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$\mathrm{A}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{rr}2 & 3 \\ -1 & -2\end{array}\right]$, then which of the following is/are correct?
$1.$ $\mathrm{AB}\left(\mathrm{A}^{-1} \mathrm{~B}^{-1}\right)$ is a unit matrix.
$2.$ $(\mathrm{AB})^{-1}=\mathrm{A}^{-1} \mathrm{~B}^{-1}$
Select the correct answer using the code given below.
Options:
$1.$ $\mathrm{AB}\left(\mathrm{A}^{-1} \mathrm{~B}^{-1}\right)$ is a unit matrix.
$2.$ $(\mathrm{AB})^{-1}=\mathrm{A}^{-1} \mathrm{~B}^{-1}$
Select the correct answer using the code given below.
Solution:
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Verified Answer
The correct answer is:
Neither 1 nor 2
Here, $A=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 3 \\ -1 & -2\end{array}\right]$
$|A|=3-(-2)=5$ and $|B|=-4-(-3)=-1$
$\Rightarrow A^{-1}=\frac{1}{5}\left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]$ and $B^{-1}=-1\left[\begin{array}{cc}-2 & -3 \\ 1 & 2\end{array}\right]$
$A B=\left[\begin{array}{ll}3 & 5 \\ 1 & 0\end{array}\right]$ and $A^{-1} B^{-1}=\frac{1}{5}\left[\begin{array}{cc}5 & 7 \\ -5 & 8\end{array}\right]$
$\Rightarrow A B\left(A^{-1} B^{-1}\right)=\frac{1}{5}\left[\begin{array}{cc}-10 & -61 \\ 5 & 7\end{array}\right] \neq 1 .$
$|A B|=0-5=-5$
$\therefore(A B)^{-1}=\frac{-1}{5}\left[\begin{array}{cc}0 & -5 \\ -1 & 3\end{array}\right] \neq A^{-1} B^{-1}$
$|A|=3-(-2)=5$ and $|B|=-4-(-3)=-1$
$\Rightarrow A^{-1}=\frac{1}{5}\left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]$ and $B^{-1}=-1\left[\begin{array}{cc}-2 & -3 \\ 1 & 2\end{array}\right]$
$A B=\left[\begin{array}{ll}3 & 5 \\ 1 & 0\end{array}\right]$ and $A^{-1} B^{-1}=\frac{1}{5}\left[\begin{array}{cc}5 & 7 \\ -5 & 8\end{array}\right]$
$\Rightarrow A B\left(A^{-1} B^{-1}\right)=\frac{1}{5}\left[\begin{array}{cc}-10 & -61 \\ 5 & 7\end{array}\right] \neq 1 .$
$|A B|=0-5=-5$
$\therefore(A B)^{-1}=\frac{-1}{5}\left[\begin{array}{cc}0 & -5 \\ -1 & 3\end{array}\right] \neq A^{-1} B^{-1}$
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