Question

Solve the augmented matrix below
2 1 5
6 3 1

Answer 1

To solve the augmented matrix

\[
\begin{bmatrix}
2 & 1 & | & 5 \\
6 & 3 & | & 1
\end{bmatrix},
\]

we can perform row operations to bring it to row echelon form, and then ultimately to reduced row echelon form.

1. **Normalize the first row:** We can divide the first row by 2 to make the leading coefficient 1.

\[
\begin{bmatrix}
1 & 0.5 & | & 2.5 \\
6 & 3 & | & 1
\end{bmatrix}
\]

2. **Eliminate the first column's second-row element:** We can subtract 6 times the first row from the second row.

First, we calculate \(6 \times \text{Row 1} = 6 \times [1, 0.5 | 2.5] = [6, 3 | 15]\).

Now subtract this from Row 2:

\[
\begin{bmatrix}
1 & 0.5 & | & 2.5 \\
(6 - 6) & (3 - 3) & | & (1 - 15)
\end{bmatrix}
\]

We get:

\[
\begin{bmatrix}
1 & 0.5 & | & 2.5 \\
0 & 0 & | & -14
\end{bmatrix}
\]

3. Now we have the following system of equations represented by the above matrix:

1. \(x + 0.5y = 2.5\)
2. \(0 = -14\)

The second equation, \(0 = -14\), is a contradiction, which suggests that there are no solutions to the system. The original system of equations is inconsistent.

Thus, the solution to the augmented matrix indicates that there are no solutions.