Question
Given matrix $A =\left[\begin{array}{l}4 \sin 30^{\circ}, \cos 0^{\circ} \\ \cos 0^{\circ}, 4 \sin 30^{\circ}\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$ If $AX = B$. Find the matrix $' X \ '$

Answer

Let the matrix $X =\left[\begin{array}{l}x \\ y\end{array}\right]$
$AX = B$
$\begin{array}{l}\Rightarrow\left[\begin{array}{c}4 \sin 30^{\circ}, \cos 0^{\circ} \\ \cos 0^{\circ}, 4 \sin 30^{\circ}\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}4\left(\frac{1}{2}\right) & 1 \\ 1 & 4\left(\frac{1}{2}\right)\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] \\ \Rightarrow\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] \\ \Rightarrow\left[\begin{array}{l}2 x+y \\ x+2 y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right]\end{array}$
$\Rightarrow 2x + y = 4 .............(1)$
$x + 2y = 5 .........(2)$
Multiplying $(1)$ by $2,$ we get
$4x + 2y = 8 ….(3)$
Subtracting $(2)$ from $(3)$, we get
$3x = 3$
$\Rightarrow x = 1$
Substituting the value of $x$ in $(1),$ we get
$2(1) + y = 4$
$\Rightarrow 2 + y = 4$
$\Rightarrow y = 2$
Hence, the matrix $X =\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}1 \\ 2\end{array}\right]$

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