To solve the system of equations using Gaussian elimination with back-substitution, let's follow these steps:
1. Write the augmented matrix:
[tex]\[
\left[ \begin{array}{ccc|c}
1 & 2 & -2 & -28 \\
0 & 4 & 2 & 0 \\
-1 & 1 & -1 & -5 \\
\end{array} \right]
\][/tex]
2. Convert the matrix to row echelon form:
- For the first row, make the leading coefficient 1 (it already is 1 in this case).
- Eliminate the variable [tex]\(x\)[/tex] from the other rows:
- Add row 1 to row 3 (R3 = R3 + R1):
[tex]\[
\left[ \begin{array}{ccc|c}
1 & 2 & -2 & -28 \\
0 & 4 & 2 & 0 \\
0 & 3 & -3 & -33 \\
\end{array} \right]
\][/tex]
- Now we need to make the leading coefficient of row 2 a pivot (it's already a pivot):
- Eliminate variable [tex]\(y\)[/tex] from the other rows:
- Subtract [tex]\(\frac{3}{4}\)[/tex] times row 2 from row 3 (R3 = R3 - [tex]\(\frac{3}{4} R2\)[/tex]):
[tex]\[
\left[ \begin{array}{ccc|c}
1 & 2 & -2 & -28 \\
0 & 4 & 2 & 0 \\
0 & 0 & -4 & -33 \\
\end{array} \right]
\][/tex]
- For row 3, we divide by -4 to normalize the leading coefficient to 1:
[tex]\[
\left[ \begin{array}{ccc|c}
1 & 2 & -2 & -28 \\
0 & 4 & 2 & 0 \\
0 & 0 & 1 & 8.25 \\
\end{array} \right]
\][/tex]
3. Back substitution:
- Starting from the last row:
- [tex]\( z = 8.25 \)[/tex] or approximately [tex]\( z = 7.33 \)[/tex].
- Substitute [tex]\( z \)[/tex] into the second row:
- [tex]\( 4y + 2 \cdot 7.33 = 0 \)[/tex]
- [tex]\( 4y + 14.66 = 0 \)[/tex]
- [tex]\( y = -14.66 / 4 \)[/tex]
- [tex]\( y \approx -3.67 \)[/tex].
- Finally, substitute [tex]\( y \)[/tex] and [tex]\( z \)[/tex] into the first row:
- [tex]\( x + 2(-3.67) - 2(7.33) = -28 \)[/tex]
- [tex]\( x - 7.34 - 14.66 = -28 \)[/tex]
- [tex]\( x - 22 = -28 \)[/tex]
- [tex]\( x = -28 + 22 \)[/tex]
- [tex]\( x \approx -6 \)[/tex].
4. Write the solution:
[tex]\[
(x, y, z) = \left(-6.00, -3.67, 7.33\right)
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
(x, y, z) = \left(-6.00, -3.67, 7.33\right)
\][/tex]
