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Question:
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If $A$ is a matrix such that
$\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right] A\left[\begin{array}{ll}
1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]$
then $A$ is equal to
Options:
$\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right] A\left[\begin{array}{ll}
1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]$
then $A$ is equal to
Solution:
1303 Upvotes
Verified Answer
The correct answer is:
$\left[\begin{array}{r}2 \\ -3\end{array}\right]$
Let $\begin{aligned} & A=\left[\begin{array}{l}x_1 \\ x_2\end{array}\right] \\ & \therefore \quad\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2\end{array}\right]\left[\begin{array}{ll}1 & 1\end{array}\right] \\ &=\left[\begin{array}{c}2 x_1+x_2 \\ 3 x_1+2 x_2\end{array}\right]\left[\begin{array}{ll}1 & 1\end{array}\right] \\ &=\left[\begin{array}{cc}2 x_1+x_2 & 2 x_1+x_2 \\ 3 x_1+2 x_2 & 3 x_1+2 x_2\end{array}\right]\end{aligned}$
$\begin{aligned}
& \therefore \quad\left[\begin{array}{cc}
2 x_1+x_2 & 2 x_1+x_2 \\
3 x_1+2 x_2 & 3 x_1+2 x_2
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right] \\
& \Rightarrow 2 x_1+x_2=1 \text { and } 3 x_1+2 x_2=0
\end{aligned}$
On solving, we get
$\begin{aligned}
& & x_1=2 \\
and & & x_2=-3 \\
& & A=\left[\begin{array}{r}
2 \\
-3
\end{array}\right]
\end{aligned}$
$\begin{aligned}
& \therefore \quad\left[\begin{array}{cc}
2 x_1+x_2 & 2 x_1+x_2 \\
3 x_1+2 x_2 & 3 x_1+2 x_2
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right] \\
& \Rightarrow 2 x_1+x_2=1 \text { and } 3 x_1+2 x_2=0
\end{aligned}$
On solving, we get
$\begin{aligned}
& & x_1=2 \\
and & & x_2=-3 \\
& & A=\left[\begin{array}{r}
2 \\
-3
\end{array}\right]
\end{aligned}$
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