# The given matrix is an augmented matrix used in the solution of a system of linear equations. Use...

## Question:

The given matrix is an augmented matrix used in the solution of a system of linear equations. Use the Gauss-Jordan method to solve the system.

{eq}\left.\begin{matrix} 1 &1 &2 \\ 0&3 &1 \\ 0&-2 &4 \end{matrix}\right|\begin{matrix} -1\\ 7\\ 0 \end{matrix} {/eq}

## Gauss-Jordan Elimination

A system of linear equations can be represented as an augmented matrix. This provides a clear and concise way to view this information, and it also provides a method for which we can solve this system. By performing elementary row operations, we can attempt to construct the identity matrix on the left side, which if possible will yield our solutions in the rightmost column.

In order to solve this system, we need to apply elementary row operations to this augmented matrix. We can swap any two rows, multiply rows by a constant, and add multiples of rows to each other. By performing these tasks with an aim to construct the identity matrix, we can follow the procedure below.

{eq}\begin{align*} \begin{bmatrix} 1 &1 &2&-1 \\ 0&3 &1 &7\\ 0&-2 &4 &0 \end{bmatrix} & R_2 = R_2 + R_3\\ \begin{bmatrix} 1 &1 &2&-1 \\ 0&1 &5 &7\\ 0&-2 &4 &0 \end{bmatrix} & R_3 = R_3 + 2R_2\\ \begin{bmatrix}1 &1 &2&-1 \\ 0&1 &5 &7\\ 0&0 &14 &14 \end{bmatrix} & R_3 = \frac{1}{14} R_3\\ \begin{bmatrix} 1 &1 &2&-1 \\ 0&1 &5 &7\\ 0&0 &1 &1\end{bmatrix} & R_2 = R_2 - 5R_3\\ \begin{bmatrix} 1 &1 &2&-1 \\ 0&1 &0 &2\\ 0&0 &1 &1\end{bmatrix} & R_1 = R_1 - R_2\\ \begin{bmatrix} 1 &0 &2&-3 \\ 0&1 &0 &2\\ 0&0 &1 &1\end{bmatrix} &R_1 = R_1 - 2R_3\\ \begin{bmatrix}1 &0 &0&-5 \\ 0&1 &0 &2\\ 0&0 &1 &1\end{bmatrix} \end{align*} {/eq}

Therefore, there is one unique solution to this system: {eq}x = -5 {/eq}, {eq}y =2 {/eq}, and {eq}z = 1 {/eq}. 