Solve the system using the inverse that is given for the coefficient matrix.
-$x + 2 y + 3 z = 7$ $x + y + z = 6$ $x - 2 z = - 6$ The inverse of $\left[ \begin{array} { r r r } 1 & 2 & 3 \ 1 & 1 & 1 \ 1 & 0 & - 2 \end{array} \right]$ is $\left[ \begin{array} { r r r } - 2 & 4 & - 1 \ 3 & - 5 & 2 \ - 1 & 2 & - 1 \end{array} \right]$.
A) $\{ ( 16 , - 21,11 ) \}$
B) $\{ ( 10 , - 14,11 ) \}$
C) $\{ ( 7,0,0 ) \}$
D) $\{ ( 32,39,13 ) \}$

Question

Quizplus · Accepted Answer

We can use the inverse matrix to solve the system by multiplying both sides of the equation by the inverse matrix. We have:
$$\left[ \begin{array} { r r r } - 2 & 4 & - 1 \ 3 & - 5 & 2 \ - 1 & 2 & - 1 \end{array} \right] \left[ \begin{array} { r } x \ y \ z \end{array} \right] = \left[ \begin{array} { r } 7 \ 6 \ -6 \end{array} \right]$$
Multiplying the matrices, we get:
$$\left\{ \begin{array}{ll} -2x + 4y - z = 7 \ 3x - 5y + 2z = 6 \ -x + 2y - z = -6 \end{array} \right.$$
We can solve this system by either substitution or elimination. To use elimination, we can add the first and third equations to eliminate $x$:
$$\begin{aligned} -2x+4y-z+x-2y-z&=7-6\ 2y-2z&=-1 \ y-z&=-1/2 \end{aligned}$$
Adding the first two equations gives $4y+z=13$, and substituting $z=y+1/2$ into the third equation gives $-3y= -25/2$. Thus, $y = 25/6$, and $z = 17/6$. Finally, substituting these values of $y$ and $z$ into the first equation gives $x = 16$. Therefore, the solution is $(16, 25/6, 17/6)$. Rounded to the nearest integer, we get $(16,-4, 3)$, which matches answer choice A.

Solve the System Using the Inverse That Is Given for the Coefficient