# Solve by finding the reduced row echelon form (rref) by hand. Label every step x - 3z = -2 3x +...

## Question:

Solve by finding the reduced row echelon form (rref) by hand. Label every step

{eq}x - 3z = -2 {/eq}

{eq}3x + y -2z = 5 {/eq}

{eq}2x + 2y + z = 4{/eq}

## Row Reduced Echelon Form

When solving a system of linear equations using Gaussian Elimination, the goal is to put that matrix into row reduced echelon form. In this form, the matrix appears to be the square identity matrix on the left side. If there are additional columns, which there are if we have exactly the same number of equations as variables, then the solutions to the system are listed in that final column

## Answer and Explanation:

Let's begin by constructing the augmented matrix that represents this linear system.

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

Let's apply row operations to this augmented matrix in the aim to put it into row reduced echelon form. This will isolate the solutions to this system of equations in the final column.

{eq}\begin{align*} \begin{bmatrix} 1 & 0 & -3 & -2\\ 3 & 1 & -2 & 5\\ 2 & 2 & 1 & 4\end{bmatrix} & R_2 = R_2 - 3R_1\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 1 & 7& 11\\ 2 & 2 & 1 & 4\end{bmatrix}& R_3 = R_3 - 2R_1\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 1 & 7& 11\\ 0 & 2 & 7 & 8\end{bmatrix}& R_3 = R_3 - 2R_2\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 1 & 7& 11\\ 0 & 0 & -7 & -14 \end{bmatrix}&R_3 = -\frac{1}{7}R_3\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 1 & 7& 11\\ 0 & 0 & 1& 2 \end{bmatrix}& R_2 = R_2 - 7R_3\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 1 & 0& -3 \\ 0 & 0 & 1& 2 \end{bmatrix}& R_1 = R_1 + 3R_3\\ \begin{bmatrix} 1 & 0 &0 &4\\ 0 & 1 & 0& -3 \\ 0 & 0 & 1& 2 \end{bmatrix}& \end{align*} {/eq}

The solution set is therefore: {eq}x = 4 {/eq}, {eq}y = -3 {/eq}, {eq}z =2 {/eq}.

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from Algebra II Textbook

Chapter 10 / Lesson 6