To solve the given system of linear equations using Gauss-Jordan elimination, we start by constructing the augmented matrix:
[tex]\[
\left[\begin{array}{ccc|c}
4 & 14 & -37 & 53 \\
2 & 16 & -42 & 66 \\
1 & 5 & -13 & 20
\end{array}\right]
\][/tex]
Next, we perform row operations to transform this matrix into Reduced Row Echelon Form (RREF).
Step-by-Step Transformation:
1. Normalize the first row to have a leading 1 by dividing the entire row by 4:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 3.5 & -9.25 & 13.25 \\
2 & 16 & -42 & 66 \\
1 & 5 & -13 & 20
\end{array}\right]
\][/tex]
2. Eliminate the first column elements below the leading 1:
- [tex]\(R2 \leftarrow R2 - 2 \cdot R1\)[/tex]
- [tex]\(R3 \leftarrow R3 - R1\)[/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & 3.5 & -9.25 & 13.25 \\
0 & 9 & -23.5 & 39.5 \\
0 & 1.5 & -3.75 & 6.75
\end{array}\right]
\][/tex]
3. Normalize the second row to have a leading 1 by dividing the second row by 9:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 3.5 & -9.25 & 13.25 \\
0 & 1 & -2.6111 & 4.3889 \\
0 & 1.5 & -3.75 & 6.75
\end{array}\right]
\][/tex]
4. Eliminate the second column elements above and below the leading 1 by appropriate row operations:
- [tex]\(R1 \leftarrow R1 - 3.5 \cdot R2\)[/tex]
- [tex]\(R3 \leftarrow R3 - 1.5 \cdot R2\)[/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & 0 & 0.7778 & -2 \\
0 & 1 & -2.6111 & 4.3889 \\
0 & 0 & 0.1667 & 1
\end{array}\right]
\][/tex]
5. Normalize the third row to have a leading 1 by dividing the third row by 0.1667:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 0 & 0.7778 & -2 \\
0 & 1 & -2.6111 & 4.3889 \\
0 & 0 & 1 & 1
\end{array}\right]
\][/tex]
6. Finally, eliminate the third column elements above the leading 1 in the third row:
- [tex]\(R1 \leftarrow R1 - 0.7778 \cdot R3\)[/tex]
- [tex]\(R2 \leftarrow R2 + 2.6111 \cdot R3\)[/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 7 \\
0 & 0 & 1 & 1
\end{array}\right]
\][/tex]
At this point, the matrix is in Reduced Row Echelon Form, and we can directly read off the solutions:
[tex]\[
x_1 = -2, \quad x_2 = 7, \quad x_3 = 1
\][/tex]
So, the unique solution to the given system of equations is:
[tex]\[
\boxed{x_1 = -2, x_2 = 7, x_3 = 1}
\][/tex]
Thus, the correct choice is A. The unique solution is [tex]\( x_1 = -2 \)[/tex], [tex]\( x_2 = 7 \)[/tex], and [tex]\( x_3 = 1 \)[/tex].
