Answer
$x_1=\frac{11}{3}$
$x_2=\frac{-17}{3}$
$x_3=\frac{-16}{3}$
$x_4=-2$
Work Step by Step
We are given:
$x_1-2x_2+3x_3-x_4=1$
$2x_1+x_3=2$
$x_1+x_2-x_4=0$
$x_2-2x_3+x_4=3$
which can also be written as:
$A=\begin{bmatrix}
1 & -2&3&1\\
2 & 0&1&0\\
1&1&0&-1 \\
0&1&-2&1
\end{bmatrix} \rightarrow \det(A)=-3$
$B_1=\begin{bmatrix}
1 & -2&3&1\\
2 & 0&1&0\\
0&1&0&-1 \\
3&1&-2&1
\end{bmatrix} \rightarrow \det(B_1)=-11$
$B_2=\begin{bmatrix}
1 & 1&3&-1\\
2 & 2&1&0\\
1&0&0&-1 \\
0&3&-2&1
\end{bmatrix} \rightarrow \det(B_2)=17$
$B_3=\begin{bmatrix}
1 & -2&1&-1\\
2 & 0&2&0\\
1&1&0&-1 \\
0&1&3&1
\end{bmatrix} \rightarrow \det(B_3)=16$
$B_4=\begin{bmatrix}
1 & -2&3&1\\
2 & 0&1&2\\
1&1&0&0 \\
0&1&-2&3
\end{bmatrix} \rightarrow \det(B_4)=6$
Use Cramer’s rule: $x_k=\frac{\det(B_k)}{\det(A)}$ to find the results:
$x_1=\frac{\det(B_1)}{\det(A)}=\frac{-11}{-3}=\frac{11}{3}$
$x_2=\frac{\det(B_2)}{\det(A)}=\frac{-17}{3}$
$x_3=\frac{\det(B_3)}{\det(A)}=\frac{-16}{3}$
$x_4=\frac{\det(B_4)}{\det(A)}=\frac{-6}{3}=-2$
