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Solve using Gauss-Jordan elimination.

[tex]\[
\begin{aligned}
4x_1 + 14x_2 - 37x_3 &= 53 \\
2x_1 + 16x_2 - 42x_3 &= 66 \\
x_1 + 5x_2 - 13x_3 &= 20
\end{aligned}
\][/tex]

Select the correct choice below and fill in the answer box(es) within your choice:

A. The unique solution is [tex]\( x_1 = \square \)[/tex], [tex]\( x_2 = \square \)[/tex], and [tex]\( x_3 = \square \)[/tex].

B. The system has infinitely many solutions. The solution is [tex]\( x_1 = \square \)[/tex], [tex]\( x_2 = \square \)[/tex], and [tex]\( x_3 = t \)[/tex]. (Simplify your answers. Type expressions using [tex]\( t \)[/tex] as the variable.)

C. The system has infinitely many solutions. The solution is [tex]\( x_1 = \square \)[/tex], [tex]\( x_2 = \square \)[/tex], and [tex]\( x_3 = t \)[/tex]. (Simplify your answers. Type an expression using [tex]\( s \)[/tex] and [tex]\( t \)[/tex] as the variables.)

D. There is no solution.

Sagot :

GinnyAnswer
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].
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