Question:

Two cars that are 600 km apart are moving towards each other. Their speeds differ by 6 km per hour and the cars are 123 km apart after $4\frac{1}{2}$ hours. Find the speed of each car.

Difficulty: Easy

Solution:

Let the speed of the two cars be $x$ and $y$ km/h.

According to the given conditions:

The speed of the two cars differ by 6 km/h

$x-y=6 \quad\quad\quad\quad\quad...~(1)$

We know that, Distace Covered $=$ Speed $\times$ Time.

Distance Covered by Car with $x$ Speed $= x \times 4\frac{1}{2} = 4\frac{1}{2}x$

Distance Covered by Car with $y$ Speed $= y \times 4\frac{1}{2} = 4\frac{1}{2}y$

Total Distance Covered by Both Cars $= 600 - 123 = 477$

Hence

$4\frac{1}{2}x+4\frac{1}{2}y=477$

$\frac{9}{2}x+\frac{9}{2}y=477$

$\frac{9}{2}\left( x+y \right)=477$

$x+y=\frac{477\times 2}{9}$

$x+y=\frac{954}{9}$

$x+y=106 \quad\quad\quad\quad\quad...~(2)$

Method 1: Matrix Inversion Method

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

$AX = B$

$\left[ \begin{matrix} 1 & -1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 6 \\ 106 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 1 & -1 \\ 1 & 1 \\ \end{matrix} \right|=1\times 1-1\times \left( -1 \right)=1+1=2~\ne ~0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} 6 \\ 106 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} 6 \\ 106 \\ \end{matrix} \right]$

= $\frac{1}{2}\left[ \begin{matrix} 1 & 1 \\ -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 6 \\ 106 \\ \end{matrix} \right]$

= $\frac{1}{2}\left[ \begin{matrix} 1~\times 6+1\times 106 \\ -1~\times 6+1\times 106 \\ \end{matrix} \right]$

= $\frac{1}{2}\left[ \begin{matrix} 6+106 \\ -6+106 \\ \end{matrix} \right]$

= $\frac{1}{2}\left[ \begin{matrix} 112 \\ 100 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$= $\left[ \begin{matrix} 56 \\ 50 \\ \end{matrix} \right]$

Therefore $x=56$ and $y=50$

Hence the speed of the two cars is 56 km/h and 50 km/h respectively.

Method 2: Cramer’s Rule

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

$AX = B$

$\left[ \begin{matrix} 1 & -1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 6 \\ 106 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 1 & -1 \\ 1 & 1 \\ \end{matrix} \right|=1\times 1-1\times \left( -1 \right)=1+1=2~\ne ~0$

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

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

${{A}_{x}}=\left[ \begin{matrix} 6 & -1 \\ 106 & 1 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 6 & -1 \\ 106 & 1 \\ \end{matrix} \right|=6~\times 1-\left( -1 \right)~\times \left( 106 \right)=6+106=112$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{112}{2}=56$

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

${{A}_{y}}=\left[ \begin{matrix} 1 & 6 \\ 1 & 106 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 1 & 6 \\ 1 & 106 \\ \end{matrix} \right|=1~\times 106-6~\times 1=106-6=100$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{100}{2}=50$

Hence the speed of the two cars is 56 km/h and 50 km/h respectively.