To solve the given linear system using Gauss-Jordan elimination, follow these detailed steps:
1. Write the system of equations in augmented matrix form:
[tex]\[
\begin{pmatrix}
-3 & 4 & | & -6 \\
5 & -1 & | & 10
\end{pmatrix}
\][/tex]
2. Normalize the first row to make the leading coefficient (first element) 1.
Divide each element in the first row by -3:
[tex]\[
\begin{pmatrix}
1 & -\frac{4}{3} & | & 2 \\
5 & -1 & | & 10
\end{pmatrix}
\][/tex]
3. Eliminate the x-term from the second row.
Use row operations to make the element in the second row, first column (5) zero. Subtract 5 times the first row from the second row:
[tex]\[
\begin{pmatrix}
1 & -\frac{4}{3} & | & 2 \\
0 & \frac{5}{3} & | & 0
\end{pmatrix}
\][/tex]
4. Normalize the second row to make the leading coefficient (second element) 1.
Divide each element in the second row by [tex]\(\frac{5}{3}\)[/tex]:
[tex]\[
\begin{pmatrix}
1 & -\frac{4}{3} & | & 2 \\
0 & 1 & | & 0
\end{pmatrix}
\][/tex]
5. Eliminate the y-term from the first row.
Use row operations to make the element in the first row, second column [tex]\(-\frac{4}{3}\)[/tex] zero. Add [tex]\(\frac{4}{3}\)[/tex] times the second row to the first row:
[tex]\[
\begin{pmatrix}
1 & 0 & | & 2 \\
0 & 1 & | & 0
\end{pmatrix}
\][/tex]
Now the augmented matrix is in reduced row-echelon form (RREF). From this matrix, we can directly read off the solutions:
[tex]\[
\begin{cases}
x = 2 \\
y = 0
\end{cases}
\][/tex]
So the solution to the system of equations is [tex]\((2, 0)\)[/tex].
Therefore, the best answer is:
A. (2,0)
