Mashaal Masha

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 $y=2$