Solve the system using the inverse that is given for the coefficient matrix.
-$x + 2 y + 3 z = 2$ $x + y + z = 1$ $- x + y + 2 z = - 1$$\quad$ The inverse of $\left[ \begin{array} { r r r } 1 & 2 & 3 \ 1 & 1 & 1 \ - 1 & 1 & 2 \end{array} \right]$ is $\left[ \begin{array} { r r r } 1 & - 1 & - 1 \ - 3 & 5 & 2 \ 2 & - 3 & - 1 \end{array} \right]$
A) $\{ ( 2 , - 3,2 ) \}$
B) $\{ ( - 3,6,1 ) \}$
C) $\{ ( 2,4,2 ) \}$
D) $\{ ( 2,9,6 ) \}$

Question

Quizplus · Accepted Answer

To solve the system using the inverse matrix, multiply the inverse matrix by the constants on the right side of the equations. The inverse matrix is given, and the constants are $[2, 1, -1]^T$. Multiplying the inverse matrix by this column vector gives the solution for $[x, y, z]^T$, which results in $[2, -3, 2]^T$.

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