Answer
$(-1,2,0,3)$
Work Step by Step
Step 1. Based on the Cramer’s Rule, with the given equations, we can define the following determinants:
$\begin{array}( \\|D|= \\ \\ \end{array}
\begin{vmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1 \\-1&0&0&1\end{vmatrix},
\begin{array}( \\|D_x|= \\ \\ \end{array}
\begin{vmatrix}1&1&0&0\\2&1&1&0\\3&0&1&1 \\4&0&0&1\end{vmatrix},
\begin{array}( \\|D_y|= \\ \\ \end{array}
\begin{vmatrix}1&1&0&0\\0&2&1&0\\0&3&1&1 \\-1&4&0&1\end{vmatrix},
\begin{array}( \\|D_z|= \\ \\ \end{array}
\begin{vmatrix}1&1&1&0\\0&1&2&0\\0&0&3&1 \\-1&0&4&1\end{vmatrix},
\begin{array}( \\|D_w|= \\ \\ \end{array}
\begin{vmatrix}1&1&0&1\\0&1&1&2\\0&0&1&3 \\-1&0&0&4\end{vmatrix}$
Step 2. Evaluate the above determinants: (you can do some row operations to create more zeros to simply the calculations)
$|D|=(1)((1)(1-0))-(-1)((1)(1-0))=2$ (column1 then column1 and row1 expansions),
$|D_x|=(1)((1)(1-0))-(1)((2)(1-0)-(1)(3-4))=-2$ (row1 then column1 and row1 expansions),
$|D_y|=(1)((2)(1-0)-(1)(3-4))-(-1)((1)(1-0))=4$ (column1 then row1 expansions),
$|D_z|=(1)((1)(3-4))-(-1)((1)(2-1))=0$ (column1 then column1 and column3 expansions),
$|D_w|=(1)((1)(4-0))-(-1)((1)(3-2)-(1)(0-1))=6$ (column1 then column1 expansions)
Step 3. Find the solutions as:
$x=\frac{|D_x|}{|D|}=-1$,
$y=\frac{|D_y|}{|D|}=2$,
$z=\frac{|D_z|}{|D|}=0$,
$w=\frac{|D_z|}{|D|}=3$
Step 4. Double check the answers by solving the equations with variable substitutions and eliminations (much easier than using the Cramer's Rule).
