Question

Given $\begin{bmatrix} 1 & -1 & 3 & 6 \ 1 & -2 & 4 & 9 \ -1 & 1 & -3 & -6 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 2 & 3 \ 0 & 1 & -1 & -3 \ 0 & 0 & 0 & 0 \end{bmatrix}$ use the reduced row echelon form above to solve the system $\begin{cases} x - y + 3z &= 6 \ x - 2y + 4z &= 9 \ -x + y - 3z &= -6 \end{cases}$ If necessary, parametrize your answer using the free variables of the system. $\begin{bmatrix} x \ y \ z \end{bmatrix} = $

          Given
$\begin{bmatrix} 1 & -1 & 3 & 6 \ 1 & -2 & 4 & 9 \ -1 & 1 & -3 & -6 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 2 & 3 \ 0 & 1 & -1 & -3 \ 0 & 0 & 0 & 0 \end{bmatrix}$
use the reduced row echelon form above to solve the system
$\begin{cases} x - y + 3z &= 6 \ x - 2y + 4z &= 9 \ -x + y - 3z &= -6 \end{cases}$
If necessary, parametrize your answer using the free variables of the system.
$\begin{bmatrix} x \ y \ z \end{bmatrix} = $

Given
< b m a t r i x >
∼
< b m a t r i x >
use the reduced row echelon form above to solve the system
x - y + 3z = 6 x - 2y + 4z = 9 -x + y - 3z = -6
If necessary, parametrize your answer using the free variables of the system.
< b m a t r i x >
=

Added by Jonathan B.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Willis James

Step 1

First, let's write the system of equations in matrix form: [1 -1 3 | 0] [1 -2 4 | 9] [1 1 3 | -6] Now, let's use row operations to reduce the matrix to its reduced row echelon form: R2 = R2 - R1 R3 = R3 - R1 [1 -1 3 | 0] [0 -1 1 | 9] [0 2 0 | -6] R3 = R3 + Show more…

Show all steps

Thanks for your feedback!

3 0 2 1 0 3 1 ? - 3 3 0 Use the reduced row echelon form above to solve the system: z - y + 3z = z - 2y + 4z = 9 x + y + 3z = -6 If necessary, parametrize your answer using the free variables of the system.