Question

2x − 3z + w = 1
x − y + 2w = 1
− 3y + z + w = 1
x + y + z = 1

Answer 1

\[D = \begin{vmatrix}2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{vmatrix}\] 
\[2\begin{vmatrix}- 1 & 0 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 0 - 3\begin{vmatrix}1 & - 1 & 2 \\ 0 & - 3 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & - 1 & 0 \\ 0 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix}\] 
\[ = 2\left[ - 1\left( 0 - 1 \right) - 0\left( 0 - 1 \right) + 2\left( - 3 - 1 \right) \right] - 3\left[ 1\left( 0 - 1 \right) + 1\left( 0 - 1 \right) + 2\left( 0 + 3 \right) \right] - 1\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) + 0\left( 0 + 3 \right) \right]\] 
\[ = - 21\] 
\[ D_1 = \begin{vmatrix}1 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 1 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{vmatrix}\] 
\[1\begin{vmatrix}- 1 & 0 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 0 - 3\begin{vmatrix}1 & - 1 & 2 \\ 1 & - 3 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & - 1 & 0 \\ 1 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix}\] 
\[ = 1\left[ - 1\left( 0 - 1 \right) - 0\left( 0 - 1 \right) + 2\left( - 3 - 1 \right) \right] - 3\left[ 1\left( 0 - 1 \right) + 1\left( 0 - 1 \right) + 2\left( 1 + 3 \right) \right] - 1\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) + 2\left( 1 + 3 \right) \right]\] 
\[ = - 21\] 
\[ D_2 = \begin{vmatrix}2 & 1 & - 3 & 1 \\ 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0\end{vmatrix}\] 
\[ = 2\begin{vmatrix}1 & 0 & 2 \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} + ( - 3)\begin{vmatrix}1 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1\end{vmatrix}\] 
\[2\left[ 1\left( 0 - 1 \right) + 2\left( 1 - 1 \right) \right] - 1\left[ 1\left( 0 - 1 \right) + 2\left( 0 - 1 \right) \right] - 3\left[ 1\left( 0 - 1 \right) - 1\left( 0 - 1 \right) + 2\left( 0 - 1 \right) \right] - 1\left[ 1\left( 1 - 1 \right) - 1\left( 0 - 1 \right) \right]\] 
\[ = 6\] 
\[ D_3 = \begin{vmatrix}2 & 0 & 1 & 1 \\ 1 & - 1 & 1 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0\end{vmatrix}\] 
\[ = 2\begin{vmatrix}- 1 & 1 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0\end{vmatrix} - 0 + 1\begin{vmatrix}1 & - 1 & 2 \\ 0 & - 3 & 1 \\ 1 & 1 & 0\end{vmatrix} - 1\begin{vmatrix}1 & - 1 & 1 \\ 0 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix}\] 
\[ = 2\left[ - 1\left( 0 - 1 \right) - 1\left( 0 - 1 \right) + 2\left( - 3 - 1 \right) \right] + 1\left[ 1\left( 0 - 1 \right) + 1\left( 0 - 1 \right) + 2\left( 0 + 3 \right) \right] - 1\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) + 1\left( 0 + 3 \right) \right]\] 
\[ = - 6\] 
\[ D_4 = \begin{vmatrix}2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 1 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 1\end{vmatrix}\] 
\[ = 2\begin{vmatrix}- 1 & 0 & 1 \\ - 3 & 1 & 1 \\ 1 & 1 & 1\end{vmatrix} - 0 - 3\begin{vmatrix}1 & - 1 & 1 \\ 0 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix} - 1\begin{vmatrix}1 & - 1 & 0 \\ 0 & - 3 & 1 \\ 1 & 1 & 1\end{vmatrix}\] 
\[ = 2\left[ - 1\left( 1 - 1 \right) + 1\left( - 3 - 1 \right) \right] - 3\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) + 1\left( 0 + 3 \right) \right] - 1\left[ 1\left( - 3 - 1 \right) + 1\left( 0 - 1 \right) \right]\] 
\[ = 3\] 
So, by Cramer's rule , we obtain
\[x = \frac{D_1}{D} = \frac{21}{21} = 1\] 
\[y = \frac{D_2}{D} = \frac{6}{- 21} = - \frac{2}{7}\] 
\[z = \frac{D_3}{D} = \frac{- 6}{- 21} = \frac{2}{7}\] 
\[w = \frac{D_4}{D} = \frac{3}{- 21} = - \frac{1}{7}\] 
\[\text{ Hence,} x = 1, y = - \frac{2}{7}, z = \frac{2}{7}, w = - \frac{1}{7}\]