To solve the given system of equations using an augmented matrix, we first need to represent the system in matrix form.
The system is:
\[ \begin{align*}
- & \quad x + 4y - z = 20 \
- & \quad 3x + 2y + z = 8 \
- & \quad 2x - 3y + 2z = -16 \end{align*} \]
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.
- We can scale Row 1, \( R_1 \), but no scaling is necessary since the first pivot is already in place.
- 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:
\[ \begin{bmatrix} 1 & 4 & -1 & | & 20 \ 0 & -10 & 4 & | & -52 \ 0 & -11 & 4 & | & -56 \end{bmatrix} \]
- Next, we can simplify \( R_2 \) by dividing by -2:
\[ R_2 \div -2 \Rightarrow [0, 5, -2 | 26] \]
- 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.