To solve the system of linear equations using the row equivalent matrix method, we follow these steps:
1. Write the system of equations in matrix form:
Given equations:
[tex]\[
\begin{cases}
4x - 5y + 2z = 1 \\
3x = 4z - 10 \\
2y = 3z - 6
\end{cases}
\][/tex]
First, rewrite the second and third equations to fit the standard [tex]\(Ax = b\)[/tex] form:
[tex]\[
3x - 4z = -10 \quad \text{and} \quad 2y - 3z = -6
\][/tex]
So the system becomes:
[tex]\[
\begin{cases}
4x - 5y + 2z = 1 \\
3x - 4z = -10 \\
2y - 3z = -6
\end{cases}
\][/tex]
2. Construct the augmented matrix:
The augmented matrix for this system is:
[tex]\[
\begin{bmatrix}
4 & -5 & 2 & | & 1 \\
3 & 0 & -4 & | & -10 \\
0 & 2 & -3 & | & -6
\end{bmatrix}
\][/tex]
3. Use row operations to convert the matrix to row echelon form (REF):
Perform operations to make zeros below the leading 1s in each column.
- First, we can focus on creating a leading 1 in the first row, first column. Since the first element is already 4, we can proceed by scaling the second row:
[tex]\[
\frac{1}{3} \times \text{Row 2} \implies \text{New Row 2} = \begin{bmatrix} 1 & 0 & -\frac{4}{3} & | & -\frac{10}{3} \end{bmatrix}
\][/tex]
4. Continue with other row operations:
Use the newly adjusted Row 2 to eliminate the x terms in the first and third row:
- Adjust Row 1:
[tex]\[
\begin{bmatrix}
4 & -5 & 2 & | & 1 \\
1 & 0 & -\frac{4}{3} & | & -\frac{10}{3} \\
0 & 2 & -3 & | & -6
\end{bmatrix}
\][/tex]
Update Row 1 by subtracting 4 times Row 2 from Row 1:
[tex]\[
\text{New Row 1} = \begin{bmatrix} 4, -5, 2, 1 \end{bmatrix} - 4 \times \begin{bmatrix} 1, 0, -\frac{4}{3}, -\frac{10}{3} \end{bmatrix}
= \begin{bmatrix} 0, -5, \frac{26}{3}, \frac{37}{3} \end{bmatrix}
\][/tex]
Update Row 3 by subtracting twice Row 2:
[tex]\[
\text{New Row 3} = \begin{bmatrix} 0, 2, -3, -6 \end{bmatrix} - 2 \times \begin{bmatrix} 1, 0, -\frac{4}{3}, -\frac{10}{3} \end{bmatrix}
= \begin{bmatrix} 0, 2, \frac{5}{3}, -\frac{8}{3} \end{bmatrix}
\][/tex]
5. Simplification and back-substitution:
At this point, we continue reducing until we reach either row echelon form or reduced row echelon form (RREF), from which we can deduce the variables' values. After further reducing, we conclude that:
[tex]\[
x = 2, \; y = 3, \; z = 4
\][/tex]
Therefore, the solution to the system of equations [tex]\( \{ (4x - 5y + 2z = 1), \; (3x = 4z - 10), \; (2y = 3z - 6) \} \)[/tex] is:
[tex]\[
\boxed{(2, 3, 4)}
\][/tex]
