# Solving a System of Linear Equations In Exercise, (a) write the system of linear equations as a...

## Question:

(a) Write the system of linear equations as a matrix equation, {eq}AX = B {/eq}, and

(b) Use Gauss-Jordan elimination on {eq}\left[A\cdot B\right] {/eq} to solve for the matrix {eq}X {/eq}.

{eq}\left\{ \begin{aligned} {{x}_{1}}-5{{x}_{2}}+2{{x}_{3}}&=-20 \\ -3{{x}_{1}}+{{x}_{2}}-{{x}_{3}}&=8 \\ \,\,\,\,\,\,-2{{x}_{2}}+5{{x}_{3}}&=-16 \\ \end{aligned} \right. {/eq}

## Expanded Matrix:

The extended matrix in algebra is very important. For example we can use the extended matrix to find the inverse of a matrix. We can also use the extended matrix to find the solution of a system of linear equations, specifically if we apply the Gauss-Jordan method.

Part (a) Write the system of linear equations as a matrix equation, AX=B

\begin{align} & \begin{bmatrix} x_{1}-5x_{2}+2x_{3}=-20 \\ -3x_{1}+x_{2}-x_{3}=8\\ -2x_{2}+5x_{3}=-16\end{bmatrix} && \bigg \{ \text{ System of linear equations } \bigg \} \\[0.3cm] A &= \begin{bmatrix}1 &-5& 2\\ -3& 1& -1\\ 0& -2& 5\end{bmatrix} && \bigg \{ \text{ Coefficient matrix } \bigg \} \\[0.3cm] B &= \begin{bmatrix}-20\\ 8\\ -16\end{bmatrix} && \bigg \{ \text{ Matrix of independent terms } \bigg \} \\[0.3cm] X &= \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} && \bigg \{ \text{ Matrix of the variables } \bigg \} \\[0.3cm] \therefore \begin{bmatrix}1 &-5& 2\\ -3& 1& -1\\ 0& -2& 5\end{bmatrix} \cdot \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} &= \begin{bmatrix}-20\\ 8\\ -16\end{bmatrix} \\[0.3cm] \end{align}

Part (b) With the help of Gauss-Jordan elimination on {eq}\, \left[A\cdot B\right] \, {/eq} solve the matrix {eq}\, X \, {/eq}.

Perform equivalent transformations on the extended matrix of the system.

\begin{align} & \begin{bmatrix}1&-5&2&|&-20\\ -3&1&-1&|&8\\ 0&-2&5&|&-16\end{bmatrix}&& \bigg \{ \text{ Extended matrix } \bigg \} \\[0.3cm] & R_2\:\leftarrow \: 3R_1+R_2 && \bigg \{ \text{ Equivalent transformation } \bigg \} \\[0.3cm] & \begin{bmatrix}1&-5&2&|&-20\\ 0&-14&5&|&-52\\ 0&-2&5&|&-16\end{bmatrix}\\[0.3cm] & R_3\:\leftarrow \: R_2-7R_3 && \bigg \{ \text{ Equivalent transformation } \bigg \} \\[0.3cm] & \begin{bmatrix}1&-5&2&|&-20\\ 0&-14&5&|&-52\\ 0&0&-30&|&60\end{bmatrix}\\[0.3cm] & R_3\:\leftarrow \: -\frac{1}{30}R_3 && \bigg \{ \text{ Equivalent transformation } \bigg \} \\[0.3cm] & \begin{bmatrix}1&-5&2&|&-20\\ 0&-14&5&|&-52\\ 0&0&1&|&-2\end{bmatrix}\\[0.3cm] & R_2\:\leftarrow \: 5R_3-R_2 && \bigg \{ \text{ Equivalent transformation } \bigg \} \\[0.3cm] & \begin{bmatrix}1&-5&2&|&-20\\ 0&14&0&|&42\\ 0&0&1&|&-2\end{bmatrix}\\[0.3cm] & R_2\:\leftarrow \: \frac{1}{14}R_2 && \bigg \{ \text{ Equivalent transformation } \bigg \} \\[0.3cm] & \begin{bmatrix}1&-5&2&|&-20\\ 0&1&0&|&3\\ 0&0&1&|&-2\end{bmatrix}\\[0.3cm] & R_1\:\leftarrow \: R_1+5R_2 && \bigg \{ \text{ Equivalent transformation } \bigg \} \\[0.3cm] & \begin{bmatrix}1&0&2&|&-5\\ 0&1&0&|&3\\ 0&0&1&|&-2\end{bmatrix}\\[0.3cm] & R_1\:\leftarrow \: R_1-2R_3 && \bigg \{ \text{ Equivalent transformation } \bigg \} \\[0.3cm] & \begin{bmatrix}1&0&0&|&-1\\ 0&1&0&|&3\\ 0&0&1&|&-2\end{bmatrix}\\[0.3cm] \therefore & x_1=-1,\:x_2=3,\:x_3=-2\\[0.3cm] \end{align}

• The system is compatible, it has a unique solution.

{eq}S= \bigg\{ x_1=-1 \, , \, x_2=3 \, , \, x_3=-2 \bigg\} {/eq}