Solve the system.
-$$\left\{ \begin{aligned}
x - y + 3 z & = 0 \
4 x + z & = 1 \
x + 5 y + z & = 16
\end{aligned} \right.$$
A) $( 0,3,1 )$
B) $( 1,3,0 )$
C) $( 1,0,3 )$
D) $\varnothing$

Question

Quizplus · Accepted Answer

The system can be solved by using row operations to transform the augmented matrix into row echelon form.$$\left[ \begin{array}{rrr|r}1 & -1 & 3 & 0 \4 & 0 & 1 & 1 \1 & 5 & 1 & 16\end{array} \right] \rightarrow \left[ \begin{array}{rrr|r}1 & -1 & 3 & 0 \0 & 4 & -11 & 1 \0 & 6 & -2 & 16\end{array} \right] \rightarrow \left[ \begin{array}{rrr|r}1 & -1 & 3 & 0 \0 & 1 & -11/4 & 1/4 \0 & 0 & 0 & 4\end{array} \right]$$The last row of the row echelon form is equivalent to the equation $0 = 4$, which is a contradiction. Therefore, the system has no solution.

Solve the System {x−y+3z=04x+z=1x+5y+z=16\left\{ \begin{aligned}x - y + 3 z & = 0 \\4 x + z & = 1 \\x + 5 y + z & = 16\end{aligned} \right.⎩⎨⎧​x−y+3z4x+zx+5y+z​=0=1=16​

Solve the System $\left\{ \begin{aligned}x - y + 3 z & = 0 \\4 x + z & = 1 \\x + 5 y + z & = 16\end{aligned} \right.$