Using Cramer's rule for a system of equations with four variables

Question

Given the following system of equations: \begin{align*} w + x + y &= 3 \\ x + y + z &= 4 \\ x + y + 2z &= 10 \\ w + x + z &= 20 \end{align*} Find $w$ using Cramer's rule.
Answer:
\begin{align*} \begin{vmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 \\ 1 & 1 & 0 & 1 \end{vmatrix} &= \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} - \begin{vmatrix} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} + \begin{vmatrix} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 1 & 1 \end{vmatrix} \\ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} &= \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} - \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 1 - (1 - 2) + (0 - 1) \\ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} &= 1 \\ \begin{vmatrix} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} &= - \begin{vmatrix} 0 & 2 \\ 1 & 1 \end{vmatrix} + \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = -(0 - 2) + (0-1) = 1 \\ \begin{vmatrix} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 1 & 1 \end{vmatrix} &= - \begin{vmatrix} 0 & 2 \\ 1 & 1 \end{vmatrix} + \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} = - ( 0 - 2) + (0 - 1) = 1 \\ \begin{vmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 \\ 1 & 1 & 0 & 1 \end{vmatrix} &= 1 - 1 + 1 = 1 \\ w &= \frac{ \begin{vmatrix} 3 & 1 & 1 & 0 \\ 4 & 1 & 1 & 1 \\ 6 & 1 & 1 & 2 \\ 20 & 1 & 0 & 1 \end{vmatrix} } { 1 } = \begin{vmatrix} 3 & 1 & 1 & 0 \\ 4 & 1 & 1 & 1 \\ 6 & 1 & 1 & 2 \\ 20 & 1 & 0 & 1 \end{vmatrix} \\ \begin{vmatrix} 3 & 1 & 1 & 0 \\ 4 & 1 & 1 & 1 \\ 6 & 1 & 1 & 2 \\ 20 & 1 & 0 & 1 \end{vmatrix} &= 3 \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} - \begin{vmatrix} 4 & 1 & 1 \\ 6 & 1 & 2 \\ 20 & 0 & 1 \end{vmatrix} + \begin{vmatrix} 4 & 1 & 1 \\ 6 & 1 & 2 \\ 20 & 1 & 1 \end{vmatrix} \end{align*} \begin{align*} \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} &= \begin{vmatrix} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & -1 & 1 \end{vmatrix} = \begin{vmatrix} 0 & 1 \\ -1 & 1 \end{vmatrix} = ( 0 - 1(-1) ) \\ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} &= 1 \\ \begin{vmatrix} 4 & 1 & 1 \\ 6 & 1 & 2 \\ 20 & 1 & 1 \end{vmatrix} &= 4 \begin{vmatrix} 1 & 2 \\ 1 & 1 \\ \end{vmatrix} - \begin{vmatrix} 6 & 2 \\ 20 & 1 \\ \end{vmatrix} + \begin{vmatrix} 6 & 1 \\ 20 & 1 \\ \end{vmatrix} = 4(1-2) - (6 - 40) + 6 - 20 \\ \begin{vmatrix} 4 & 1 & 1 \\ 6 & 1 & 2 \\ 20 & 1 & 1 \end{vmatrix} &= 4(-1) - 6 + 40 + 6 - 20 = 16 \\ \begin{vmatrix} 4 & 1 & 1 \\ 6 & 1 & 2 \\ 20 & 1 & 1 \end{vmatrix} &= \begin{vmatrix} 4 & 1 & 1 \\ 0 & -\frac{1}{2} & \frac{1}{2} \\ 0 & -4 & -4 \\ \end{vmatrix} = 4 \begin{vmatrix} -\frac{1}{2} & \frac{1}{2} \\ -4 & -4 \\ \end{vmatrix} = 4( 2 + 2 ) = 16 \\ \begin{vmatrix} 3 & 1 & 1 & 0 \\ 4 & 1 & 1 & 1 \\ 6 & 1 & 1 & 2 \\ 20 & 1 & 0 & 1 \end{vmatrix} &= 3( 1 ) - 16 + 16 = 3 \\ W &= \frac{3}{1} \\ W &= 3 \end{align*} I have good reason to solution this system of equations is: $(w,x,y,z) = (5,9,-11,6)$

Where did I go wrong?

Answer 1

It should be: $$w= \frac{ \begin{vmatrix} 3 & 1 & 1 & 0 \\ 4 & 1 & 1 & 1 \\ 10 & 1 & 1 & 2 \\ 20 & 1 & 0 & 1 \end{vmatrix} } { 1 }=$$

$$ = 3 \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} - \begin{vmatrix} 4 & 1 & 1 \\ 10 & 1 & 2 \\ 20 & 0 & 1 \end{vmatrix} + \begin{vmatrix} 4 & 1 & 1 \\ 10 & 1 & 2 \\ 20 & 1 & 1 \end{vmatrix}= $$ $$=3(1+2+0-1-1-0)-(4+40+0-20-10-0)+$$ $$+(4+40+10-20-10-8)=3-14+16=5.$$