Question

W+2x+2y+z=-2
W+3x-2y-z=-6
-2w-x+3y+3z=6
W+4x+y-2z=-14

Answer 1

The matrics of coeficients of the equations are

`\left[\begin{array}{cccc}1&2&2&1\\1&3&-2&-1\\-2&-1&3&3\\1&4&1&-2\end{array}\right]\left[\begin{array}{c}w&x&y&z\end{array}\right] = \left[\begin{array}{c}-2&-6&6&-14\end{array}\right] \ \ \ \ \ \ \ \left\begin{array}{ccc}-R_1+R_2 \rightarrow R_2\\2R_1+R_3\rightarrow R_3\\-R_1+R_4 \rightarrow R_4\end{array}\right`

`\left[\begin{array}{cccc}1&2&2&1\\0&1&-4&-2\\0&3&7&5\\0&2&-1&-3\end{array}\right]\left[\begin{array}{c}w&x&y&z\end{array}\right] = \left[\begin{array}{c}-2&-4&2&-12\end{array}\right] \ \ \ \ \ \ \ \left\begin{array}{ccc}-2R_2+R_1 \rightarrow R_1\\-3R_2+R_3\rightarrow R_3\\-2R_2+R_4 \rightarrow R_4\end{array}\right`

`\left[\begin{array}{cccc}1&0&10&5\\0&1&-4&-2\\0&0&19&11\\0&0&7&1\end{array}\right]\left[\begin{array}{c}w&x&y&z\end{array}\right] =\left[\begin{array}{c}6&-4&14&-4\end{array}\right] \ \ \ \ \ \ \ \left\begin{array}{c}(1)/(19)R_3\rightarrow R_3\end{array}\right\\ \left[\begin{array}{cccc}1&0&10&5\\0&1&-4&-2\\0&0&1&(11)/(19) \\0&0&7&1\end{array}\right]\left[\begin{array}{c}w&x&y&z\end{array}\right]=\left[\begin{array}{c}6&-4&(14)/(19) &-4\end{array}\right] \ \ \ \ \ \ \ \left\begin{array}{ccc}-10R_3+R_1 \rightarrow R_1\\4R_3+R_2\rightarrow R_2\\-7R_3+R_4 \rightarrow R_4\end{array}\right`

`\left[\begin{array}{cccc}1&0&0&- (15)/(19) \\0&1&0& (6)/(19) \\0&0&1& (11)/(19) \\0&0&0&- (58)/(19) \end{array}\right]\left[\begin{array}{c}w&x&y&z\end{array}\right] =\left[\begin{array}{c} -(26)/(19) &- (20)/(19) & (14)/(19) &- (174)/(19) \end{array}\right] \ \ \ \ \ \ \ \left\begin{array}{c}-(19)/(58)R_4\rightarrow R_4\end{array}\right`

`\left[\begin{array}{cccc}1&0&0&- (15)/(19) \\0&1&0& (6)/(19) \\0&0&1& (11)/(19) \\0&0&0&1\end{array}\right] \left[\begin{array}{c}w&x&y&z\end{array}\right] =\left[\begin{array}{c} -(26)/(19) &- (20)/(19) & (14)/(19) &3\end{array}\right] \ \ \ \ \ \ \ \left\begin{array}{ccc} (15)/(19) R_4+R_1 \rightarrow R_1\\ -(6)/(19) R_4+R_2\rightarrow R_2\\ -(11)/(19) R_4+R_3 \rightarrow R_3\end{array}\right`

`\left[\begin{array}{cccc}1&0&0&0 \\0&1&0&0 \\0&0&1&0\\0&0&0&1\end{array}\right] \left[\begin{array}{c}w&x&y&z\end{array}\right] =\left[\begin{array}{c}1 &-2 &-1&3\end{array}\right]`

Therefore, w = 1, x = -2, y = -1 and z = 3