Use Strassen’s algorithm to compute the matrix product

$$\begin{pmatrix} 1 & 3 \\ 7 & 5 \\ \end{pmatrix} \begin{pmatrix} 6 & 8 \\ 4 & 2 \\ \end{pmatrix}$$.

The sums:

\begin {aligned} & S_1 = B_{12} - B_{22} = 8 - 2 = 6 \\ & S_2 = A_{11} + A_{12} = 1 + 3 = 4 \\ & S_3 = A_{21} + A_{22} = 7 + 5 = 12 \\ & S_4 = B_{21} - B_{11} = 4 - 6 = -2 \\ & S_5 = A_{11} + A_{22} = 1 + 5 = 6 \\ & S_6 = B_{11} + B_{22} = 6 + 2 = 8 \\ & S_7 = A_{12} - A_{22} = 3 - 5 = -2 \\ & S_8 = B_{21} + B_{22} = 4 + 2 = 6 \\ & S_9 = A_{11} - A_{21} = 1 - 7 = -6 \\ & S_{10} = B_{11} + B_{12} = 6 + 8 = 14 \end {aligned}

The products:

\begin {aligned} & P_1 = A_{11} \cdot S_1 = 6 \\ & P_2 = S_2 \cdot B{22} = 8 \\ & P_3 = S_3 \cdot B{11} = 72 \\ & P_4 = A_{22} \cdot S_4 = -10 \\ & P_5 = S_5 \cdot S_6 = 48 \\ & P_6 = S_7 \cdot S_8 = -12 \\ & P_7 = S_9 \cdot S_{10} = -84 \end {aligned}

The result sub-matrix sums:

\begin {aligned} & C_{11} = P_5 + P_4 - P_2 + P_6 = 18 \\ & C_{12} = P_1 + P_2 = 14 \\ & C_{21} = P_3 + P_4 = 62 \\ & C_{22} = P_5 + P_1 - P_3 - P_7 = 66 \end {aligned}

So, the answer is $$\begin{pmatrix} 18 & 14 \\ 62 & 66 \\ \end{pmatrix}$$.