Answer
Augmented matrix:
$$
\begin{bmatrix}
1&2&4&-2\\
0&1&5&2\\
-2&-4&-3&9\\
\end{bmatrix}
$$
Solution:
$$
\begin{bmatrix}
0\\-3\\1\end{bmatrix}
$$
Work Step by Step
We write the augmented matrix by adding $b$ as the fourth column of a matrix whose first three columns are $A$:
$$
\begin{bmatrix}
1&2&4&-2\\
0&1&5&2\\
-2&-4&-3&9\\
\end{bmatrix}
$$
To solve the system, we use row reduction.
First, add twice the first row to the third row:
$$
\begin{bmatrix}
1&2&4&-2\\
0&1&5&2\\
0&0&5&5\\
\end{bmatrix}
$$
Divide the third row by $5$ to put the matrix into row echelon form:
$$
\begin{bmatrix}
1&2&4&-2\\
0&1&5&2\\
0&0&1&1\\
\end{bmatrix}
$$
Now, we put it into reduced echelon form. First, add multiples of the third row to the first and second rows so that there is only one nonzero entry in the third column of the matrix:
$$
\begin{bmatrix}
1&2&0&-6\\
0&1&0&-3\\
0&0&1&1\\
\end{bmatrix}
$$
We do the same for the second column, subtracting twice the second row from the first row:
$$
\begin{bmatrix}
1&0&0&0\\
0&1&0&-3\\
0&0&1&1\\
\end{bmatrix}
$$
This leaves us with our solution $x$ to the equation $Ax=b$:
$$
x=
\begin{bmatrix}
0\\-3\\1\end{bmatrix}
$$
