Solve the following system of equations by using technology to reduce an augmented matrix to reduced row echelon form. What is the value of y?

Question

⎧⎩⎨−2x−6y+z=−33−3x−y−6z=34−4x+4y−4z=20

Answer 1

To solve the system of equations using an augmented matrix and reduce it to reduced row echelon form (RREF), we start by setting up the augmented matrix from the given equations:

\[ \begin{align*} -2x - 6y + z &= -33 \ -3x - y - 6z &= 34 \ -4x + 4y - 4z &= 20 \end{align*} \]

This system can be expressed in augmented matrix form as:

\[ \begin{bmatrix} -2 & -6 & 1 & | & -33 \ -3 & -1 & -6 & | & 34 \ -4 & 4 & -4 & | & 20 \end{bmatrix} \]

Next, we row reduce this matrix to RREF. You can use technology such as a graphing calculator, software like MATLAB or Python, or online tools to perform the row reduction.

Let's perform the row operations step-by-step:

1. Multiply row 1 by -1/2 to make the first pivot \(1\):

   \[ R_1 = \frac{1}{-2} R_1 \implies \begin{bmatrix} 1 & 3 & -\frac{1}{2} & | & \frac{33}{2} \ -3 & -1 & -6 & | & 34 \ -4 & 4 & -4 & | & 20 \end{bmatrix} \]

2. Add \(3 R_1\) to \(R_2\) and add \(4 R_1\) to \(R_3\) to eliminate the first column entries below the pivot:

   \[ \begin{bmatrix} 1 & 3 & -\frac{1}{2} & | & \frac{33}{2} \ 0 & 8 & -\frac{15}{2} & | & \frac{172}{2} \ 0 & 16 & -6 & | & 80 \end{bmatrix} \]

3. Multiply row 2 by 1/8 to make the pivot in the second column \(1\):

   \[ R_2 = \frac{1}{8} R_2 \implies \begin{bmatrix} 1 & 3 & -\frac{1}{2} & | & \frac{33}{2} \ 0 & 1 & -\frac{15}{16} & | & 21.5 \ 0 & 16 & -6 & | & 80 \end{bmatrix} \]

4. Subtract \(16 R_2\) from \(R_3\):

   \[ R_3 = R_3 - 16 R_2 \implies \begin{bmatrix} 1 & 3 & -\frac{1}{2} & | & \frac{33}{2} \ 0 & 1 & -\frac{15}{16} & | & 21.5 \ 0 & 0 & 1.5 & | & -36 \end{bmatrix} \]

5. Next, simplify \(R_3\) by multiplying it by \(\frac{2}{3}\):

   \[ R_3 = \frac{2}{3} R_3 \implies \begin{bmatrix} 1 & 3 & -\frac{1}{2} & | & \frac{33}{2} \ 0 & 1 & -\frac{15}{16} & | & 21.5 \ 0 & 0 & 1 & | & -24 \end{bmatrix} \]

6. Back substitute to eliminate entries above pivot \(z\) (which corresponds to \(R_3\)):

   From \(R_3\): \(z = -24\)

   Substitute into \(R_2\): \[ y - \frac{15}{16}(-24) = 21.5 \implies y + 22.5 = 21.5 \implies y = -1 \]

   Substitute into \(R_1\): \[ x + 3(-1) - \frac{1}{2}(-24) = \frac{33}{2} \implies x - 3 + 12 = \frac{33}{2} \] Solve for \(x\): \[ x + 9 = \frac{33}{2} \implies x = \frac{33}{2} - 9 = \frac{33}{2} - \frac{18}{2} = \frac{15}{2} \]

The solution to the system is:

• \(x = \frac{15}{2}\)
• \(y = -1\)
• \(z = -24\)

Thus, the value of \(y\) is:

\[ \boxed{-1} \]