### Table of Contents

Question:

Use matrices, if possible, to solve the following systems of linear equations by:

 (A) the matrix inversion method (B) the Cramer’s rule

 (i)        $2x - 2y = 4$             $3x + 2y = 6$ (ii)       $2x + y = 3$                   $6x + 5y = 1$ (iii)      $4x + 2y = 8$             $3x - y = -1$ (iv)      $3x - 2y = -6$             $5x - 2y = -10$ (v)        $3x - 2y = 4$                $-6x + 4y = 7$ (vi)       $4x + y = 9$                   $-3x - y = -5$ (vii)      $2x - 2y = 4$                 $-5x - 2y = -10$ (viii)    $3x - 4y = 4$             $x + 2y = 8$
Difficulty: Easy

Solution:

(i)        $2x - 2y = 4$

$3x + 2y = 6$

(A) the matrix inversion method

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & -2 \\ 3 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{matrix} 4 \\ 6 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right| = \left| \begin{matrix} 2 & -2 \\ 3 & 2 \\ \end{matrix} \right| = 2\times 2-3\times \left( -2 \right) = 4+6 = 10\ne 0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{A}^{-1}}\left[ \begin{matrix} 4 \\ 6 \\ \end{matrix} \right]$

= $\frac{1}{\left| A \right|}~Adj~A\left[ \begin{matrix} 4 \\ 6 \\ \end{matrix} \right]$

= $\frac{1}{10}\left[ \begin{matrix} 2 & 2 \\ -3 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 4 \\ 6 \\ \end{matrix} \right]$

= $\frac{1}{10}\left[ \begin{matrix} 2\times 4+2\times 6 \\ -3\times 4+2\times 6 \\ \end{matrix} \right]$

= $\frac{1}{10}\left[ \begin{matrix} 8+12 \\ -12+12 \\ \end{matrix} \right]$

= $\frac{1}{10}\left[ \begin{matrix} 20 \\ 0 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 2 \\ 0 \\ \end{matrix} \right]$

Therefore $x=2$ and $y=0$

(B) the Cramer’s rule

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & -2 \\ 3 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{matrix} 4 \\ 6 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|$ = $\left| \begin{matrix} 2 & -2 \\ 3 & 2 \\ \end{matrix} \right| = 2\times 2-3\times \left( -2 \right) = 4+6 = 10\ne 0$

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

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

${{A}_{x}}$ = $\left[ \begin{matrix} 4 & -2 \\ 6 & 2 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|$ = $\left| \begin{matrix} 4 & -2 \\ 6 & 2 \\ \end{matrix} \right|=4~\times 2-6~\times \left( -2 \right)=8+12=20\ne 0$

$x$ = $\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{20}{10}=2$

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

${{A}_{y}}$ = $\left[ \begin{matrix} 2 & 4 \\ 3 & 6 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|$ = $\left| \begin{matrix} 2 & 4 \\ 3 & 6 \\ \end{matrix} \right|=2~\times 6-3~\times 4=12-12=0$

$y$ = $\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{0}{10}=0$

Therefore $x=2$ and $y=0$

(ii)       $2x + y = 3$

$6x + 5y = 1$

(A) the matrix inversion method

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & 1 \\ 6 & 5 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 3 \\ 1 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 2 & 1 \\ 6 & 5 \\ \end{matrix} \right|=2~\times 5-1~\times 6=10-6=4\ne 0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} 3 \\ 1 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} 3 \\ 1 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 5 & -1 \\ -6 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 3 \\ 1 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 5\times 3+\left( -1 \right)\times 1 \\ -6\times 3+2\times 1 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 14 \\ -16 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} \frac{7}{2} \\ -4 \\ \end{matrix} \right]$

Therefore $x=\frac{7}{2}$ and $y=-4$

(B) the Cramer’s rule

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & 1 \\ 6 & 5 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 3 \\ 1 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 2 & 1 \\ 6 & 5 \\ \end{matrix} \right|=2~\times 5-1~\times 6=10-6=4\ne 0$

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

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

${{A}_{x}}$ = $\left[ \begin{matrix} 3 & 1 \\ 1 & 5 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 3 & 1 \\ 1 & 5 \\ \end{matrix} \right|=3~\times 5-1~\times 1=15-1=14$

$x$ = $\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{14}{4}=\frac{7}{2}$

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

${{A}_{y}}$ = $\left[ \begin{matrix} 2 & 3 \\ 6 & 1 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 2 & 3 \\ 6 & 1 \\ \end{matrix} \right|=2~\times 1-3~\times 6=2-18=-16$

$y$ = $\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{-16}{4}=-4$

Therefore $x=\frac{7}{2}$ and $y=-4$

(iii)      $4x + 2y = 8$

$3x - y = -1$

(A) the matrix inversion method

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 4 & 2 \\ 3 & -1 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 8 \\ -1 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 4 & 2 \\ 3 & -1 \\ \end{matrix} \right|=4~\times \left( -1 \right)-2~\times 3=-4-6=-10\ne 0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{A}^{-1}}\left[ \begin{matrix} 8 \\ -1 \\ \end{matrix} \right]$

= $\frac{1}{\left| A \right|}~Adj~A$ $\left[ \begin{matrix} 8 \\ -1 \\ \end{matrix} \right]$

= $\frac{1}{-10}\left[ \begin{matrix} -1 & -2 \\ -3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 8 \\ -1 \\ \end{matrix} \right]$

= $\frac{1}{-10}\left[ \begin{matrix} \left( -1 \right)\times 8+\left( -2 \right)\times \left( -1 \right) \\ -3\times 8+4\times \left( -1 \right) \\ \end{matrix} \right]$

= $\frac{1}{-10}\left[ \begin{matrix} -8+2 \\ -24-4 \\ \end{matrix} \right]$

= $\frac{1}{-10}\left[ \begin{matrix} -6 \\ -28 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{-6}{-10} \\ \frac{-28}{-10} \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} \frac{3}{5} \\ \frac{14}{5} \\ \end{matrix} \right]$

Therefore $x=\frac{3}{5}$ and $y=\frac{14}{5}$

(B) the Cramer’s rule

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 4 & 2 \\ 3 & -1 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 8 \\ -1 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 4 & 2 \\ 3 & -1 \\ \end{matrix} \right|=4~\times \left( -1 \right)-2~\times 3=-4-6=-10\ne 0$

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

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

${{A}_{x}}$ = $\left[ \begin{matrix} 8 & 2 \\ -1 & -1 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 8 & 2 \\ -1 & -1 \\ \end{matrix} \right|=8~\times \left( -1 \right)-2~\times \left( -1 \right)=-8+2=-6$

$x$ = $\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{-6}{-10}=\frac{3}{5}$

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

${{A}_{y}}$ = $\left[ \begin{matrix} 4 & 8 \\ 3 & -1 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 4 & 8 \\ 3 & -1 \\ \end{matrix} \right|=4~\times \left( -1 \right)-3~\times 8=-4-24=-28$

$y$ = $\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{-28}{-10}=\frac{14}{5}$

Therefore $x=\frac{3}{5}$ and $y=\frac{14}{5}$

(iv)      $3x - 2y = -6$

$5x - 2y = -10$

(A) the matrix inversion method

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 3 & -2 \\ 5 & -2 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -6 \\ -10 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 3 & -2 \\ 5 & -2 \\ \end{matrix} \right|=3~\times \left( -2 \right)-5~\times \left( -2 \right)=-6+10=4\ne 0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} -6 \\ -10 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} -6 \\ -10 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} -2 & 2 \\ -5 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} -6 \\ -10 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} \left( -2 \right)\times \left( -6 \right)+\left( 2 \right)\times \left( -10 \right) \\ \left( -5 \right)\times \left( -6 \right)+3\times \left( -10 \right) \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 12-20 \\ 30-30 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} -8 \\ 0 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -2 \\ 0 \\ \end{matrix} \right]$

Therefore $x=-2$ and $y=0$

(B) the Cramer’s rule

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 3 & -2 \\ 5 & -2 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -6 \\ -10 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 3 & -2 \\ 5 & -2 \\ \end{matrix} \right|=3~\times \left( -2 \right)-5~\times \left( -2 \right)=-6+10=4\ne 0$

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

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

${{A}_{x}}=\left[ \begin{matrix} -6 & -2 \\ -10 & -2 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} -6 & -2 \\ -10 & -2 \\ \end{matrix} \right|=-6~\times \left( -2 \right)-\left( -10 \right)~\times \left( -2 \right)=12-20=-8$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{-8}{4}=-2$

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

${{A}_{y}}=\left[ \begin{matrix} 3 & -6 \\ 5 & -10 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 3 & -6 \\ 5 & -10 \\ \end{matrix} \right|=3~\times \left( -10 \right)-5~\times \left( -6 \right)=-30+30=0$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{0}{4}=0$

Therefore $x=-2$ and $y=0$

(v)        $3x - 2y = 4$

$-6x + 4y = 7$

(A) the matrix inversion method

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 3 & -2 \\ -6 & 4 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 \\ 7 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 3 & -2 \\ -6 & 4 \\ \end{matrix} \right|=3~\times 4-\left( -6 \right)~\times \left( -2 \right)=12-12=0$

The coefficient matrix $A$ is non-singular therefore the solution of the linear equations does not exist.

(B) the Cramer’s rule

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 3 & -2 \\ -6 & 4 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 \\ 7 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 3 & -2 \\ -6 & 4 \\ \end{matrix} \right|=3~\times 4-\left( -6 \right)~\times \left( -2 \right)=12-12=0$

The coefficient matrix $A$ is non-singular therefore the solution of the linear equations does not exist.

(vi)       $4x + y = 9$

$-3x - y = -5$

(A) the matrix inversion method

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 4 & 1 \\ -3 & -1 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 9 \\ -5 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 4 & 1 \\ -3 & -1 \\ \end{matrix} \right|=4~\times \left( -1 \right)-~1\times \left( -3 \right)=-4+3=-1\ne 0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 9 \\ -5 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} 9 \\ -5 \\ \end{matrix} \right]$

= $\frac{1}{-1}\left[ \begin{matrix} -1 & -1 \\ 3 & 4 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 9 \\ -5 \\ \end{matrix} \right]$

= $\frac{1}{-1}\left[ \begin{matrix} \left( -1 \right)\times 9+\left( -1 \right)\times \left( -5 \right) \\ 3\times 9+4\times \left( -5 \right) \\ \end{matrix} \right]$

= $-1\left[ \begin{matrix} -9+5 \\ 27-20 \\ \end{matrix} \right]$

= $-1\left[ \begin{matrix} -4 \\ 7 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 \\ -7 \\ \end{matrix} \right]$

Therefore $x=4$ and $y=-7$

(B) the Cramer’s rule

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 4 & 1 \\ -3 & -1 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 9 \\ -5 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 4 & 1 \\ -3 & -1 \\ \end{matrix} \right|=4~\times \left( -1 \right)-~1\times \left( -3 \right)=-4+3=-1\ne 0$

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

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

${{A}_{x}}=\left[ \begin{matrix} 9 & 1 \\ -5 & -1 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 9 & 1 \\ -5 & -1 \\ \end{matrix} \right|=9~\times \left( -1 \right)-\left( -5 \right)~\times 1=-9+5=-4$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{-4}{-1}=4$

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

${{A}_{y}}=\left[ \begin{matrix} 4 & 9 \\ -3 & -5 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 4 & 9 \\ -3 & -5 \\ \end{matrix} \right|=4~\times \left( -5 \right)-\left( -3 \right)~\times 9=-20+27=7$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{7}{-1}=-7$

Therefore $x=4$ and $y=-7$

(vii)      $2x - 2y = 4$

$-5x - 2y = -10$

(A) the matrix inversion method

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & -2 \\ -5 & -2 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 \\ -10 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -2 \\ -5 & -2 \\ \end{matrix} \right|=2~\times \left( -2 \right)-\left( -5 \right)~\times \left( -2 \right)=-4-10=-14\ne 0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 4 \\ -10 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} 4 \\ -10 \\ \end{matrix} \right]$

= $\frac{1}{-14}\left[ \begin{matrix} -2 & 2 \\ 5 & 2 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 4 \\ -10 \\ \end{matrix} \right]$

= $\frac{1}{-14}\left[ \begin{matrix} \left( -2 \right)\times 4+2\times \left( -10 \right) \\ 5\times 4+2\times \left( -10 \right) \\ \end{matrix} \right]$

= $\frac{1}{-14}\left[ \begin{matrix} -8-20 \\ 20-20 \\ \end{matrix} \right]$

= $\frac{1}{-14}\left[ \begin{matrix} -28 \\ 0 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 2 \\ 0 \\ \end{matrix} \right]$

Therefore $x=2$ and $y=0$

(B) the Cramer’s rule

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & -2 \\ -5 & -2 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 \\ -10 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -2 \\ -5 & -2 \\ \end{matrix} \right|=2~\times \left( -2 \right)-\left( -5 \right)~\times \left( -2 \right)=-4-10=-14\ne 0$

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

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

${{A}_{x}}=\left[ \begin{matrix} 4 & -2 \\ -10 & -2 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 4 & -2 \\ -10 & -2 \\ \end{matrix} \right|=4~\times \left( -2 \right)-\left( -10 \right)~\times \left( -2 \right)=-8-20=-28$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{-28}{-14}=2$

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

${{A}_{y}}=\left[ \begin{matrix} 2 & 4 \\ -5 & -10 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 2 & 4 \\ -5 & -10 \\ \end{matrix} \right|=2~\times \left( -10 \right)-\left( -5 \right)\times 4=-20+20=0$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{0}{-14}=0$

Therefore $x=2$ and $y=0$

(viii)    $3x - 4y = 4$

$x + 2y = 8$

(A) the matrix inversion method

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 3 & -4 \\ 1 & 2 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 \\ 8 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 3 & -4 \\ 1 & 2 \\ \end{matrix} \right|=3~\times 2-\left( -4 \right)~\times 1=6+4=10\ne 0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 4 \\ 8 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} 4 \\ 8 \\ \end{matrix} \right]$

= $\frac{1}{10}\left[ \begin{matrix} 2 & 4 \\ -1 & 3 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 4 \\ 8 \\ \end{matrix} \right]$

= $\frac{1}{10}\left[ \begin{matrix} 2\times 4+4\times 8 \\ \left( -1 \right)\times 4+3\times 8 \\ \end{matrix} \right]$

= $\frac{1}{10}\left[ \begin{matrix} 8+32 \\ -4+24 \\ \end{matrix} \right]$

= $\frac{1}{10}\left[ \begin{matrix} 40 \\ 20 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 \\ 2 \\ \end{matrix} \right]$

Therefore $x=4$ and $y=2$

(B) the Cramer’s rule

Write the system of linear equations in matrix form.

$AX = B$

$\left[ \begin{matrix} 3 & -4 \\ 1 & 2 \\ \end{matrix} \right]$ $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 4 \\ 8 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 3 & -4 \\ 1 & 2 \\ \end{matrix} \right|=3~\times 2-\left( -4 \right)~\times 1=6+4=10\ne 0$

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

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

${{A}_{x}}=\left[ \begin{matrix} 4 & -4 \\ 8 & 2 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 4 & -4 \\ 8 & 2 \\ \end{matrix} \right|=4~\times 2-8~\times \left( -4 \right)=8+32=40$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{40}{10}=4$

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

${{A}_{y}}=\left[ \begin{matrix} 3 & 4 \\ 1 & 8 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 3 & 4 \\ 1 & 8 \\ \end{matrix} \right|=3~\times 8-~1~\times 4=24-4=20$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{20}{10}=2$

Therefore $x=4$ and $y=2$

Question:

The length of a rectangle is 4 times its width. The perimeter of the rectangle is 150 cm. Find the dimensions of the rectangle.

Difficulty: Easy

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

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

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Question:

The third angle of an isosceles triangle is $16^{\circ}$ less than the sum of the two equal angles. Find three angles of the triangle.

Difficulty: Easy

Solution:

In an isosceles triangle, two angles are of equal measure. Let each equal angle be of $x$ degrees and the third angle be of $y$ degrees.

According to the given conditions:

The third angle $y$ is $16^{\circ}$ less than $2x$ (the sum of the two equal angles).

$2x-16=y$

$2x-y=16 \quad\quad\quad\quad\quad...~(1)$

The sum of all the three angles of a triangle is $180^{\circ}$

$x+x+y=180$

$2x+y=180 \quad\quad\quad\quad\quad...~(2)$

Method 1: Matrix Inversion Method

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

$AX = B$

$\left[ \begin{matrix} 2 & -1 \\ 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -1 \\ 2 & 1 \\ \end{matrix} \right|=2\times 1-\left( -1 \right)\times 2=2+2=4~\ne ~0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 1 & 1 \\ -2 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 1~\times 16+1\times 180 \\ \left( -2 \right)\times 16+2\times 180 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 16+180 \\ -32+360 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 196 \\ 328 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{matrix} 49 \\ 82 \\ \end{matrix} \right]$

Therefore $x=49$ and $y=82$

Hence the angles of the isosceles triangle are $49^{\circ}$, $49^{\circ}$ and $82^{\circ}$.

Method 2: Cramer’s Rule

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

$AX = B$

$\left[ \begin{matrix} 2 & -1 \\ 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -1 \\ 2 & 1 \\ \end{matrix} \right|=2\times 1-\left( -1 \right)\times 2=2+2=4~\ne ~0$

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

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

${{A}_{x}}=\left[ \begin{matrix} 16 & -1 \\ 180 & 1 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 16 & -1 \\ 180 & 1 \\ \end{matrix} \right|=16\times 1-\left( -1 \right)\times 180=16+180=196$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{196}{4}=49$

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

${{A}_{y}}=\left[ \begin{matrix} 2 & 16 \\ 2 & 180 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 2 & 16 \\ 2 & 180 \\ \end{matrix} \right|=2\times 180-16~\times 2=360-32=328$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{328}{4}=82$

Hence the angles of the isosceles triangle are $49^{\circ}$, $49^{\circ}$ and $82^{\circ}$.

Question:

One acute angle of a right triangle is 12° more than twice the other acute angle. Find the acute angles of the right triangle.

Difficulty: Easy

Solution:

In a Right Triangle one angle is of $90^{\circ}$ and the other two angles are acute angles. Let the two acute angles be of $x$ and $y$ degrees.

According to the given conditions:

One acute angle $x$ is $12^{\circ}$ more than twice the other acute angle $y$.

$2x+12=y$

$2x-y=-12 \quad\quad\quad\quad\quad...~(1)$

In a Right Triangle the sum of the two acute angles is $90^{\circ}$.

$x+y=90 \quad\quad\quad\quad\quad...~(2)$

Method 1: Matrix Inversion Method

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

$AX = B$

$\left[ \begin{matrix} 2 & -1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

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

$X=A^{-1}B$

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -1 \\ 1 & 1 \\ \end{matrix} \right|=2\times 1-\left( -1 \right)\times 1=2+1=3~\ne ~0$

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

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

= $\frac{1}{3}\left[ \begin{matrix} 1 & 1 \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

= $\frac{1}{3}\left[ \begin{matrix} 1~\times \left( -12 \right)+1\times 90 \\ -1~\times \left( -12 \right)+2\times 90 \\ \end{matrix} \right]$

= $\frac{1}{3}\left[ \begin{matrix} -12+90 \\ 12+180 \\ \end{matrix} \right]$

= $\frac{1}{3}\left[ \begin{matrix} 78 \\ 192 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]= \left[ \begin{matrix} 26 \\ 64 \\ \end{matrix} \right]$

Therefore $x=26$ and $y=64$

Hence the acute angles of the Right Triangle are $26^{\circ}$ and $64^{\circ}$

Method 2: Cramer’s Rule

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

$AX = B$

$\left[ \begin{matrix} 2 & -1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -12 \\ 90 \\ \end{matrix} \right]$

According to Cramer's Rule:

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

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

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -1 \\ 1 & 1 \\ \end{matrix} \right|=2\times 1-\left( -1 \right)\times 1=2+1=3~\ne ~0$

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

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

${{A}_{x}}=\left[ \begin{matrix} -12 & -1 \\ 90 & 1 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} -12 & -1 \\ 90 & 1 \\ \end{matrix} \right|=\left( -12 \right)\times 1-\left( -1 \right)\times 90=-12+90=78$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{78}{3}=26$

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

${{A}_{y}}=\left[ \begin{matrix} 2 & -12 \\ 1 & 90 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 2 & -12 \\ 1 & 90 \\ \end{matrix} \right|=90\times 2-\left( -12 \right)~\times 1=180+12=192$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{192}{3}=64$

Hence the acute angles of the Right Triangle are $26^{\circ}$ and $64^{\circ}$

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.

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