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

## Question:

Write and row reduce the augmented matrix to find the general solution of

{eq}2x - z = 2 \\ 6x + 5y + 3z = 7 \\ 2x - y = 4 {/eq}

## Row Reduced Form

If we perform row operations on a matrix, the goal is to put the matrix into row reduced form. This means that the matrix matches the identify matrix as much as possible. If this is a matrix representing a linear system, we can use the row reduced matrix to identify the solution if we have the identity matrix on the left and a column of numbers representing the solution on the right.

Let's define the augmented matrix that represents this linear system. To do so, we need to construct a square matrix that consists of our coefficients on each variable. We will then add one more column to the right containing the numbers after the equals signs.

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

We can perform elementary row operations on this matrix with the goal of getting this into reduced echelon form. If we can do this, then we can isolate the solution or solutions to this system.

{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_3\\ \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}

Therefore, we have the following solution to this system.

{eq}x = 1.5\\ y = -1\\ z = 1 {/eq}