Answer
$(\frac{1}{2},\frac{1}{4},\frac{1}{4},-1)$
Work Step by Step
Step 1. Based on the Cramer’s Rule, with the given equations, we can define the following determinants:
$\begin{array}( \\|D|= \\ \\ \end{array}
\begin{vmatrix}1&1&1&1\\2&0&0&1\\0&1&-1&0 \\1&0&2&0\end{vmatrix},
\begin{array}( \\|D_x|= \\ \\ \end{array}
\begin{vmatrix}0&1&1&1\\0&0&0&1\\0&1&-1&0 \\1&0&2&0\end{vmatrix},
\begin{array}( \\|D_y|= \\ \\ \end{array}
\begin{vmatrix}1&0&1&1\\2&0&0&1\\0&0&-1&0 \\1&1&2&0\end{vmatrix},
\begin{array}( \\|D_z|= \\ \\ \end{array}
\begin{vmatrix}1&1&0&1\\2&0&0&1\\0&1&0&0 \\1&0&1&0\end{vmatrix},
\begin{array}( \\|D_w|= \\ \\ \end{array}
\begin{vmatrix}1&1&1&0\\2&0&0&0\\0&1&-1&0 \\1&0&2&1\end{vmatrix}$
Step 2. Evaluate the above determinants:
$|D|=-(1)(2(2-0))+(1)(1(2-0)-1(0+1)+1(0-1))=-4$ (column4 then row1 expansions),
$|D_x|=-(1)(-(1)(-1-1))=-2$ (column1 then row2 expansions),
$|D_y|=(1)(-(-1)(1-2))=-1$ (column2 then row3 expansions),
$|D_z|=-(1)(-(1)(1-2))=-1$ (column3 then row3 expansions),
$|D_w|=(1)(-(2)(-1-1))=4$ (column4 then row2 expansions)
Step 3. Find the solutions as:
$x=\frac{|D_x|}{|D|}=\frac{1}{2}$,
$y=\frac{|D_y|}{|D|}=\frac{1}{4}$,
$z=\frac{|D_z|}{|D|}=\frac{1}{4}$,
$w=\frac{|D_w|}{|D|}=-1$
