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?
Best Answer
It should be: $$w= \frac{ \begin{vmatrix} 3 & 1 & 1 & 0 \\ 4 & 1 & 1 & 1 \\ 10 & 1 & 1 & 2 \\ 20 & 1 & 0 & 1 \end{vmatrix} } { 1 }=$$
$$ = 3 \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} - \begin{vmatrix} 4 & 1 & 1 \\ 10 & 1 & 2 \\ 20 & 0 & 1 \end{vmatrix} + \begin{vmatrix} 4 & 1 & 1 \\ 10 & 1 & 2 \\ 20 & 1 & 1 \end{vmatrix}= $$ $$=3(1+2+0-1-1-0)-(4+40+0-20-10-0)+$$ $$+(4+40+10-20-10-8)=3-14+16=5.$$