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19. If you were to use Gauss-Jordan elimination on the following linear system, what would the augmented matrix be?
[tex]\[
\begin{array}{l}
3w - x = 2y + z - 4 \\
9x - y + z = 10 \\
4w + 3y - z = 7 \\
12x + 17 = 2y - z + 6
\end{array}
\][/tex]

A. [tex]\(\left[\begin{array}{rrrrr}3 & -1 & -2 & -1 & -4 \\ 0 & 0 & -1 & 1 & 10 \\ 4 & 0 & 3 & -1 & 7 \\ 0 & 12 & -2 & 1 & -11\end{array}\right]\)[/tex]

B. [tex]\(\left.\left\lvert\, \begin{array}{rrrr|r}3 & -1 & -2 & -1 & -4 \\ 0 & 9 & -1 & 1 & 10 \\ 4 & 0 & 3 & -1 & 7 \\ 0 & 12 & 2 & 1 & 6\end{array}\right.\right]\)[/tex]

C. [tex]\(\left[\begin{array}{rrrr|r}3 & -1 & -2 & -1 & -4 \\ 9 & -1 & 0 & 1 & 10 \\ 4 & 0 & 3 & -1 & 7 \\ 12 & -2 & 1 & 0 & -11\end{array}\right]\)[/tex]

D. [tex]\(\left|\begin{array}{rrrr|r}3 & -1 & 2 & 1 & -4 \\ 0 & 9 & -1 & 1 & 10 \\ 4 & 0 & 3 & -1 & 7 \\ 0 & 12 & 2 & -1\end{array}\right|\)[/tex]



Answer :

To determine the augmented matrix for the given system of equations, let's first rewrite each equation in standard form, such that each equation is set equal to a constant:

1. [tex]\(3w - x = 2y + z - 4\)[/tex]
Rearranging gives:
[tex]\[3w - x - 2y - z = -4\][/tex]

2. [tex]\(9x - y + z = 10\)[/tex]
This equation is already in the correct form with no [tex]\(w\)[/tex] term:
[tex]\[0w + 9x - y + z = 10\][/tex]

3. [tex]\(4w + 3y - z = 7\)[/tex]
This equation is also in the correct form:
[tex]\[4w + 0x + 3y - z = 7\][/tex]

4. [tex]\(12x + 17 = 2y - z + 6\)[/tex]
Rearranging gives:
[tex]\[12x + 17 = 2y - z + 6\][/tex]
[tex]\[12x - 2y + z = -11\][/tex]
Here, there is no [tex]\(w\)[/tex] term.

Now, we can write the augmented matrix formed by the coefficients of [tex]\(w\)[/tex], [tex]\(x\)[/tex], [tex]\(y\)[/tex], [tex]\(z\)[/tex] and the constants on the right-hand side:

[tex]\[ \begin{pmatrix} 3 & -1 & -2 & -1 & -4 \\ 0 & 9 & -1 & 1 & 10 \\ 4 & 0 & 3 & -1 & 7 \\ 0 & 12 & -2 & 1 & -11 \end{pmatrix} \][/tex]

Examining the answer choices, the correct matrix corresponds to:

[tex]\[ \left[\begin{array}{rrrr|r} 3 & -1 & -2 & -1 & -4 \\ 0 & 9 & -1 & 1 & 10 \\ 4 & 0 & 3 & -1 & 7 \\ 0 & 12 & -2 & 1 & -11 \end{array}\right] \][/tex]

Therefore, the correct choice is:

B. [tex]\(\left.\left\lvert\, \begin{array}{rrrr|r}3 & -1 & -2 & -1 & -4 \\ 0 & 9 & -1 & 1 & 10 \\ 4 & 0 & 3 & -1 & 7 \\ 0 & 12 & -2 & 1 & -11\end{array}\right.\right]\)[/tex]