Correct Answer - Option 4 : 1, 2, -1, -2
Concept:
Gauss-Jordan Method:
- This method is used to simplify an augmented matrix to its reduced-row echelon form.
- The following are the steps that must be followed while using the Gauss-Jordan Elimination method -
- Write the system as an augmented matrix.
-
Look at the first entry in the first row. Make this entry into a 1 and all other entries in that column 0s. This is called pivoting the matrix about this element.
-
Once this is done, move down the diagonal to the second entry of the second row and pivot about this entry.
- Continue until the whole matrix is in row-reduced form.
- Solve for the values of variables.
Calculation:
Given,
5x1 + x2 + x3 + x4 = 4 ...(1)
x1 + 7x2 + x3 + x4 = 12 ...(2)
x1 + x2 + 6x3 + x4 = - 5 ....(3)
x1 + x2 + x3 + 4x4 = - 6 .....(4)
This system of equations can be written in the form of matrices as follows -
\(\left[ {\begin{array}{*{20}{c}} 5&1&1&{ 1}\\ 1&7&1&1\\ 1&{ 1}&{ 6}&1 \\1&1&1&4\\ \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}\\{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { 4}\\ {12}\\ { - 5}\\{-6} \end{array}} \right]\)
⇒ [A][X] = [D] ....(5)
Here,
\([A] = \left[ {\begin{array}{*{20}{c}} 5&1&1&{ 1}\\ 1&7&1&1\\ 1&{ 1}&{ 6}&1 \\1&1&1&4\\ \end{array}} \right]\)
\([D] = \left[ {\begin{array}{*{20}{c}} { 4}\\ {12}\\ { -5}\\-6 \end{array}} \right]\)
\([X] = \left[ {\begin{array}{*{20}{c}} {x_1}\\ {x_2}\\ {x_3}\\{x_4} \end{array}} \right]\)
Step 1: Writing Augment Matrix
\(\therefore[A~\vdots ~D] = \left[ {\begin{array}{*{20}{c}} 5&1&1&1\\ 1&7&1&1\\ 1&1&6&1\\1&1&1&4 \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { 4}\\ {12}\\ { - 5}\\-6 \end{array}} \right]\)
Step 2: Applying elementary transformations or pivoting of the augment matrix
a) R1 → \(\frac{1}{5}\)R1
\( = \left[ {\begin{array}{*{20}{c}} 1&0.20&{0.20}&0.20\\ 1&7&1&1\\ 1&{ 1}&{ 6}&1\\1&1&1&4\end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { 0.80}\\ {12}\\ { - 5} \\-6\end{array}} \right]\)
b) R2 → R2 - R1; R3 → R3 - R1; R4 → R4 - R1
\( = \left[ {\begin{array}{*{20}{c}} 1&0.2&{0.2}&0.2\\ 0&6.8&0.8&0.8\\ 0&{ 0.8}&{ 5.8}&0.8\\0&0.8&0.8&3.8 \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { 0.8}\\ {11.2}\\ { -5.8}\\-6.8 \end{array}} \right]\)
c) R2 → \(\frac{1}{6.8}\)R2
\( = \left[ {\begin{array}{*{20}{c}} 1&0.2&{0.2}&0.2\\ 0&1&\frac{2}{17}&\frac{2}{17}\\ 0&{ 0.8}&{ 5.8}&0.8\\0&0.8&0.8&3.8 \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { 0.8}\\ {\frac{28}{17}}\\ { -5.8}\\-6.8 \end{array}} \right]\)
d) R3 → R3 - 0.8R2; R4 → R4 - 0.8R2
\( = \left[ {\begin{array}{*{20}{c}} 1&0.2&{0.2}&0.2\\ 0&1&\frac{2}{17}&\frac{2}{17}\\ 0&{ 0}&{ \frac{97}{17}}&\frac{12}{17}\\0&0&\frac{12}{17}&\frac{63}{17} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} {\frac{4}{5}}\\ {\frac{28}{17}}\\ { -\frac{121}{17}}\\-\frac{138}{17} \end{array}} \right]\)
e) R3 → \(\frac{17}{97}\)R3
\( = \left[ {\begin{array}{*{20}{c}} 1&0.2&{0.2}&0.2\\ 0&1&\frac{2}{17}&\frac{2}{17}\\ 0&{ 0}&{ 1}&\frac{12}{97}\\0&0&\frac{12}{17}&\frac{63}{17} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} {\frac{4}{5}}\\ {\frac{28}{17}}\\ { -\frac{121}{97}}\\-\frac{138}{17} \end{array}} \right]\)
f) R4 → R4 - \(\frac{12}{17}\)R3
\( = \left[ {\begin{array}{*{20}{c}} 1&0.2&{0.2}&0.2\\ 0&1&\frac{2}{17}&\frac{2}{17}\\ 0&{ 0}&{ 1}&\frac{12}{97}\\0&0&0&\frac{351}{97} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} {\frac{4}{5}}\\ {\frac{28}{17}}\\ { -\frac{121}{97}}\\-\frac{702}{97} \end{array}} \right]\)
Step 3: Solving for variables -
Now, the reduced equations can be written as -
x1 + 0.2x2 + 0.2x3 + 0.2x4 = \(\frac{4}{5}\) ...(6)
x2 + \(\frac{2}{17}\)x3 + \(\frac{2}{17}\)x4 = \(\frac{28}{17}\) ...(7)
x3 + \(\frac{12}{97}\)x4 = \(-\frac{121}{97}\) ...(8)
\(\frac{351}{97}\)x4 = \(-\frac{702}{97}\) ....(9)
Solving (6), (7), (8), and (9)
x4 = -2
x3 = -1
x2 = 2
x1 = 1