Question

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.﻿{x+y+z=4x−y−z=0x−y+z=2\begin{cases}
x + y + z = 4 \
x - y - z = 0 \
x - y + z = 2
\end{cases}⎩⎨⎧​x+y+z=4x−y−z=0x−y+z=2​

Accepted Answer

Write the system of equations as an augmented matrix. For the system $$ \begin{cases} x + y + z = 4 \ x - y - z = 0 \ x - y + z = 2 \end{cases} $$, the augmented matrix is:

$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 4 \
1 & -1 & -1 & 0 \
1 & -1 & 1 & 2
\end{array}ight]
$$ Use Gaussian elimination to create zeros below the first pivot (the element in the first row, first column). Subtract the first row from the second and third rows:

- Row 2 = Row 2 - Row 1
- Row 3 = Row 3 - Row 1 After these row operations, the matrix will have zeros in the first column below the pivot. Next, focus on the second row and use it to create a zero below its pivot (second row, second column) by manipulating the third row accordingly. Once the matrix is in upper triangular form (all zeros below the main diagonal), use back-substitution to solve for the variables starting from the last row upwards. Alternatively, you can continue with Gauss-Jordan elimination by creating zeros above and below each pivot to get the matrix into reduced row echelon form, from which the solutions for $$x$$, $$y$$, and $$z$$ can be read directly.

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$\begin{cases}\)x + y + z = 4 \(\x\) - y - z = 0 \(\x\) - y + z = 2\(\end{cases}$

Verified video answer for a similar problem:

Key Concepts

Systems of Linear Equations

Gaussian Elimination with Back-Substitution

Gauss-Jordan Elimination

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.{x+y+z=4x−y−z=0x−y+z=2\(\begin{cases}\)x + y + z = 4 \(\x\) - y - z = 0 \(\x\) - y + z = 2\(\end{cases}\)⎩⎨⎧x+y+z=4x−y−z=0x−y+z=2