MOLT:
Suppose \(S:C^3 \rightarrow C^4\) is defined by \(S \bigg ( \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} \bigg ) = \begin{bmatrix} 3x_1 - 2x_2 + 5x_3\\ x_1 + x_2 + x_3\\ 9x_1 - 2x_2 + 5x_3\\ 4x_2\\ \end{bmatrix}\)
Then
\[ C_1 = S(e_1) = S \bigg ( \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} \bigg ) = \begin{bmatrix} 3\\ 1\\ 9\\ 0 \end{bmatrix} \]
\[C_2 = S(e_2) = S \bigg ( \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix} \bigg ) = \begin{bmatrix} -2\\ 1\\ -2\\ 4 \end{bmatrix} \] \[C_3 = S(e_3) = S \bigg ( \begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix} \bigg ) = \begin{bmatrix} 5\\ 1\\ 5\\ 0 \end{bmatrix} \\ \] so define
\[ C = [C_1|C_2|C_3] = \begin{bmatrix} 3 & -2 & 5\\ 1 & 1 & 1\\ 9 & -2 & 5\\ 0 & 4 & 0 \end{bmatrix} \]
\(S(w)\) will be calculated using both direct substitution and vector-matrix product of \(C(w)\). This should result in the same vector for each method.
Computing \(S(w)\):
\[S \bigg (\begin{bmatrix} -3 \\ 1 \\4 \end{bmatrix} \bigg) = \begin{bmatrix} 3(-3) - 2(1) + 5(4)\\ -3 + 1 + 4\\ 9(-3) - 2(1) + 5(4)\\ 4(1)\\ \end{bmatrix} = \begin{bmatrix} 9\\ 2\\ -9\\ 4\\ \end{bmatrix}\]
Computing matrix-vector product \(Cw\):
\[Cw = \begin{bmatrix} 3 & -2 & 5\\ 1 & 1 & 1\\ 9 & -2 & 5\\ 0 & 4 & 0 \end{bmatrix} \begin{bmatrix} -3 \\ 1 \\4 \end{bmatrix} = \begin{bmatrix} 3(-3) - 2(1) + 5(4)\\ -3 + 1 + 4\\ 9(-3) - 2(1) + 5(4)\\ 0(-3) + 4(1) + 0(4)\\ \end{bmatrix} = \begin{bmatrix} 9\\ 2\\ -9\\ 4\\ \end{bmatrix}\]
Therefore \(S(w) = Cw\)