Question:
If $A=\left[\begin{array}{ll}2 & 4 \\ 4 & 3\end{array}\right], X=\left[\begin{array}{l}n \\ 1\end{array}\right], B=\left[\begin{array}{c}8 \\ 11\end{array}\right]$ and $A X=B$, then find $n$.
Solution:
Here,
$\left[\begin{array}{ll}2 & 4 \\ 4 & 3\end{array}\right]\left[\begin{array}{l}n \\ 1\end{array}\right]=\left[\begin{array}{c}8 \\ 11\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}2 n+4 \\ 4 n+3\end{array}\right]=\left[\begin{array}{c}8 \\ 11\end{array}\right]$
$\Rightarrow 2 n+4=8$
$\Rightarrow 2 n=4$
$\Rightarrow n=2$
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