Question:

Two sides of a rectangle differ by 3.5 cm. Find the dimensions of the rectangle if its perimeter is 67 cm.

Difficulty: Easy

Solution:

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

According to the given conditions:

The two sides $x$ and $y$ of the rectangle differ by $3.5$.

$x-y=3.5$

$10x-10y=35 \quad\quad\quad\quad\quad...~(1)$

Perimeter = 67 cm

$2(x + y) = 67$

$2x + 2y = 67 \quad\quad\quad\quad\quad...~(2)$

Method 1: Matrix Inversion Method

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

$AX = B$

$\left[ \begin{matrix} 10 & -10 \\ 2 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 10 & -10 \\ 2 & 2 \\ \end{matrix} \right|=10\times 2-\left( -10 \right)\times 2=20+20=40\ne 0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$

= $\frac{Adj~\text{A}}{\left| \text{A} \right|}~\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$

= $\frac{1}{40}\left[ \begin{matrix} 2 & 10 \\ -2 & 10 \\ \end{matrix} \right]\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$

= $\frac{1}{40}\left[ \begin{matrix} 2~\times 35+10\times 67 \\ -2~\times 35+10\times 67 \\ \end{matrix} \right]$

= $\frac{1}{40}\left[ \begin{matrix} 70+670 \\ -70+670 \\ \end{matrix} \right]$

= $\frac{1}{40}\left[ \begin{matrix} 740 \\ 600 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$= $\left[ \begin{matrix} 18.5 \\ 15 \\ \end{matrix} \right]$

Therefore $x=18.5$ and $y=15$

Length of the Rectangle = $x$ = 18.5cm

Width of the Rectangle = $y$ = 15 cm

Method 2: Cramer’s Rule

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

$AX = B$

$\left[ \begin{matrix} 10 & -10 \\ 2 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 10 & -10 \\ 2 & 2 \\ \end{matrix} \right|=10\times 2-\left( -10 \right)\times 2=20+20=40\ne 0$

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

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

${{A}_{x}}=\left[ \begin{matrix} 35 & -10 \\ 67 & 2 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 35 & -10 \\ 67 & 2 \\ \end{matrix} \right|=35\times 2-\left( -10 \right)\times 67=70+670=740$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{740}{40}=18.5$

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

${{A}_{y}}=\left[ \begin{matrix} 10 & 35 \\ 2 & 67 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 10 & 35 \\ 2 & 67 \\ \end{matrix} \right|=10\times 67-35\times 2=670-70=600$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{600}{40}=15$

Length of the Rectangle = $x$ = 18.5 cm

Width of the Rectangle = $y$ = 15 cm