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To solve the given system of equations using Gauss-Jordan elimination, we should first write the augmented matrix corresponding to the system:
[tex]\[ \begin{pmatrix} 4 & 10 & -24 & -22 \\ 3 & 25 & -61 & -23 \\ 1 & 5 & -12 & -6 \end{pmatrix} \][/tex]
We now proceed with the Gauss-Jordan elimination method:
1. Make the leading coefficient of the first row 1 by dividing the first row by 4:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 3 & 25 & -61 & -23 \\ 1 & 5 & -12 & -6 \end{pmatrix} \][/tex]
2. Eliminate the x1 term from the second and third rows using the first row:
- For the second row: [tex]\( R2 = R2 - 3 \cdot R1 \)[/tex]
[tex]\[ R2 = \begin{pmatrix} 3 & 25 & -61 & -23 \end{pmatrix} - 3 \cdot \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 3 - 3 \cdot 1 & 25 - 3 \cdot 2.5 & -61 + 3 \cdot 6 & -23 + 3 \cdot 5.5 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 17.5 & -43 & -6.5 \end{pmatrix} \][/tex]
- For the third row: [tex]\( R3 = R3 - R1 \)[/tex]
[tex]\[ R3 = \begin{pmatrix} 1 & 5 & -12 & -6 \end{pmatrix} - \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
The updated matrix is:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 0 & 17.5 & -43 & -6.5 \\ 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
3. Make the leading coefficient of the second row 1 by dividing the second row by 17.5:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 0 & 1 & -2.4571 & -0.3714 \\ 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
4. Eliminate the x2 term from the first and third rows using the second row:
- For the first row: [tex]\( R1 = R1 - 2.5 \cdot R2 \)[/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} - 2.5 \cdot \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} \][/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \end{pmatrix} \][/tex]
- For the third row: [tex]\( R3 = R3 - 2.5 \cdot R2 \)[/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 2.5 & -6 & -0.5 \end{pmatrix} - 2.5 \cdot \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 0 & 0.1425 & 0.4285 \end{pmatrix} \][/tex]
The updated matrix is:
[tex]\[ \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \\ 0 & 1 & -2.4571 & -0.3714 \\ 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
5. Make the leading coefficient of the third row 1 (it's already 1), and then we have to eliminate the x3 term from the first and second rows using the third row:
- For the first row: [tex]\( R1 = R1 - 0.1429 \cdot R3 \)[/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \end{pmatrix} - 0.1429 \cdot \begin{pmatrix} 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0 & -4 + (- 5) \end{pmatrix} \][/tex]
- For the second row: [tex]\( R2 = R2 + 2.4571 \cdot R3 \)[/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} + 2.4571 \cdot \begin{pmatrix} 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 1 & 0 & 7 \end{pmatrix} \][/tex]
So, we finally have:
[tex]\[ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 3 \end{pmatrix} \][/tex]
So the solution to the system is:
[tex]\[ x_1 = -5, \quad x_2 = 7, \quad x_3 = 3. \][/tex]
Thus, the correct choice is:
A. The unique solution is [tex]\(x_1 = -5\)[/tex] , [tex]\(x_2 = 7\)[/tex], and [tex]\(x_3 = 3\)[/tex].
[tex]\[ \begin{pmatrix} 4 & 10 & -24 & -22 \\ 3 & 25 & -61 & -23 \\ 1 & 5 & -12 & -6 \end{pmatrix} \][/tex]
We now proceed with the Gauss-Jordan elimination method:
1. Make the leading coefficient of the first row 1 by dividing the first row by 4:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 3 & 25 & -61 & -23 \\ 1 & 5 & -12 & -6 \end{pmatrix} \][/tex]
2. Eliminate the x1 term from the second and third rows using the first row:
- For the second row: [tex]\( R2 = R2 - 3 \cdot R1 \)[/tex]
[tex]\[ R2 = \begin{pmatrix} 3 & 25 & -61 & -23 \end{pmatrix} - 3 \cdot \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 3 - 3 \cdot 1 & 25 - 3 \cdot 2.5 & -61 + 3 \cdot 6 & -23 + 3 \cdot 5.5 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 17.5 & -43 & -6.5 \end{pmatrix} \][/tex]
- For the third row: [tex]\( R3 = R3 - R1 \)[/tex]
[tex]\[ R3 = \begin{pmatrix} 1 & 5 & -12 & -6 \end{pmatrix} - \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
The updated matrix is:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 0 & 17.5 & -43 & -6.5 \\ 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
3. Make the leading coefficient of the second row 1 by dividing the second row by 17.5:
[tex]\[ \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \\ 0 & 1 & -2.4571 & -0.3714 \\ 0 & 2.5 & -6 & -0.5 \end{pmatrix} \][/tex]
4. Eliminate the x2 term from the first and third rows using the second row:
- For the first row: [tex]\( R1 = R1 - 2.5 \cdot R2 \)[/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 2.5 & -6 & -5.5 \end{pmatrix} - 2.5 \cdot \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} \][/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \end{pmatrix} \][/tex]
- For the third row: [tex]\( R3 = R3 - 2.5 \cdot R2 \)[/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 2.5 & -6 & -0.5 \end{pmatrix} - 2.5 \cdot \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 0 & 0.1425 & 0.4285 \end{pmatrix} \][/tex]
The updated matrix is:
[tex]\[ \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \\ 0 & 1 & -2.4571 & -0.3714 \\ 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
5. Make the leading coefficient of the third row 1 (it's already 1), and then we have to eliminate the x3 term from the first and second rows using the third row:
- For the first row: [tex]\( R1 = R1 - 0.1429 \cdot R3 \)[/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0.1429 & -4.5571 \end{pmatrix} - 0.1429 \cdot \begin{pmatrix} 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
[tex]\[ R1 = \begin{pmatrix} 1 & 0 & 0 & -4 + (- 5) \end{pmatrix} \][/tex]
- For the second row: [tex]\( R2 = R2 + 2.4571 \cdot R3 \)[/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 1 & -2.4571 & -0.3714 \end{pmatrix} + 2.4571 \cdot \begin{pmatrix} 0 & 0 & 1 & 3.0053 \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & 1 & 0 & 7 \end{pmatrix} \][/tex]
So, we finally have:
[tex]\[ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 3 \end{pmatrix} \][/tex]
So the solution to the system is:
[tex]\[ x_1 = -5, \quad x_2 = 7, \quad x_3 = 3. \][/tex]
Thus, the correct choice is:
A. The unique solution is [tex]\(x_1 = -5\)[/tex] , [tex]\(x_2 = 7\)[/tex], and [tex]\(x_3 = 3\)[/tex].
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