Question 8.1.4: Gaussian Elimination with Back-Substitution Use matrices to ...

Step 1 Write the augmented matrix for the system.

\begin{matrix} \text{Linear System} \\ \begin{cases} 2w + x + 3y – z = 6 \\w – x + 2y – 2z = -1\\w – x – y + z = -4\\-w + 2x – 2y – z = -7.\end{cases}\end{matrix}                \begin{matrix} \text{Augmented Matrix} \\\begin{bmatrix} \begin{matrix}  \begin{array} {c c c c | c} 2 & 1&3&-1&6\\1 &-1&2&-2&-1\\1 & -1&-1&1&-4 \\ -1 &2&-2&-1&-7 \end{array} \end{matrix}\end{bmatrix} \end{matrix}

Step 2 Use matrix row operations to simplify the matrix to row-echelon form, with 1s down the main diagonal from upper left to lower right, and 0s below the 1s in the first, second, and third columns. Our first step in achieving this goal is to get”1 in the top position of the first column. To do this, we interchange row 1 and row 2: R_1↔R_2.

Now we use the 1 at the top of the first column to get 0s below it.

We move on to the second column. We can obtain 1 in the desired position by multiplying the numbers in the second row by \frac{1}{3}, the reciprocal of 3.

Now we use the 1 in the second row, second column position to get 0s below it.

We move on to the third column. We can obtain 1 in the desired position by multiplying the numbers in the third row by -\frac{1}{3}, the reciprocal of -3.

Now we use the 1 in the third column to get 0 below it.

We move on to the fourth column. Because we want 1s down the diagonal from upper left to lower right, we want 1 where there is now -\frac{11}{3}. We can obtain 1 in this position by multiplying the numbers in the fourth row by -\frac{11}{3}.

We now have the desired matrix in row-echelon form, with 1s down the main diagonal and 0s below the 1s. An equivalent row-echelon matrix can be obtained using a graphing utility and the \fbox{ref} command on the augmented matrix.

Step 3  Write the system of linear equations corresponding to the matrix in step 2 and use back-substitution to find the system’s solution. The system represented by the final matrix in step 2 is

\begin{bmatrix} \begin{matrix}  \begin{array} {c c c c | c} 1 & -1&2&-2&-1\\0 &1&-\frac{1}{3} &1&\frac{8}{3} \\0 & 0&1&-1&1\\ 0 &0&0&1&3 \end{array} \end{matrix} \end{bmatrix}\rightarrow \begin{cases}1w – 1x + 2y – 2z = -1 \\0w + 1x -\frac{1}{3}y+ 1z =\frac{8}{3} \\0w + 0x + 1y – 1z = 1\\0w + 0x + 0y + 1z = 3\end{cases}      or    \begin{cases}w – x + 2y – 2z = -1\\ x -\frac{1}{3}y+ z =\frac{8}{3} \\y – z = 1\\z = 3.\end{cases}

We immediately see that the value for z is 3. We can now use back-substitution to find the values for y, x, and w.

Let’s agree to write the solution for the system in the alphabetical order of the variables from left to right, namely (w, x, y, z). Thus, the solution set is {(-2, 1, 4, 3)}. We can verify the solution by substituting the value for each variable into the original system of equations and obtaining four true statements.