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If $a_{i j}=\frac{1}{2}(3 i-2 j)$ and $A=\left[a_{i j}\right]_{2 \times 2}$, then $A$ is equal to
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The correct answer is:
$\left[\begin{array}{cc}1 / 2 & -1 / 2 \\ 2 & 1\end{array}\right]$
$\Rightarrow a_{11}=1 / 2, \quad a_{12}=-1 / 2$ and $a_{21}=2, a_{22}=1$
$\therefore \quad A=\left[a_{i j}\right]_{2 \times 2}=$ $\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$
$\therefore \quad A=\left[\begin{array}{cc}1 / 2 & -1 / 2 \\ 2 & 1\end{array}\right]$
$\therefore \quad A=\left[a_{i j}\right]_{2 \times 2}=$ $\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$
$\therefore \quad A=\left[\begin{array}{cc}1 / 2 & -1 / 2 \\ 2 & 1\end{array}\right]$
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