Solve the equation \(\begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ -6 & 5 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 3 \\ 14 \\ -7 \end{pmatrix}\) by using forward substitution.

This equation is already setup for us to apply forward substition because the first matrix is a lower triangular matrix and the second contains only unknown variables. So we may simply solve the three different subequations sequentially:

\[x_1 = 3\] \[\begin{align*} 4x_1 + x_2 &= 14 \\ 12 + x_2 &= 14 \\ x_2 &= 2 \end{align*}\] \[\begin{align*} -6x_1 + 5x_2 + x_3 &= -7 \\ -18 + 10 + x_3 &= -7 \\ x_3 &= 1 \\ \end{align*}\]

So we have:

\[\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\]