow A^{-1}=\frac{1}{|A|} \cdot(A d j A)=\frac{1}{1} \cdot\left[\begin{array}{lll}C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
Now, $A^2=\left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right]$
and $A^3=A^2 \cdot A=\left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right] \times\left[\begin{array}{ccc}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]=A^{-1}$.
$C_{21}=-1, C_{22}=3, C_{23}=3$
$C_{31}=0, C_{32}=-4, C_{33}=-3$
$\left.\Rightarrow \operatorname{det} A \neq A|=| \begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array} \right\rvert\,=1$
$\Rightarrow A^{-1}=\frac{1}{|A|} \cdot(A d j A)=\frac{1}{1} \cdot\left[\begin{array}{lll}C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
Now, $A^2=\left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right]$
and $A^3=A^2 \cdot A=\left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right] \times\left[\begin{array}{ccc}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]=A^{-1}$.
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