Question

The length of a rectangle is twice its width. The perimeter is 30. Find its dimensions.(Use either Matrix inversion method orCramer’sRule)

Answer 1

Let the width of the rectangle be $x$ and the length of rectangle be $y.$

Since length is twice its width,

$\longrightarrow y=2x$

$\longrightarrow 2x-y=0\quad\quad\dots(1)$

And since perimeter is 30,

$\longrightarrow 2(x+y)=30$

$\longrightarrow x+y=15\quad\quad\dots(2)$

Thus we form a matrix equation from (1) and (2) as,

$\longrightarrow \left[\begin{array}{cc}2&-1\\1&1\end{array}\right]\,\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{c}0&15\end{array}\right]$

Method 1:- Matrix Inversion Method

Let,

• $A=\left[\begin{array}{cc}2&-1\\1&1\end{array}\right]$

• $X=\left[\begin{array}{c}x&y\end{array}\right]$

• $B=\left[\begin{array}{c}0&15\end{array}\right]$

Then our equation becomes,

$\longrightarrow A\cdot X=B\quad\quad\dots(3)$

We can't divide by a matrix but we can multiply by its inverse.

Since order of $A$ is $2\times2,$ so will be that of $A^{-1},$ whose no. of columns is equal to no. of rows of $B,$ whose order is $2\times1.$

We see no. of rows of $A$ is not equal to no. of columns of $B.$ Hence,

$\longrightarrow X=A^{-1}\cdot B$

Let's find $A^{-1}.$

$\longrightarrow A^{-1}=\dfrac{1}{|A|}\,\left[\begin{array}{cc}1&1\\-1&2\end{array}\right]$

$\longrightarrow A^{-1}=\dfrac{1}{2-(-1)}\,\left[\begin{array}{cc}1&1\\-1&2\end{array}\right]$

$\longrightarrow A^{-1}=\dfrac{1}{3}\,\left[\begin{array}{cc}1&1\\-1&2\end{array}\right]$

$\longrightarrow A^{-1}=\left[\begin{array}{cc}\frac{1}{3}&\frac{1}{3}\\\\-\frac{1}{3}&\frac{2}{3}\end{array}\right]$

Hence (3) becomes,

$\longrightarrow \left[\begin{array}{c}x\\\\y\end{array}\right]=\left[\begin{array}{cc}\frac{1}{3}&\frac{1}{3}\\\\-\frac{1}{3}&\frac{2}{3}\end{array}\right]\cdot\left[\begin{array}{cc}0\\\\15\end{array}\right]$

$\longrightarrow \left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}5\\10\end{array}\right]$

Therefore,

$\longrightarrow\underline{\underline{x=5}}$

$\longrightarrow\underline{\underline{y=10}}$

Method 2:- Application of Cramer's Rule

Cramer's rule states that, the solution to the matrix equation,

$\longrightarrow \left[\begin{array}{cc}a&b\\p&q\end{array}\right]\cdot\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{c}c\\r\end{array}\right]$

is given by,

$\longrightarrow x=\dfrac{\left|\begin{array}{cc}c&b\\r&q\end{array}\right|}{\left|\begin{array}{cc}a&b\\p&q\end{array}\right|}$

$\longrightarrow x=\dfrac{cq-br}{aq-bp}$

and,

$\longrightarrow y=\dfrac{\left|\begin{array}{cc}a&c\\p&r\end{array}\right|}{\left|\begin{array}{cc}a&b\\p&q\end{array}\right|}$

$\longrightarrow y=\dfrac{ar-cp}{aq-bp}$

According to the question,

• $\left[\begin{array}{cc}a&b\\p&q\end{array}\right]=\left[\begin{array}{cc}2&-1\\1&1\end{array}\right]$

• $\left[\begin{array}{c}c\\r\end{array}\right]=\left[\begin{array}{c}0\\15\end{array}\right]$

So,

• $\left[\begin{array}{cc}c&b\\r&q\end{array}\right]=\left[\begin{array}{cc}0&-1\\15&1\end{array}\right]$

And,

• $\left[\begin{array}{cc}a&c\\p&r\end{array}\right]=\left[\begin{array}{cc}2&0\\1&15\end{array}\right]$

Hence,

$\longrightarrow x=\dfrac{0\times1+1\times15}{2\times1+1\times1}$

$\longrightarrow x=\dfrac{0+15}{2+1}$

$\longrightarrow\underline{\underline{x=5}}$

And,

$\longrightarrow y=\dfrac{ar-cp}{aq-bp}$

$\longrightarrow y=\dfrac{2\times15-0\times1}{2\times1+1\times1}$

$\longrightarrow y=\dfrac{30-0}{2+1}$

$\longrightarrow\underline{\underline{y=10}}$

Thus solved!