Chapter 7.5, Problem 59E

a.

To determine

To write: System of equations as a matrix equation.

Answer to Problem 59E

The matrix equation of given system of equations are

  $\begin{array}{l}AX=B\\ \left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}4\\ 0\end{array}\right]\end{array}$

Explanation of Solution

Given:

The system of equations is :

  $\begin{array}{l}-x_1+x_2=4\\ -2x_1+x_2=0\end{array}$

Concept Used:

A system of equation

  $\begin{array}{l}a{x}_1+b{x}_2=c\\ d{x}_1+e{x}_2=f\end{array}$

can be written in $AX=B$ form as $\left[\begin{array}{cc}a& b\\ d& e\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}c\\ f\end{array}\right]$

Where $A=\left[\begin{array}{cc}a& b\\ d& e\end{array}\right],X=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right],B=\left[\begin{array}{c}c\\ f\end{array}\right]$

Calculation:

The system of equations is

  $\begin{array}{l}-x_1+x_2=4\\ -2x_1+x_2=0\\ AX=B\\ \left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}4\\ 0\end{array}\right]\\ \text{Where, }A=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right],X=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\text{ and }B=\left[\begin{array}{c}4\\ 0\end{array}\right]\end{array}$

Conclusion:

The matrix equation of given system of equations is

  $\begin{array}{l}AX=B\\ \left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}4\\ 0\end{array}\right]\end{array}$

b.

To determine

To find: use Gauss-Jordan elimination to solve for matrix $X$ .

Answer to Problem 59E

The matrix $X=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}4\\ 8\end{array}\right]$

Explanation of Solution

Given:

The system of equations is :

  $\begin{array}{l}-x_1+x_2=4\\ -2x_1+x_2=0\end{array}$

Concept Used:

Gauss - Jordan elimination method is used to solve system of linear equation by converting augmented matrix in reduced row echelon form.

Calculation:

The system of equations is :

  $\begin{array}{l}-x_1+x_2=4\\ -2x_1+x_2=0\end{array}$

Augmented matrix : $\left[\begin{array}{cccc}-1& 1& ⋮& 4\\ -2& 1& ⋮& 0\end{array}\right]$

Now reducing the augmented matrix in reduced row echelon form.

  $\begin{array}{l}\left[\begin{array}{cccc}-1& 1& ⋮& 4\\ -2& 1& ⋮& 0\end{array}\right]\\ R_1\to \left(-R_1\right)\\ \left[\begin{array}{cccc}1& -1& ⋮& -4\\ -2& 1& ⋮& 0\end{array}\right]\\ R_2\to R_2+2R_1\\ \left[\begin{array}{cccc}1& -1& ⋮& -4\\ 0& -1& ⋮& -8\end{array}\right]\\ R_2\to -\left(R_2\right)\\ \left[\begin{array}{cccc}1& -1& ⋮& -4\\ 0& 1& ⋮& 8\end{array}\right]\\ R_1\to R_1+R_2\\ \left[\begin{array}{cccc}1& 0& ⋮& 4\\ 0& 1& ⋮& 8\end{array}\right]\end{array}$

Now, system of linear equation according to reduced row echelon form

  $\begin{array}{l}{x}_1=4\\ x_2=8\end{array}$

Conclusion:

Therefore, the matrix $X=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}4\\ 8\end{array}\right]$

c.

To determine

Explanation of Solution

Given:

The system of equations is :

  $\begin{array}{l}-x_1+x_2=4\\ -2x_1+x_2=0\end{array}$

Graph:

  

Interpretation:

From graph it is clear that two lines intersect at point $\left(4,8\right)$

Therefore, it is verified that the system of equation

  $\begin{array}{l}-x_1+x_2=4\\ -2x_1+x_2=0\end{array}$

Has solution

  $\begin{array}{l}{x}_1=4\\ x_2=8\end{array}$

Therefore, the matrix $X=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}4\\ 8\end{array}\right]$