Question

Solve the following system of equations by using technology to reduce an augmented matrix to row echelon form. What is the solution to the system?
⎪x+4y−z=20
3x+2y+z=8
2x−3y+2z=−16
x= , y= , z=

Answer 1

To solve the given system of equations using an augmented matrix, we first need to represent the system in matrix form.

The system is:

\[
\begin{align*}
1. & \quad x + 4y - z = 20 \\
2. & \quad 3x + 2y + z = 8 \\
3. & \quad 2x - 3y + 2z = -16
\end{align*}
\]

This corresponds to the augmented matrix:

\[
\begin{bmatrix}
1 & 4 & -1 & | & 20 \\
3 & 2 & 1 & | & 8 \\
2 & -3 & 2 & | & -16
\end{bmatrix}
\]

Next, we can perform row operations to reduce this matrix to row echelon form.

1. We can scale Row 1, \( R_1 \), but no scaling is necessary since the first pivot is already in place.
2. We can eliminate \( x \) from \( R_2 \) and \( R_3 \).
- Row 2: \( R_2 - 3R_1 \)
- Row 3: \( R_3 - 2R_1 \)

Carrying out these operations:

\[
R_2 = [3, 2, 1 | 8] - 3[1, 4, -1 | 20] \Rightarrow [0, -10, 4 | -52]
\]
\[
R_3 = [2, -3, 2 | -16] - 2[1, 4, -1 | 20] \Rightarrow [0, -11, 4 | -56]
\]

This gives us:

\[
\begin{bmatrix}
1 & 4 & -1 & | & 20 \\
0 & -10 & 4 & | & -52 \\
0 & -11 & 4 & | & -56
\end{bmatrix}
\]

3. Next, we can simplify \( R_2 \) by dividing by -2:

\[
R_2 \div -2 \Rightarrow [0, 5, -2 | 26]
\]

4. We can also update \( R_3 \):

Subtract \( \frac{11}{10} R_2 \) from \( R_3 \):

\[
R_3 = [0, -11, 4 | -56] - \frac{11}{10}[0, 5, -2 | 26] \Rightarrow [0, 0, 0 | -56 + 28.6] \Rightarrow [0, 0, 0 | -27.4] \text{ (assuming exact numbers)}
\]

So now our matrix looks like:

\[
\begin{bmatrix}
1 & 4 & -1 & | & 20 \\
0 & 5 & -2 & | & 26 \\
0 & 0 & 0 & | & -27.4
\end{bmatrix}
\]

Since \( 0 = -27.4 \) suggests an inconsistency in the equations, we have no solution for this system.

We can summarize:

- The lines corresponding to these equations do not intersect at any point.
- Therefore, the system has no solution.

Final answer: **The system is inconsistent, hence there is no solution.**