Question A.6: Find the product of two matrices A and B by partitioning A =......
Find the product of two matrices \overline{A} and \overline{B} by partitioning
\overline{A} = \left|\begin{matrix} 1 &2& 3 \\ 2& 0& 1 \\ 1& 3& 6 \end{matrix} \right| \ \ \ \overline{B} = \left|\begin{matrix} 1& 2& 1& 0 \\ 2& 3& 5& 1 \\ 4& 6& 1& 2 \end{matrix} \right|
is given by
A matrix can be inverted by partition. In this case, each of the diagonal submatrices must be square. Consider a square matrix partitioned into four submatrices:
\overline{A} = \left|\begin{matrix} \overline{A}_{1}&\overline{A}_{2} \\ \overline{A}_{3}&\overline{A}_{4} \end{matrix} \right| (A.60)
The diagonal submatrices \overline{A}_{1} and \overline{A}_{4} are square, though these can be of different dimensions. Let the inverse of \overline{A} be
\overline{A}^{-1} = \left|\begin{matrix} \overline{A}^{\prime \prime}_{1}&\overline{A}^{\prime \prime}_{2} \\ \overline{A}^{\prime \prime}_{3}&\overline{A}^{\prime \prime}_{4} \end{matrix} \right| (A.61)
then
\overline{A}^{-1} \overline{A}= \left|\begin{matrix} \overline{A}^{\prime \prime}_{1}&\overline{A}^{\prime \prime}_{2} \\ \overline{A}^{\prime \prime}_{3}&\overline{A}^{\prime \prime}_{4} \end{matrix} \right| \left|\begin{matrix} \overline{A}_{1}&\overline{A}_{2} \\ \overline{A}_{3}&\overline{A}_{4} \end{matrix} \right| = \left|\begin{matrix} 1&0 \\ 0&1 \end{matrix} \right| (A.62)
The following relations can be derived from this identity:
\overline{A}^{\prime \prime}_{1}=[ \overline{A}_{1}- \overline{A}_{2} \overline{A}^{-1}_{4} \overline{A}_{3}]^{-1}
\overline{A}^{\prime \prime}_{2}=- \overline{A}^{\prime \prime}_{1} \overline{A}_{2}\overline{A}^{-1}_{4} (A.63)
\overline{A}^{\prime \prime}_{4}=[-\overline{A}_{3}\overline{A}^{-1}_{1}\overline{A}_{2}+\overline{A}_{4}]^{-1}
\overline{A}^{\prime \prime}_{3}=- \overline{A}^{\prime \prime}_{4} \overline{A}_{3}\overline{A}^{-1}_{1}