Question
Find the values of $a$ and below
$
\left[\begin{array}{cc}
a+3 & b^2+2 \\
0 & -6
\end{array}\right]=\left[\begin{array}{cc}
2 a+1 & 3 b \\
0 & b^2-5 b
\end{array}\right]
$

Answer

$
\left[\begin{array}{cc}
a+3 & b^2+2 \\
0 & -6
\end{array}\right]=\left[\begin{array}{cc}
2 a+1 & 3 b \\
0 & b^2-5 b
\end{array}\right]
$
comparing the corresponding elements
$
\begin{aligned}
& a+3=2 a+1 \\
& \Rightarrow 2 a-a=3-1 \\
& \Rightarrow a=2 \\
& b^2+2=3 b \\
& \Rightarrow b^2-3 b+2=0 \\
& \Rightarrow b^2-b-2 b+2=0 \\
& \Rightarrow b(b-1)-2(b-1)=0 \\
& \Rightarrow(b-1)(b-2)=0 .
\end{aligned}
$
Either $b-1=0$,
then $b=1$
or
$
b-2=0 \text {, }
$
then $b=2$
Hence $a=2, b=2$ or 1 .

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