Answer
$x_{1}$ = $\frac{1}{4}$
$x_{2}$ = $\frac{11}{4}$
$x_{3}$ = $\frac{3}{8}$
Work Step by Step
Cramer's Rule is used to find the solutions to a system of equations.
We first have to turn the system of equations into matrices.
A = $\begin{bmatrix}
1 & 1 & 0 \\
-3 & 0 & 2\\
0 & 1 & -2
\end{bmatrix}$
We then replace the first column with the column of solutions.
$A_{1}$(b) = $\begin{bmatrix}
3 & 1 & 0 \\
0 & 0 & 2\\
2 & 1 & -2
\end{bmatrix}$
We then replace the second column with the column of solutions.
$A_{2}$(b) = $\begin{bmatrix}
1 & 3 & 0 \\
-3 & 0 & 2\\
0 & 2 & -2
\end{bmatrix}$
We then replace the third column with the column of solutions.
$A_{3}$(b) = $\begin{bmatrix}
1 & 1 & 3 \\
-3 & 0 & 0\\
0 & 1 & 2
\end{bmatrix}$
We then need to find the determinant of each matrix.
det(A) = -8
det($A_{1}$(b)) = -2
det($A_{2}$(b)) = -22
det($A_{3}$(b)) = -3
We then use the formula:
$x_{1}$ = $\frac{det(A_{1}(b))}{det(A)}$ = $\frac{-2}{-8}$ = $\frac{1}{4}$
$x_{2}$ = $\frac{det(A_{2}(b))}{det(A)}$ = $\frac{-22}{-8}$ = $\frac{11}{4}$
$x_{3}$ = $\frac{det(A_{3}(b))}{det(A)}$ = $\frac{-3}{-8}$ = $\frac{3}{8}$
The solutions to the system are:
$x_{1}$ = $\frac{1}{4}$
$x_{2}$ = $\frac{11}{4}$
$x_{3}$ = $\frac{3}{8}$