# Write and row reduce the augmented matrix to find the general solution: 2x - z = 2, 6x + 5y + 3z...

## Question:

Write and row reduce the augmented matrix to find the general solution: 2x - z = 2, 6x + 5y + 3z = 7, 2x - y = 4.

## Gaussian Elimination

If we have a system of linear equations, we can construct and solve an augmented matrix using row operations. This involves multiplying rows by constants, swapping rows around, and adding multiples of rows to each other. The goal is to get this matrix in the form of the identity matrix, with the solution to the system in the final column.

An augmented matrix that represents a linear system is one where each element in the matrix represents the coefficient on a variable and the number after the equals sign represents the element in the final column.

{eq}\begin{bmatrix}2 & 0 & -1 & 2\\ 6 & 5 & 3 & 7\\ 2 & -1 & 0 & 4\end{bmatrix} {/eq}

Let's perform row operations on this matrix in order to find the solutions.

{eq}\begin{align*} \begin{bmatrix}2 & 0 & -1 & 2\\ 6 & 5 & 3 & 7\\ 2 & -1 & 0 & 4\end{bmatrix} & R_2 = R_2 - 3R_1\\ \begin{bmatrix}2 & 0 & -1 & 2\\ 0& 5 & 6 & 1\\ 2 & -1 & 0 & 4\end{bmatrix} & R_3 = R_3 - R_1\\ \begin{bmatrix}2 & 0 & -1 & 2\\ 0& 5 & 6 & 1\\ 0 & -1 & 1 & 2\end{bmatrix} &R_3 = R_2 + 5R_3\\ \begin{bmatrix}2 & 0 & -1 & 2\\ 0& 5 & 6 & 1\\ 0 & 0 & 11 & 11\end{bmatrix} & R_3 = \frac{1}{11} R_3\\ \begin{bmatrix}2 & 0 & -1 & 2\\ 0& 5 & 6 & 1\\ 0 & 0 & 1 & 1\end{bmatrix} &R_2 = R_2 -6R_3\\ \begin{bmatrix}2 & 0 & -1 & 2\\ 0& 5 & 0 & -5\\ 0 & 0 & 1 & 1\end{bmatrix} & R_2 =\frac{1}{5} R_2\\ \begin{bmatrix}2 & 0 & -1 & 2\\ 0& 1 & 0 & -1\\ 0 & 0 & 1 & 1\end{bmatrix} &R_1 = R_1 + R_2\\ \begin{bmatrix}2 & 0 & 0 & 3\\ 0& 1 & 0 & -1\\ 0 & 0 & 1 & 1\end{bmatrix} & R_1 = \frac{1}{2}R_1\\ \begin{bmatrix}1 & 0 & 0 & 1.5\\ 0& 1 & 0 & -1\\ 0 & 0 & 1 & 1\end{bmatrix} \end{align*} {/eq}

The solutions is therefore: {eq}x = 1.5, y = -1, z = 1 {/eq}.