Chapter 8.2, Problem 67E

a.

To determine

To find:Write the system of linear equations as a matrix equation $AX=B$ .

Answer to Problem 67E

The given system of linear equations as a matrix equation is

  $\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}5\\ 10\end{array}\right]$

Explanation of Solution

Given information:

The system of linear equations is

  $\{\begin{array}{c}2x_1+3x_2=5\\ x_1+4x_2=10\end{array}$ .

Calculations:

Writing the given system of linear equations as a matrix equation is

  $\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}5\\ 10\end{array}\right]$

Where

  $A=\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]$ the coefficient matrix

  $X=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$ the variable matrix

  $B=\left[\begin{array}{c}5\\ 10\end{array}\right]$ the constant matrix.

b.

To determine

To find:The solution using Gauss-Jordan elimination on $\left[A⋮B\right]$ .

Answer to Problem 67E

The solution of the given system of equations using Gauss-Jordan elimination is $x_1=-2,x_2=3.$

Explanation of Solution

Given information:

The system of linear equations is

  $\{\begin{array}{c}2x_1+3x_2=5\\ x_1+4x_2=10\end{array}$

And its matrix equation is

  $\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}5\\ 10\end{array}\right]$ .

Calculations:

Writing the augmented matrix $\left[A⋮B\right]$ is

  $\left[\begin{array}{cccc}2& 3& :& 5\\ 1& 4& :& 10\end{array}\right]$

Solving it using Gauss-Jordan Elimination method, we have

  $\begin{array}{l}\left[\begin{array}{cccc}1& 4& :& 10\\ 2& 3& :& 5\end{array}\right]\begin{array}{c}\leftarrow \\ \end{array}\begin{array}{c}{R}_1\leftrightarrow R_2\\ \end{array}\\ \left[\begin{array}{cccc}1& 4& :& 10\\ 0& -5& :& -15\end{array}\right]\begin{array}{c}\\ \leftarrow \end{array}\begin{array}{c}\\ R_2-2R_1\end{array}\\ \left[\begin{array}{cccc}1& 4& :& 10\\ 0& 1& :& 3\end{array}\right]\begin{array}{c}\\ \leftarrow \end{array}\begin{array}{c}\\ R_2/-5\end{array}\\ \left[\begin{array}{cccc}1& 0& :& -2\\ 0& 1& :& 3\end{array}\right]\begin{array}{c}\leftarrow \\ \end{array}\begin{array}{c}{R}_1-4R_2\\ \end{array}\end{array}$

Conclusion:

Thus, the solution of the given system of equations is $x_1=-2,x_2=3.$