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Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]=\left[a_{i j}\right]$, where $i, j=1,2$, If its inverse matrix is $\left[b_{i j}\right]$, what is $b_{22} ?$
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The correct answer is:
$-\frac{1}{2}$
Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
$|A|=4 \times 1-2 \times 3=4-6=-2$
$\therefore \quad A^{-1}=-\frac{1}{2}\left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right]$
$\Rightarrow\left[b_{i j}\right]=-\frac{1}{2}\left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right]=\left[\begin{array}{lc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$
$\Rightarrow b_{22}=\frac{1}{2}$
$|A|=4 \times 1-2 \times 3=4-6=-2$
$\therefore \quad A^{-1}=-\frac{1}{2}\left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right]$
$\Rightarrow\left[b_{i j}\right]=-\frac{1}{2}\left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right]=\left[\begin{array}{lc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$
$\Rightarrow b_{22}=\frac{1}{2}$
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