Solve this

Question:

$2 x-3 z+w=1$

$x-y+2 w=1$

$-3 y+z+w=1$

$x+y+z=1$

Solution:

$D=\left|\begin{array}{cccc}2 & 0 & -3 & 1 \\ 1 & -1 & 0 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{array}\right|$

$2\left|\begin{array}{ccc}-1 & 0 & 2 \\ -3 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|-0-3\left|\begin{array}{ccc}1 & -1 & 2 \\ 0 & -3 & 1 \\ 1 & 1 & 0\end{array}\right|-1\left|\begin{array}{ccc}1 & -1 & 0 \\ 0 & -3 & 1 \\ 1 & 1 & 1\end{array}\right|$

$=2[-1(0-1)-0(0-1)+2(-3-1)]-3[1(0-1)+1(0-1)+2(0+3)]-1[1(-3-1)+1(0-1)+0(0+3)]$

$=-21$

$D_{1}=\left|\begin{array}{cccc}1 & 0 & -3 & 1 \\ 1 & -1 & 0 & 2 \\ 1 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{array}\right|$

$1\left|\begin{array}{ccc}-1 & 0 & 2 \\ -3 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|-0-3\left|\begin{array}{ccc}1 & -1 & 2 \\ 1 & -3 & 1 \\ 1 & 1 & 0\end{array}\right|-1\left|\begin{array}{ccc}1 & -1 & 0 \\ 1 & -3 & 1 \\ 1 & 1 & 1\end{array}\right|$

$=1[-1(0-1)-0(0-1)+2(-3-1)]-3[1(0-1)+1(0-1)+2(1+3)]-1[1(-3-1)+1(0-1)+2(1+3)]$

$=-21$

$D_{2}=\left|\begin{array}{cccc}2 & 1 & -3 & 1 \\ 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0\end{array}\right|$

$=2\left|\begin{array}{lll}1 & 0 & 2 \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|-1\left|\begin{array}{lll}1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|+$ $(-3)\left|\begin{array}{lll}1 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|-1\left|\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|$

$2[1(0-1)+2(1-1)]-1[1(0-1)+2(0-1)]-3[1(0-1)-1(0-1)+2(0-1)]-1[1(1-1)-1(0-1)]$

$=6$

$D_{3}=\left|\begin{array}{cccc}2 & 0 & 1 & 1 \\ 1 & -1 & 1 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{array}\right|$

$=2\left|\begin{array}{ccc}-1 & 1 & 2 \\ -3 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|-0+1\left|\begin{array}{ccc}1 & -1 & 2 \\ 0 & -3 & 1 \\ 1 & 1 & 0\end{array}\right|-1\left|\begin{array}{ccc}1 & -1 & 1 \\ 0 & -3 & 1 \\ 1 & 1 & 1\end{array}\right|$

$=2[-1(0-1)-1(0-1)+2(-3-1)]+1[1(0-1)+1(0-1)+2(0+3)]-1[1(-3-1)+1(0-1)+1(0+3)]$

$=-6$

$D_{4}=\left|\begin{array}{cccc}2 & 0 & -3 & 1 \\ 1 & -1 & 0 & 1 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 1\end{array}\right|$

$=2\left|\begin{array}{ccc}-1 & 0 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|-0-3\left|\begin{array}{ccc}1 & -1 & 1 \\ 0 & -3 & 1 \\ 1 & 1 & 1\end{array}\right|-1\left|\begin{array}{ccc}1 & -1 & 0 \\ 0 & -3 & 1 \\ 1 & 1 & 1\end{array}\right|$

$=2[-1(1-1)+1(-3-1)]-3[1(-3-1)+1(0-1)+1(0+3)]-1[1(-3-1)+1(0-1)]$

$=3$

So, by Cramer's rule, we obtain

$x=\frac{D_{1}}{D}=\frac{21}{21}=1$

$y=\frac{D_{2}}{D}=\frac{6}{-21}=-\frac{2}{7}$

$z=\frac{D_{3}}{D}=\frac{-6}{-21}=\frac{2}{7}$

$w=\frac{D_{4}}{D}=\frac{3}{-21}=-\frac{1}{7}$

Hence, $x=1, y=-\frac{2}{7}, z=\frac{2}{7}, w=-\frac{1}{7}$