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If the matrix $A$ is such that $\left(\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right) A=\left(\begin{array}{cc}1 & 1 \\ 0 & -1\end{array}\right)$, then what is $A$ equal to?
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The correct answer is:
$\left(\begin{array}{cc}1 & 4 \\ 0 & -1\end{array}\right)$
$\because\left(\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right) \mathrm{A}=\left(\begin{array}{rr}1 & 1 \\ 0 & -1\end{array}\right)$
Let $\mathrm{B} \quad=\left(\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right)$ and $|\mathrm{B}|=1$
$\therefore \mathrm{B}^{-1}=\left(\begin{array}{rr}1 & -3 \\ 0 & 1\end{array}\right)$ $\quad\left(\because \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\right)$
$\therefore \mathrm{A}=\left(\begin{array}{rr}1 & -3 \\ 0 & 1\end{array}\right)\left(\begin{array}{rr}1 & 1 \\ 0 & -1\end{array}\right)=\left(\begin{array}{rr}1 & 4 \\ 0 & -1\end{array}\right)$
Let $\mathrm{B} \quad=\left(\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right)$ and $|\mathrm{B}|=1$
$\therefore \mathrm{B}^{-1}=\left(\begin{array}{rr}1 & -3 \\ 0 & 1\end{array}\right)$ $\quad\left(\because \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\right)$
$\therefore \mathrm{A}=\left(\begin{array}{rr}1 & -3 \\ 0 & 1\end{array}\right)\left(\begin{array}{rr}1 & 1 \\ 0 & -1\end{array}\right)=\left(\begin{array}{rr}1 & 4 \\ 0 & -1\end{array}\right)$
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