Question

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. ﻿{2w+x−y=3w−3x+2y=−43w+x−3y+z=1w+2x−4y−z=−2\begin{cases}
2w + x - y = 3 \
w - 3x + 2y = -4 \
3w + x - 3y + z = 1 \
w + 2x - 4y - z = -2
\end{cases}⎩⎨⎧​2w+x−y=3w−3x+2y=−43w+x−3y+z=1w+2x−4y−z=−2​

Accepted Answer

Write the system of equations as an augmented matrix. Each row corresponds to an equation, and each column corresponds to the coefficients of variables w, x, y, z, and the constants on the right side:

$$\begin{bmatrix} 2 & 1 & -1 & 0 & | & 3 \ 1 & -3 & 2 & 0 & | & -4 \ 3 & 1 & -3 & 1 & | & 1 \ 1 & 2 & -4 & -1 & | & -2 \end{bmatrix}$$ Use row operations to create zeros below the leading 1 in the first column (pivot position). For example, use the second row (which has a 1 in the first column) to eliminate the w-terms in the first, third, and fourth rows by appropriate row replacements. Move to the second column and create a leading 1 in the second row, second column position if necessary. Then use this pivot to eliminate the x-terms in the rows below it by row operations, creating zeros below the pivot. Continue this process for the third and fourth columns, creating leading 1s (pivots) and zeros below them, transforming the matrix into an upper triangular (row echelon) form. Once in row echelon form, use back substitution to solve for the variables starting from the bottom row up, expressing each variable in terms of constants or previously found variables to find the complete solution.

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. $\begin{cases}\) 2w + x - y = 3 \\ w - 3x + 2y = -4 \\ 3w + x - 3y + z = 1 \\ w + 2x - 4y - z = -2 \(\end{cases}$

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Concetti chiave

Systems of Linear Equations

Gaussian Elimination

Row Operations and Row-Echelon Form