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If A is any $2 \times 2$ matrix such that $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right] \mathrm{A}=\left[\begin{array}{cc}-1 & 0 \\ 6 & 3\end{array}\right]$
then what is A equal to?
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then what is A equal to?
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Verified Answer
The correct answer is:
$\left[\begin{array}{cc}-5 & -2 \\ 2 & 1\end{array}\right]$
Let $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]=\mathrm{B}$
Then $\mathrm{BA}=\left[\begin{array}{cc}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow A=B^{-1}\left[\begin{array}{ll}-1 & 0 \\ -6 & 3\end{array}\right]$
$|\mathrm{B}|=3$
$\operatorname{adj} \mathrm{B}=\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right]$
$\mathrm{B}^{-1}=\frac{1}{3}\left[\begin{array}{rr}3 & -2 \\ 0 & 1\end{array}\right]$
$\Rightarrow \mathrm{A}=\frac{1}{3}\left[\begin{array}{rr}3 & -2 \\ 0 & 1\end{array}\right]\left[\begin{array}{rr}-1 & 0 \\ 6 & 3\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}-3-12 & -6 \\ 6 & 3\end{array}\right]$
$=\left[\begin{array}{rr}-5 & -2 \\ 2 & 1\end{array}\right]$
Aliter:
$\operatorname{Let} A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
then $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]=\left[\begin{array}{ll}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}\mathrm{a}+2 \mathrm{c} & \mathrm{b}+2 \mathrm{~d} \\ 0+3 \mathrm{c} & 0+3 \mathrm{~d}\end{array}\right]=\left[\begin{array}{ll}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow 3 \mathrm{c}=6$ or $\mathrm{c}=2$
$3 \mathrm{~d}=3$ or $\mathrm{d}=1$
$\mathrm{a}+2 \times 2=-1$ or $\mathrm{a}=-5$
$\mathrm{~b}+2 \times 1=0, \mathrm{~b}=-2$
So, $\mathrm{A}=\left[\begin{array}{cc}-5 & -2 \\ 2 & 1\end{array}\right]$
Then $\mathrm{BA}=\left[\begin{array}{cc}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow A=B^{-1}\left[\begin{array}{ll}-1 & 0 \\ -6 & 3\end{array}\right]$
$|\mathrm{B}|=3$
$\operatorname{adj} \mathrm{B}=\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right]$
$\mathrm{B}^{-1}=\frac{1}{3}\left[\begin{array}{rr}3 & -2 \\ 0 & 1\end{array}\right]$
$\Rightarrow \mathrm{A}=\frac{1}{3}\left[\begin{array}{rr}3 & -2 \\ 0 & 1\end{array}\right]\left[\begin{array}{rr}-1 & 0 \\ 6 & 3\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}-3-12 & -6 \\ 6 & 3\end{array}\right]$
$=\left[\begin{array}{rr}-5 & -2 \\ 2 & 1\end{array}\right]$
Aliter:
$\operatorname{Let} A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
then $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]=\left[\begin{array}{ll}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow\left[\begin{array}{ll}\mathrm{a}+2 \mathrm{c} & \mathrm{b}+2 \mathrm{~d} \\ 0+3 \mathrm{c} & 0+3 \mathrm{~d}\end{array}\right]=\left[\begin{array}{ll}-1 & 0 \\ 6 & 3\end{array}\right]$
$\Rightarrow 3 \mathrm{c}=6$ or $\mathrm{c}=2$
$3 \mathrm{~d}=3$ or $\mathrm{d}=1$
$\mathrm{a}+2 \times 2=-1$ or $\mathrm{a}=-5$
$\mathrm{~b}+2 \times 1=0, \mathrm{~b}=-2$
So, $\mathrm{A}=\left[\begin{array}{cc}-5 & -2 \\ 2 & 1\end{array}\right]$
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