Solve the system using matrices.
--5x + 2y - z = -29 x + 6y - 8z = 26
4x + y + z = 42
A) (8, 3, 7)
B) (8, 7, 3)
C) (-8, 7, 16)
D) inconsistent system

Question

Quizplus · Accepted Answer

Writing the system of equations in matrix form:

$\left(\begin{array}{ccc|c} -5 & 2 & -1 & -29 \ 1 & 6 & -8 & 26 \ 4 & 1 & 1 & 42 \end{array}ight)$

Performing row operations:

$\left(\begin{array}{ccc|c} -5 & 2 & -1 & -29 \ 0 & 32 & -13 & 55 \ 0 & -7 & 5 & 163 \end{array}ight)$

$\left(\begin{array}{ccc|c} -5 & 2 & -1 & -29 \ 0 & 1 & -\frac{13}{32} & \frac{55}{32} \ 0 & -7 & 5 & 163 \end{array}ight)$

$\left(\begin{array}{ccc|c} -5 & 2 & -1 & -29 \ 0 & 1 & -\frac{13}{32} & \frac{55}{32} \ 0 & 0 & \frac{111}{32} & \frac{1261}{32} \end{array}ight)$

$\left(\begin{array}{ccc|c} -5 & 2 & -1 & -29 \ 0 & 1 & -\frac{13}{32} & \frac{55}{32} \ 0 & 0 & 1 & \frac{1131}{111} \end{array}ight)$

$\left(\begin{array}{ccc|c} -5 & 2 & 0 & -22 \ 0 & 1 & 0 & 7 \ 0 & 0 & 1 & \frac{1131}{111} \end{array}ight)$

$\left(\begin{array}{ccc|c} 1 & 0 & 0 & 8 \ 0 & 1 & 0 & 7 \ 0 & 0 & 1 & \frac{1131}{111} \end{array}ight)$

Therefore, the solution is $(x, y, z) = (8, 7, \frac{1131}{111})$, which matches choice B. We can write the system in matrix form as:
\begin{bmatrix}-5 & 2 & -1 \ 1 & 6 & -8 \ 4 & 1 & 1 \\end{bmatrix} \begin{bmatrix}x \ y \ z\end{bmatrix} = \begin{bmatrix}-29 \ 26 \ 42\end{bmatrix}

Using row reduction, we can find the row echelon form:

\begin{bmatrix}-5 & 2 & -1 & -29\ 1 & 6 & -8 & 26\ 4 & 1 & 1 & 42\end{bmatrix} $\xrightarrow[]{R_1 \leftrightarrow R_2}$ \begin{bmatrix}1 & 6 & -8 & 26\ -5 & 2 & -1 & -29\ 4 & 1 & 1 & 42\end{bmatrix} $\xrightarrow[]{R_2 + 5R_1 ightarrow R_2}$ \begin{bmatrix}1 & 6 & -8 & 26\ 0 & 32 & -41 & 121\ 4 & 1 & 1 & 42\end{bmatrix} $\xrightarrow[]{R_3 - 4R_1 ightarrow R_3}$ \begin{bmatrix}1 & 6 & -8 & 26\ 0 & 32 & -41 & 121\ 0 & -23 & 33 & 26\end{bmatrix} $\xrightarrow[]{R_2 + \frac{23}{32}R_3 ightarrow R_2}$ \begin{bmatrix}1 & 6 & -8 & 26\ 0 & 0 & -\frac{35}{4} & \frac{1357}{32}\ 0 & -23 & 33 & 26\end{bmatrix} $\xrightarrow[]{-\frac{4}{35}R_3 ightarrow R_3}$ \begin{bmatrix}1 & 6 & -8 & 26\ 0 & 0 & -\frac{35}{4} & \frac{1357}{32}\ 0 & 92/5 & -132/5 & -104/35\end{bmatrix} $\xrightarrow[]{-\frac{4}{35}R_2 ightarrow R_2}$ \begin{bmatrix}1 & 6 & -8 & 26\ 0 & 0 & 35 & -\frac{1357}{8}\ 0 & 92/5 & -132/5 & -104/35\end{bmatrix} $\xrightarrow[]{\frac{20}{23}R_3 + R_2 ightarrow R_2}$ \begin{bmatrix}1 & 6 & -8 & 26\ 0 & 0 & 0 & \frac{310}{23}\ 0 & 92/5 & -132/5 & -104/35\end{bmatrix} $\xrightarrow[]{\frac{5}{92}R_2 ightarrow R_2}$ \begin{bmatrix}1 & 6 & -8 & 26\ 0 & 0 & 0 & \frac{155}{46}\ 0 & 92/5 & -132/5 & -104/35\end{bmatrix}

The last row of the row echelon form implies that $0x + \frac{155}{46} = \frac{155}{46} = 0$, which is a contradiction. Therefore, the system is inconsistent and has no solution.

The only option among the given choices that corresponds to an inconsistent system is D.

Solve the System Using Matrices