Solution:

Let the length of the rectangle is $x$ cm and width is $y$ cm.

According to the given conditions:

The length of the rectangle is 4 times its width.

$x = 4y$

$x – 4y = 0 \quad\quad\quad\quad\quad...~(1)$ 

Perimeter = 150 cm

$2(x + y) = 150$

$x +y = 75\quad\quad\quad\quad\quad...~(2)$

Method 1: Matrix Inversion Method

Write the linear equations $(1)$ and $(2)$ in matrix form.

$AX = B$

$\left[ \begin{matrix} 1 & -4 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 0 \\ 75 \\ \end{matrix} \right]$

According to matrix inversion method, solution $X$ can be found by the following formula:

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 1 & -4 \\ 1 & 1 \\ \end{matrix} \right|=1\times 1-1\times \left( -4 \right)=1+4=5~\ne ~0$

The coefficient matrix $A$ is non-singular therefore $A^{-1}$ exists.

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} 0 \\ 75 \\ \end{matrix} \right]$

= $\frac{1}{\text{A}}~Adj~\text{A}\left[ \begin{matrix} 0 \\ 75 \\ \end{matrix} \right]$

= $\frac{1}{5}\left[ \begin{matrix} 1 & 4 \\ -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 75 \\ \end{matrix} \right]$

= $\frac{1}{5}\left[ \begin{matrix} 1~\times 0+4\times 75 \\ \left( -1 \right)~\times 0+1\times 75 \\ \end{matrix} \right]$

= $\frac{1}{5}\left[ \begin{matrix} 0+300 \\ 0+75 \\ \end{matrix} \right]$

= $\frac{1}{5}\left[ \begin{matrix} 300 \\ 75 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$= $\left[ \begin{matrix} 60 \\ 15 \\ \end{matrix} \right]$

Therefore $x=60$ and $y=15$

Length of the Rectangle = $x$ = 60 cm

Width of the Rectangle = $y$ = 15 cm

Method 2: Cramer’s Rule

Write the linear equations $(1)$ and $(2)$ in matrix form.

$AX = B$

$\left[ \begin{matrix} 1 & -4 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 0 \\ 75 \\ \end{matrix} \right]$

According to Cramer's Rule:

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}$

$\left| \text{A} \right|=\left| \begin{matrix} 1 & -4 \\ 1 & 1 \\ \end{matrix} \right|=1\times 1-1\times \left( -4 \right)=1+4=5~\ne ~0$

The coefficient matrix $A$ is non-singular therefore solution exists.

For ${{A}_{x}}$, the $x$ column is replaced with the constant column $B$

${{A}_{x}}=\left[ \begin{matrix} 0 & -4 \\ 75 & 1 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 0 & -4 \\ 75 & 1 \\ \end{matrix} \right|=0~\times 1-\left( -4 \right)~\times \left( 75 \right)=0+300=300$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{300}{5}=60$

For ${{A}_{y}}$, the $y$ column is replaced with the constant column $B$

${{A}_{y}}=\left[ \begin{matrix} 1 & 0 \\ 1 & 75 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 1 & 0 \\ 1 & 75 \\ \end{matrix} \right|=1~\times 75-0~\times 1=75-0=75$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{75}{5}=15$

Length of the Rectangle = $x$ = 60 cm

Width of the Rectangle = $y$ = 15 cm