Question

Solve the augmented matrix below
4 2 10
1 -1 -13

Answer 1

To solve the augmented matrix, we have the following system of equations:

\[
\begin{pmatrix}
4 & 2 & | & 10 \\
1 & -1 & | & -13
\end{pmatrix}
\]

This corresponds to the equations:
1. \( 4x + 2y = 10 \)
2. \( x - y = -13 \)

We will perform row operations to obtain the reduced row echelon form (RREF).

### Step 1: Normalize the second row

First, we can multiply the second row by 4 to set it for elimination:

\[
R_1: 4x + 2y = 10
\]
\[
R_2: 4x - 4y = -52 \quad (\text{multiply the second equation by 4})
\]

Now, the augmented matrix looks like this:

\[
\begin{pmatrix}
4 & 2 & | & 10 \\
4 & -4 & | & -52
\end{pmatrix}
\]

### Step 2: Eliminate \( R_1 \) from \( R_2 \)

Now, subtract the first row from the second row:

\[
R_2 = R_2 - R_1
\]

This gives us:

\[
4x - 4y - (4x + 2y) = -52 - 10
\]
\[
-6y = -62
\]
\[
y = \frac{-62}{-6} = \frac{31}{3}
\]

Next, we will update the first equation to find \( x \). Substituting \( y = \frac{31}{3} \):

Substituting \( y \) back into the first equation \((4x + 2y = 10)\):

\[
4x + 2\left(\frac{31}{3}\right) = 10
\]
\[
4x + \frac{62}{3} = 10
\]

### Step 3: Isolate \( x \)

Multiply through by 3 to eliminate the fraction:

\[
12x + 62 = 30
\]
\[
12x = 30 - 62
\]
\[
12x = -32
\]
\[
x = \frac{-32}{12} = \frac{-8}{3}
\]

### Solution

The solution to the system of equations is:

\[
x = -\frac{8}{3}, \quad y = \frac{31}{3}
\]

Thus, the final answer is:

\[
\boxed{\left(-\frac{8}{3}, \frac{31}{3}\right)}
\]