Answered

Use the Gaussian elimination method to solve the system of linear equations. If the system has infinitely many solutions, write the solution set with \( z \) arbitrary.

[tex]\[
\begin{aligned}
x + y - z & = 16 \\
-x - 2y + 4z & = -41 \\
-5x - y + 2z & = -34
\end{aligned}
\][/tex]

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

A. There is one solution. The solution set is \(\{ (\square, \square, \square) \} \). (Simplify your answers.)

B. There are infinitely many solutions. The solutions are \(\{ (\square, \square, z) \}\), where \( z \) is any real number. \(\square\) (Simplify your answer.)

C. The system is inconsistent. The solution set is [tex]\(\varnothing\)[/tex].



Answer :

GinnyAnswer
To solve the system of linear equations using the Gaussian elimination method, we need to arrange the equations in an augmented matrix and then perform row operations to bring it to row echelon form (REF) and then reduce it further to reduced row echelon form (RREF). The system of equations is:

[tex]\[ \begin{aligned} x + y - z & = 16 \\ -x - 2y + 4z & = -41 \\ -5x - y + 2z & = -34 \end{aligned} \][/tex]

Step 1: Write the augmented matrix for the system.
[tex]\[ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 16 \\ -1 & -2 & 4 & -41 \\ -5 & -1 & 2 & -34 \end{array}\right] \][/tex]

Step 2: Perform row operations to convert the matrix to Row Echelon Form.

First, we use the first row to eliminate the entries below the first pivot (1 in the first row, first column).

- Add Row 1 to Row 2:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 16 \\ 0 & -1 & 3 & -25 \\ -5 & -1 & 2 & -34 \end{array}\right] \][/tex]

- Add 5 times Row 1 to Row 3:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 16 \\ 0 & -1 & 3 & -25 \\ 0 & 4 & -3 & 46 \end{array}\right] \][/tex]

Step 3: Use the second row to eliminate entries below the second pivot (-1 in the second row, second column).

- Multiply Row 2 by -1:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 16 \\ 0 & 1 & -3 & 25 \\ 0 & 4 & -3 & 46 \end{array}\right] \][/tex]

- Subtract 4 times Row 2 from Row 3:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 16 \\ 0 & 1 & -3 & 25 \\ 0 & 0 & 9 & -54 \end{array}\right] \][/tex]

Step 4: Convert the matrix to Reduced Row Echelon Form.

- Divide Row 3 by 9:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 16 \\ 0 & 1 & -3 & 25 \\ 0 & 0 & 1 & -6 \end{array}\right] \][/tex]

- Add 3 times Row 3 to Row 2:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 16 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & -6 \end{array}\right] \][/tex]

- Add Row 3 to Row 1:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 1 & 0 & 10 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & -6 \end{array}\right] \][/tex]

Now we need to eliminate the non-zero entry in the first row (1, 2):

- Subtract Row 2 from Row 1:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & -6 \end{array}\right] \][/tex]

The reduced matrix corresponds to the equations:
[tex]\[ \begin{aligned} x & = 3 \\ y & = 7 \\ z & = -6 \end{aligned} \][/tex]

Thus, we find that there is one unique solution:
[tex]\[ \boxed{A. \ \text{There is one solution. The solution set is} \ \{(3, 7, -6)\}.} \][/tex]