Question

Example 3:
solve
the
following system by
Gauss - Jordan
method
5x+y+z+w=4;x+7y+z+w=12;x+y+6z+w=-5;x+y+z+4w=-6​

Answer 1

Answer:

i don't know answer please forgive my make me brainlist

Answer 2

Given: $5x+y+z+w=4; x+7y+z+w=12; x+y+6z+w=-5; x+y+z=4w=-6$

We have to solve the given system by using the Gauss-Jordan method.

The given question is solved in the following way;

The given system in matrix form will be,

$\left[\begin{array}{cccc|c}5 & 1 & 1 & 1 & 4 \\1 & 7 & 1 & 1 & 12 \\1 & 1 & 6 & 1 & -5 \\1 & 1 & 1 & 4 & -6\end{array}\right]\\\text { Divide row } 1 \text { by } 5: R_{1}=\frac{R_{1}}{5} \text {. }\\\left[\begin{array}{cccc|c}1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{4}{5} \\1 & 7 & 1 & 1 & 12 \\1 & 1 & 6 & 1 & -5 \\1 & 1 & 1 & 4 & -6\end{array}\right]\\\text { Subtract row } 1 \text { from row } 2: R_{2}=R_{2}-R_{1} \text {. }$

$\left[\begin{array}{cccc|c}1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{4}{5} \\0 & \frac{34}{5} & \frac{4}{5} & \frac{4}{5} & \frac{56}{5} \\1 & 1 & 6 & 1 & -5 \\1 & 1 & 1 & 4 & -6\end{array}\right]\\\text { Subtract row } 1 \text { from row } 3: R_{3}=R_{3}-R_{1} \text {. }$

$\left[\begin{array}{lccc|c}1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{4}{5} \\0 & \frac{34}{5} & \frac{4}{5} & \frac{4}{5} & \frac{56}{5} \\0 & \frac{4}{5} & \frac{29}{5} & \frac{4}{5} & -\frac{29}{5} \\1 & 1 & 1 & 4 & -6\end{array}\right]\\\text { Subtract row } 1 \text { from row } 4: R_{4}=R_{4}-R_{1} \text {. }$

$\left[\begin{array}{cccc|c}1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{4}{5} \\0 & \frac{34}{5} & \frac{4}{5} & \frac{4}{5} & \frac{56}{5} \\0 & \frac{4}{5} & \frac{29}{5} & \frac{4}{5} & -\frac{29}{5} \\0 & \frac{4}{5} & \frac{4}{5} & \frac{19}{5} & -\frac{34}{5}\end{array}\right]\\\text { Multiply row } 2 \text { by } \frac{5}{34}: R_{2}=\frac{5 R_{2}}{34} \text {. }$

$\left[\begin{array}{cccc|c}1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{4}{5} \\0 & 1 & \frac{2}{17} & \frac{2}{17} & \frac{28}{17} \\0 & \frac{4}{5} & \frac{29}{5} & \frac{4}{5} & -\frac{29}{5} \\0 & \frac{4}{5} & \frac{4}{5} & \frac{19}{5} & -\frac{34}{5}\end{array}\right]\\\text { Subtract row } 2 \text { multiplied by } \frac{1}{5} \text { from row 1: } R_{1}=R_{1}-\frac{R_{2}}{5} \text {. }$

$\left[\begin{array}{cccc|c}1 & 0 & \frac{3}{17} & \frac{3}{17} & \frac{8}{17} \\0 & 1 & \frac{2}{17} & \frac{2}{17} & \frac{28}{17} \\0 & \frac{4}{5} & \frac{29}{5} & \frac{4}{5} & -\frac{29}{5} \\0 & \frac{4}{5} & \frac{4}{5} & \frac{19}{5} & -\frac{34}{5}\end{array}\right]\\\text { Subtract row } 2 \text { multiplied by } \frac{4}{5} \text { from row } 3 \text { : } R_{3}=R_{3}-\frac{4 R_{2}}{5} \text {. }$

$\left[\begin{array}{cccc|c}1 & 0 & \frac{3}{17} & \frac{3}{17} & \frac{8}{17} \\0 & 1 & \frac{2}{17} & \frac{2}{17} & \frac{28}{17} \\0 & 0 & \frac{97}{17} & \frac{12}{17} & -\frac{121}{17} \\0 & 0 & \frac{12}{17} & \frac{63}{17} & -\frac{138}{17}\end{array}\right]\\\text { Multiply row } 3 \text { by } \frac{17}{97}: R_{3}=\frac{17 R_{3}}{97} \text {. }$

$\left[\begin{array}{cccc|c}1 & 0 & \frac{3}{17} & \frac{3}{17} & \frac{8}{17} \\0 & 1 & \frac{2}{17} & \frac{2}{17} & \frac{28}{17} \\0 & 0 & 1 & \frac{12}{97} & -\frac{121}{97} \\0 & 0 & \frac{12}{17} & \frac{63}{17} & -\frac{138}{17}\end{array}\right]\\\text { Subtract row } 3 \text { multiplied by } \frac{3}{17} \text { from row } 1: R_{1}=R_{1}-\frac{3 R_{3}}{17} \text {. }$

$\left[\begin{array}{cccc|c}1 & 0 & 0 & \frac{15}{97} & \frac{67}{97} \\0 & 1 & \frac{2}{17} & \frac{2}{17} & \frac{28}{17} \\0 & 0 & 1 & \frac{12}{97} & -\frac{121}{97} \\0 & 0 & \frac{12}{17} & \frac{63}{17} & -\frac{138}{17}\end{array}\right]\\\text { Subtract row } 3 \text { multiplied by } \frac{2}{17} \text { from row } 2: R_{2}=R_{2}-\frac{2 R_{3}}{17} \text {. }$

$\left[\begin{array}{cccc|c}1 & 0 & 0 & \frac{15}{97} & \frac{67}{97} \\0 & 1 & 0 & \frac{10}{97} & \frac{174}{97} \\0 & 0 & 1 & \frac{12}{97} & -\frac{121}{97} \\0 & 0 & \frac{12}{17} & \frac{63}{17} & -\frac{138}{17}\end{array}\right]\\\text { Subtract row } 3 \text { multiplied by } \frac{12}{17} \text { from row } 4: R_{4}=R_{4}-\frac{12 R_{3}}{17} \text {. }$

$\left[\begin{array}{cccc|c}1 & 0 & 0 & \frac{15}{97} & \frac{67}{97} \\0 & 1 & 0 & \frac{10}{97} & \frac{174}{97} \\0 & 0 & 1 & \frac{12}{97} & -\frac{121}{97} \\0 & 0 & 0 & \frac{351}{97} & -\frac{702}{97}\end{array}\right]\\\text { Multiply row } 4 \text { by } \frac{97}{351}: R_{4}=\frac{97 R_{4}}{351} \text {. }$

$\left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & 1 \\0 & 1 & 0 & \frac{10}{97} & \frac{174}{97} \\0 & 0 & 1 & \frac{12}{97} & -\frac{121}{97} \\0 & 0 & 0 & 1 & -2\end{array}\right]\\\text { Subtract row } 4 \text { multiplied by } \frac{10}{97} \text { from row 2: } R_{2}=R_{2}-\frac{10 R_{4}}{97} \text {. }$

Hence,

$A=\left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & 1 \\0 & 1 & 0 & 0 & 2 \\0 & 0 & 1 & 0 & -1 \\0 & 0 & 0 & 1 & -2\end{array}\right] \text$