Question
If $\left[\begin{array}{ll}a & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}4 & 3 \\ -3 & 2\end{array}\right]=\left[\begin{array}{cc}b & 11 \\ 4 & c\end{array}\right]$ find $a, b$ and $c$
✓
Answer
$
\begin{aligned}
& {\left[\begin{array}{ll}
a & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{cc}
4 & 3 \\
-3 & 2
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
4 a-3 & 3 a+2 \\
4+0 & 3+0
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
4 a-3 & 3 a+2 \\
4 & 3
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 3 a+2=11 \\
& \Rightarrow 3 a=11-2=9 \\
& \therefore a=\frac{9}{3}=3 \\
& 4 a-3=b \\
& \Rightarrow b=4 \times 3-3 \\
& =12-3 \\
& =9 \\
& 3=c
\end{aligned}
$
Hence $a=3, b=9, c=3$.
\begin{aligned}
& {\left[\begin{array}{ll}
a & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{cc}
4 & 3 \\
-3 & 2
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
4 a-3 & 3 a+2 \\
4+0 & 3+0
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
4 a-3 & 3 a+2 \\
4 & 3
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 3 a+2=11 \\
& \Rightarrow 3 a=11-2=9 \\
& \therefore a=\frac{9}{3}=3 \\
& 4 a-3=b \\
& \Rightarrow b=4 \times 3-3 \\
& =12-3 \\
& =9 \\
& 3=c
\end{aligned}
$
Hence $a=3, b=9, c=3$.
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