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Solve using Gauss-Jordan elimination.

[tex]\[
\begin{aligned}
4x_1 + 10x_2 - 24x_3 &= -22 \\
3x_1 + 25x_2 - 61x_3 &= -23 \\
x_1 + 5x_2 - 12x_3 &= -6
\end{aligned}
\][/tex]

1. Write the system of equations as an augmented matrix.

[tex]\[
\left[\begin{array}{ccc|c}
4 & 10 & -24 & -22 \\
3 & 25 & -61 & -23 \\
1 & 5 & -12 & -6
\end{array}\right]
\][/tex]

2. Solve the system. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

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

(Simplify your answers.)

B. The system has infinitely many solutions. The solution is [tex]\( x_1 = \)[/tex] [tex]\(\square\)[/tex] , [tex]\( x_2 = \)[/tex] [tex]\(\square\)[/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]\( x_1 = \)[/tex] [tex]\(\square\)[/tex] , [tex]\( x_2 = s \)[/tex] , and [tex]\( x_3 = t \)[/tex].

(Simplify your answers. Type expressions using [tex]\( s \)[/tex] and [tex]\( t \)[/tex] as the variables.)

D. There is no solution.



Answer :

GinnyAnswer
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].

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