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,2x1 + 4x2 - 6x3 = - 8 ---(1)x1 + 3x2 + x3 = 10 ---(2)2x1 - 4x2 - 2x3 = - 12 ---(3)This system of equations can be written in the form of matrices as follows -\(\left[ {\BEGIN{array}{*{20}{c}} 2&4&{ - 6}\\ 1&3&1\\ 2&{ - 4}&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 8}\\ {10}\\ { - 12} \end{array}} \right]\)⇒ [A][X]=[D] ---(4)Here,\([A] = \left[ {\begin{array}{*{20}{c}} 2&4&{ - 6}\\ 1&3&1\\ 2&{ - 4}&{ - 2} \end{array}} \right]\)\([D] = \left[ {\begin{array}{*{20}{c}} { - 8}\\ {10}\\ { - 12} \end{array}} \right]\)\([X] = \left[ {\begin{array}{*{20}{c}} {x_1}\\ {x_2}\\ {x_3} \end{array}} \right]\)STEP 1: Writing Augment Matrix\(\therefore[A~\vdots ~D] = \left[ {\begin{array}{*{20}{c}} 2&4&{ - 6}\\ 1&3&1\\ 2&{ - 4}&{ - 2} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { - 8}\\ {10}\\ { - 12} \end{array}} \right]\)Step 2: Applying elementary transformations or pivoting of the augment MATRIXA) R1 → \(\frac{1}{2}\)R1\( = \left[ {\begin{array}{*{20}{c}} 1&2&{ -3}\\ 1&3&1\\ 2&{ - 4}&{ - 2} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { - 4}\\ {10}\\ { - 12} \end{array}} \right]\)b) R2 → R2 - R1; R3 → R3 - 2R1\( = \left[ {\begin{array}{*{20}{c}} 1&2&{ -3}\\ 0&1&4\\ 0&{ - 8}&{ 4} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { - 4}\\ {14}\\ { - 4} \end{array}} \right]\)c) R3 → R3 + 8R2\( = \left[ {\begin{array}{*{20}{c}} 1&2&{ -3}\\ 0&1&4\\ 0&{ 0}&{ 36} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { - 4}\\ {14}\\ { 108} \end{array}} \right]\)Step 3: Solving for variables -Now, the reduced equations can be written as -x1 + 2x2 - 3x3 = - 4 ---(5)x2 + 4x3 = 14 ---(6)36x3 = 108 ---(7)Solving (5), (6), and (7)x3 = 3x2 = 2x1 = 1