Question

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.﻿{w+x+y+z=42w+x−2y−z=0w−2x−y−2z=−23w+2x+y+3z=4\begin{cases}
w + x + y + z = 4 \
2w + x - 2y - z = 0 \
w - 2x - y - 2z = -2 \
3w + 2x + y + 3z = 4
\end{cases}⎩⎨⎧​w+x+y+z=42w+x−2y−z=0w−2x−y−2z=−23w+2x+y+3z=4​

Accepted Answer

Write the system of equations as an augmented matrix. The coefficients of the variables $$ w, x, y, z $$ and the constants on the right side form the matrix:

$$\left[\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 4 \ 2 & 1 & -2 & -1 & 0 \ 1 & -2 & -1 & -2 & -2 \ 3 & 2 & 1 & 3 & 4 \end{array}ight]$$ Use Gaussian elimination to transform the matrix into an upper triangular form (row echelon form). Start by using the first row to eliminate the $$ w -$$terms in rows 2, 3, and 4 by performing row operations such as:

- Replace row 2 with (row 2) - 2*(row 1)
- Replace row 3 with (row 3) - (row 1)
- Replace row 4 with (row 4) - 3*(row 1) Next, use the second row to eliminate the $$ x -$$terms in rows 3 and 4. This involves finding a suitable multiplier to subtract a multiple of row 2 from rows 3 and 4 to create zeros below the pivot in the second column. Then, use the third row to eliminate the $$ y -$$term in row 4 by subtracting a multiple of row 3 from row 4 to get a zero in the third column of the fourth row. Once the matrix is in upper triangular form, use back-substitution to solve for the variables starting from the last row and moving upwards. This means solving for $$ z $$ first, then $$ y $$, then $$ x $$, and finally $$ w .$$

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$\begin{cases}\) w + x + y + z = 4 \\ 2w + x - 2y - z = 0 \\ w - 2x - y - 2z = -2 \\ 3w + 2x + y + 3z = 4 \(\end{cases}$

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Concetti chiave

Systems of Linear Equations

Matrix Representation of Systems

Gaussian Elimination and Gauss-Jordan Elimination