Question

Solve the System of Equations using an Augmented Matrix (Hint: use your calculator)(1 point)
Responses

x = -4, y = -3, z = -2
x = -4, y = -3, z = -2

x = -4, y = 3, z = -2
x = -4, y = 3, z = -2

x = 4, y = 3, z = 2
x = 4, y = 3, z = 2

x = 4, y = -3, z = -2
x = 4, y = -3, z = -2

Answer 1

x + 2y + 6z=22
3x + 4y + z=26
6x - y -z =19

Answer 2

whats the answer

Answer 3

To solve a system of equations using an augmented matrix, we first need the system of equations itself. However, since the equations are not provided in your message, I won't be able to perform the calculations.

Typically, an augmented matrix is created from a system of linear equations. For example, for a system of equations:
\[
\begin{align*}
x + 2y + 3z &= 1 \\
4x + y - z &= -2 \\
-2x + 5y + 7z &= 3
\end{align*}
\]

The corresponding augmented matrix would look like this:
\[
\begin{bmatrix}
1 & 2 & 3 & | & 1 \\
4 & 1 & -1 & | & -2 \\
-2 & 5 & 7 & | & 3
\end{bmatrix}
\]

From there, you can use row operations (Gaussian elimination) to solve the system and find the values of \(x\), \(y\), and \(z\).

Since you provided possible solutions:
1. \(x = -4, y = -3, z = -2\)
2. \(x = -4, y = 3, z = -2\)
3. \(x = 4, y = 3, z = 2\)
4. \(x = 4, y = -3, z = -2\)

If one of these matches the solution you obtained from your calculations, that would be the answer. Please provide the original equations if you'd like me to help you step through the matrix operations or verify the results.

Answer 4

To solve the system of equations using an augmented matrix, we first write the given equations in standard form:

1. \( x + 2y + 6z = 22 \)
2. \( 3x + 4y + z = 26 \)
3. \( 6x - y - z = 19 \)

Next, we form the augmented matrix from this system:

\[
\begin{bmatrix}
1 & 2 & 6 & | & 22 \\
3 & 4 & 1 & | & 26 \\
6 & -1 & -1 & | & 19
\end{bmatrix}
\]

Now, we can perform row operations to bring this matrix to row echelon form (or reduced row echelon form). Below are the steps:

1. **R1 to R1:** Keep \(R_1\) as is.
2. **R2 to R2 - 3R1:** \(R_2 \leftarrow R_2 - 3R_1\)
3. **R3 to R3 - 6R1:** \(R_3 \leftarrow R_3 - 6R_1\)

Calculating the new rows:
- \(R_2: [3 - 3(1), 4 - 3(2), 1 - 3(6), 26 - 3(22)] = [0, -2, -17, -16]\)
- \(R_3: [6 - 6(1), -1 - 6(2), -1 - 6(6), 19 - 6(22)] = [0, -13, -37, -125]\)

Now, the augmented matrix looks like:

\[
\begin{bmatrix}
1 & 2 & 6 & | & 22 \\
0 & -2 & -17 & | & -16 \\
0 & -13 & -37 & | & -125
\end{bmatrix}
\]

4. Next, we can simplify \(R_2\) by dividing it by -2:

\[
R_2 \leftarrow \frac{R_2}{-2}
\]

Now we have:

\[
\begin{bmatrix}
1 & 2 & 6 & | & 22 \\
0 & 1 & \frac{17}{2} & | & 8 \\
0 & -13 & -37 & | & -125
\end{bmatrix}
\]

5. Now, we will continue to eliminate \(y\) from \(R_3\) using \(R_2\):

\(R_3 \leftarrow R_3 + 13R_2\)

Calculating the new \(R_3\):

- \(R_3: [0, -13 + 13(1), -37 + 13\left(\frac{17}{2}\right), -125 + 13(8)] = [0, 0, -\frac{5}{2}, -1]\)

The augmented matrix now simplifies to:

\[
\begin{bmatrix}
1 & 2 & 6 & | & 22 \\
0 & 1 & \frac{17}{2} & | & 8 \\
0 & 0 & -\frac{5}{2} & | & -1
\end{bmatrix}
\]

6. Solving for \(z\) in \(R_3\):

From \(-\frac{5}{2}z = -1\):
\[
z = \frac{2}{5}
\]

7. Now substitute \(z\) back into \(R_2\):

From \(y + \frac{17}{2}z = 8\):
\[
y + \frac{17}{2}\left(\frac{2}{5}\right) = 8 \\
y + \frac{17}{5} = 8 \\
y = 8 - \frac{17}{5} = \frac{40}{5} - \frac{17}{5} = \frac{23}{5}
\]

8. Now substitute \(y\) and \(z\) back into \(R_1\):

From \(x + 2y + 6z = 22\):
\[
x + 2\left(\frac{23}{5}\right) + 6\left(\frac{2}{5}\right) = 22 \\
x + \frac{46}{5} + \frac{12}{5} = 22 \\
x + \frac{58}{5} = 22 \\
x = 22 - \frac{58}{5} = \frac{110}{5} - \frac{58}{5} = \frac{52}{5}
\]

So, the solution is:

\[
x = \frac{52}{5}, \quad y = \frac{23}{5}, \quad z = \frac{2}{5}
\]

Now let's check possible solutions provided earlier:

1. \( x = -4, y = -3, z = -2 \)
2. \( x = -4, y = 3, z = -2 \)
3. \( x = 4, y = 3, z = 2 \)
4. \( x = 4, y = -3, z = -2 \)

None of the provided options match the calculated values \(x = \frac{52}{5}, y = \frac{23}{5}, z = \frac{2}{5}\).

Therefore, the solution is:

\[
x = \frac{52}{5}, \quad y = \frac{23}{5}, \quad z = \frac{2}{5}
\]

Answer 5

The solution to the system of equations

1. \( x + 2y + 6z = 22 \)
2. \( 3x + 4y + z = 26 \)
3. \( 6x - y - z = 19 \)

is:

\[
x = \frac{52}{5}, \quad y = \frac{23}{5}, \quad z = \frac{2}{5}
\]

### In decimal form, this is approximately:
- \( x \approx 10.4 \)
- \( y \approx 4.6 \)
- \( z \approx 0.4 \)

Since none of the provided multiple-choice answers match these results, it appears that the correct answer is not among the options you provided. It is essential to verify your system of equations if you expect to find a match.