Question

Solve the augmented matrix below
-8 -10 -24
6 5 2

Answer 1

To solve the augmented matrix given by

\[
\begin{bmatrix}
-8 & -10 & | & -24 \\
6 & 5 & | & 2
\end{bmatrix}
\]

we will perform row operations to reduce it to row echelon form or reduced row echelon form.

### Step 1: Create a Leading 1

First, we can swap the two rows to make calculations easier. Let's switch Row 1 (R1) and Row 2 (R2):

\[
\begin{bmatrix}
6 & 5 & | & 2 \\
-8 & -10 & | & -24
\end{bmatrix}
\]

### Step 2: Make the Leading Coefficient of R1 a 1

To make the leading coefficient of R1 a 1, we divide the entire first row by 6:

\[
R1 \rightarrow \frac{1}{6} R1
\]

\[
\begin{bmatrix}
1 & \frac{5}{6} & | & \frac{1}{3} \\
-8 & -10 & | & -24
\end{bmatrix}
\]

### Step 3: Eliminate the First Column of R2

Next, we want to eliminate the leading coefficient in R2. We can add 8 times the first row to the second row:

\[
R2 \rightarrow R2 + 8R1
\]

Calculating this step:

\[
R2 = -8 + 8(1) = 0 \\
R2 = -10 + 8\left(\frac{5}{6}\right) = -10 + \frac{40}{6} = -10 + \frac{20}{3} = -\frac{30}{3} + \frac{20}{3} = -\frac{10}{3} \\
R2 = -24 + 8\left(\frac{1}{3}\right) = -24 + \frac{8}{3} = -\frac{72}{3} + \frac{8}{3} = -\frac{64}{3}
\]

Thus, the augmented matrix becomes:

\[
\begin{bmatrix}
1 & \frac{5}{6} & | & \frac{1}{3} \\
0 & -\frac{10}{3} & | & -\frac{64}{3}
\end{bmatrix}
\]

### Step 4: Create a Leading 1 in R2

Now we divide R2 by \(-\frac{10}{3}\):

\[
R2 \rightarrow -\frac{3}{10}R2
\]

This results in:

\[
\begin{bmatrix}
1 & \frac{5}{6} & | & \frac{1}{3} \\
0 & 1 & | & \frac{64}{10} \cdot \frac{-3}{10} = \frac{64}{10} \cdot \frac{-3}{10} = -\frac{192}{100} = -\frac{96}{50} = -\frac{48}{25}
\end{bmatrix}
\]

The matrix now looks like:

\[
\begin{bmatrix}
1 & \frac{5}{6} & | & \frac{1}{3} \\
0 & 1 & | & -\frac{48}{25}
\end{bmatrix}
\]

### Step 5: Eliminate Above the Leading 1 in R2

Now, we will eliminate the \(\frac{5}{6}\) in R1 by subtracting \(\frac{5}{6}\) times R2 from R1:

\[
R1 \rightarrow R1 - \frac{5}{6}R2
\]

Calculating this step:

\[
R1 = 1 - \frac{5}{6}(0) = 1 \\
R1 = \frac{5}{6} - \frac{5}{6}(1) = \frac{5}{6} - \frac{5}{6} = 0 \\
R1 = \frac{1}{3} - \frac{5}{6}\left(-\frac{48}{25}\right) = \frac{1}{3} + \frac{240}{150} = \frac{1}{3} + \frac{16}{1} = 1/3 + 16/1 =
\]

This leads to:

\[
\begin{bmatrix}
1 & 0 & | & \frac{1}{3} + \frac{80}{100} = 16\frac{1}{3}\\
0 & 1 & | & -\frac{48}{25}
\end{bmatrix}
\]

### Final Result

Now the augmented matrix is in reduced row echelon form, from which we can read the solution:

\[
x = \frac{1}{3} + 16 = \frac{49}{3}, \quad y = -\frac{48}{25}
\]

Thus, the solution to the system represented by the original augmented matrix is:

\[
\begin{align*}
x & = \frac{49}{3}, \\
y & = -\frac{48}{25}.
\end{align*}
\]