Question
If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ find $x$ and $y$ so that $A^2-x A+y \mid$
✓
Answer
Given
$
\begin{aligned}
& A^2=\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
4+3 & 6+6 \\
2+2 & 3+4
\end{array}\right] \\
& =\left[\begin{array}{ll}
7 & 12 \\
4 & 7
\end{array}\right] \\
& \because A ^2= xA + yl \\
& \Rightarrow\left[\begin{array}{cc}
7 & 12 \\
4 & 7
\end{array}\right]=x\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]+y\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
7 & 12 \\
4 & 7
\end{array}\right]=\left[\begin{array}{cc}
2 x & 3 x \\
x & 2 x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
0 & y
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 x+y & 3 x \\
x & 2 x+y
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 3 x=12 \\
& \Rightarrow x=4 \\
& 2 x+y=7 \\
& \Rightarrow 2 x 4+y=7 \\
& \Rightarrow 8+y=7 \\
& \Rightarrow y=7-8=-1
\end{aligned}
$
Hence $x=4, y=-1$.
$
\begin{aligned}
& A^2=\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
4+3 & 6+6 \\
2+2 & 3+4
\end{array}\right] \\
& =\left[\begin{array}{ll}
7 & 12 \\
4 & 7
\end{array}\right] \\
& \because A ^2= xA + yl \\
& \Rightarrow\left[\begin{array}{cc}
7 & 12 \\
4 & 7
\end{array}\right]=x\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]+y\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
7 & 12 \\
4 & 7
\end{array}\right]=\left[\begin{array}{cc}
2 x & 3 x \\
x & 2 x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
0 & y
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 x+y & 3 x \\
x & 2 x+y
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 3 x=12 \\
& \Rightarrow x=4 \\
& 2 x+y=7 \\
& \Rightarrow 2 x 4+y=7 \\
& \Rightarrow 8+y=7 \\
& \Rightarrow y=7-8=-1
\end{aligned}
$
Hence $x=4, y=-1$.
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