Inverse Matrices

Inverse Matrices I

Inverse II (Theory)

Inverse of Partitioned Matrices

Suppose we are given a square n×n matrix that could be partitioned as follows:
\[ {\bf A} = \begin{bmatrix} {\bf P} & {\bf Q} \\ {\bf R} & {\bf S} \end{bmatrix} , \]
where P is an r×r matrix, Q is an r×s matrix, R is an s×r matrix, and S is an s×s matrix such that r + s = n.

Assume that A−1 exists and can be partitioned in a similar way as

\[ {\bf A}^{-1} = \begin{bmatrix} {\bf X} & {\bf Y} \\ {\bf Z} & {\bf W} \end{bmatrix} , \]
where matrices Xr×r, Yr×s, Zs×r, and Ws×s are to be found. Observe that X, Y, Z, and W are of the same dimensions as P, Q, R, and S, respectively. Since A−1A = I, the identity matrix, we have
\[ {\bf A}\, {\bf A}^{-1} = \begin{bmatrix} {\bf P} & {\bf Q} \\ {\bf R} & {\bf S} \end{bmatrix}\, \begin{bmatrix} {\bf X} & {\bf Y} \\ {\bf Z} & {\bf W} \end{bmatrix} = \begin{bmatrix} {\bf I}_r & {\bf 0} \\ {\bf 0} & {\bf I}_s \end{bmatrix} , \]
where Ir and Is are identity matrices of the order r and s, respectively.

Applying the definition of the product of matrices to partitioned matrices we obtain the following equations

\[ \begin{split} {\bf P\,X} + {\bf Q\,Z} &= {\bf I}_r , \\ {\bf P\,Y} + {\bf Q\,W} &= {\bf 0} , \\ {\bf R\,X} + {\bf S\,Z} &= {\bf 0} , \\ {\bf R\,Y} + {\bf S\,W} &= {\bf I}_s . \end{split} \]
Solving for X, Y, Z, and W, gives
\[ \begin{split} {\bf W} &= \left[ {\bf S} - {\bf R\,P}^{-1}{\bf Q} \right]^{-1} , \\ {\bf Y} &= -{\bf P}^{-1} {\bf Q\,W} , \\ {\bf Z} &= -{\bf W\,R\,P}^{-1} , \\ {\bf X} &= {\bf P}^{-1} - {\bf P}^{-1}{\bf Q\,Z} . \end{split} \]
Therefore, we get
\begin{equation} \label{EqInverse.4} \begin{bmatrix} {\bf P} & {\bf Q} \\ {\bf R} & {\bf S} \end{bmatrix}^{-1} = \begin{bmatrix} \left( {\bf P} - {\bf Q} \,{\bf S}^{-1} {\bf R} \right)^{-1} & -\left( {\bf P} - {\bf Q} \,{\bf S}^{-1} {\bf R} \right)^{-1} {\bf Q}\,{\bf S}^{-1} \\ - {\bf S}^{-1} {\bf R} \left( {\bf P} - {\bf Q} \,{\bf S}^{-1} {\bf R} \right)^{-1} & {\bf S}^{-1} + {\bf S}^{-1} {\bf R} \left( {\bf P} - {\bf Q} \,{\bf S}^{-1} {\bf R} \right)^{-1} {\bf Q}\, {\bf S}^{-1} \end{bmatrix} . \end{equation}
Example 16: Upon partitioning the matrix, find its inverse. \[ {\bf A} = \begin{bmatrix} -8&-5& 6&1&-9 \\ -5& 4& -9& -8& 4 \\ 7& -9 & 5&3&0 \\ 7&-8 & 3&5&-1 \\ -9&4&0&-1&-5 \end{bmatrix} \] First, we check whether the given matrix is invertible by calculating its determinant. Since it is not zero, we conclude that matrix A is not singular.
Clear[A, Ai, At, B, Bi, B2, a, b, c, d, e, f, g, h, i, p, q, r, t, u, v, w, x, y, z, K, P, Q, R, S, W];
A = {{-8, -5, 6, 1, -9}, {-5, 4, -9, -8, 4}, {7, -9, 5, 3, 0}, {7, -8, 3, 5, -1}, {-9, 4, 0, -1, -5}};
Det[A]
-5
Of course, Mathematica is capable to find its inverse. However, our objective is to use partition of the matrix.
Inverse[A]
{{-(1054/5), 187/5, 1638/5, -(61/5), 2059/5}, {-85, 15, 132, -5, 166}, {22, -4, -34, 1, -43}, {1001/5, -(178/5), -(1557/5), 59/ 5, -(1956/5)}, {1357/5, -(241/5), -(2109/5), 78/5, -(2652/5)}}
\[ {\bf A}^{-1} = \begin{bmatrix} -(1054/5)& 187/5& 1638/5& -(61/ 5)& 2059/5 \\ -85& 15& 132& -5& 166 \\ 22& -4& -34& 1& -43 \\ 1001/5& -(178/5)& -(1557/5)& 59/5& -(1956/5) \\ 1357/5& -(241/5)& -(2109/5)& 78/5& -(2652/5) \end{bmatrix} . \] We make a partition the given matrix as follows: \[ {\bf A} = \begin{bmatrix} {\bf P} & {\bf Q} \\ {\bf R} & {\bf S} \end{bmatrix} , \] where \[ {\bf P} = \begin{bmatrix} -8& -5& \phantom{-}6 \\ -5& \phantom{-}4& -9 \\ \phantom{-}7& -9 & \phantom{-}5 \end{bmatrix} , \qquad {\bf Q} = \begin{bmatrix} \phantom{-}1& -9 \\ -8&\phantom{-}4 \\ \phantom{-}3&\phantom{-}0 \end{bmatrix} , \] \[ {\bf R} = \begin{bmatrix} \phantom{-}7&-8&3 \\ -9&\phantom{-}4&0 \end{bmatrix} , \quad {\bf S} = \begin{bmatrix} \phantom{-}5&-1 \\ -1&-5 \end{bmatrix} . \] We enter these matrices into computer.
P = {{-8, -5, 6}, {-5, 4, -9}, {7, -9, 5}};
Q = {{1, -9}, {-8, 4}, {3, 0}};
R = {{7, -8, 3}, {-9, 4, 0}};
S = {{5, -1}, {-1, -5}};
First, we calculate the common multiple in Eq.(1): \[ {\bf W} = \left( {\bf P} - {\bf Q} \,{\bf S}^{-1} {\bf R} \right)^{-1} = \begin{bmatrix} -\frac{1054}{5}& \frac{187}{5}& \frac{1638}{5} \\ -85& 15& 132 \\ \phantom{-}22& -4& -34 \end{bmatrix} . \]
W = Inverse[P - Q.Inverse[S].R]
{{-(1054/5), 187/5, 1638/5}, {-85, 15, 132}, {22, -4, -34}}
Then we calculate other terms in formula (1):
-W.Q.Inverse[S]
{{-(61/5), 2059/5}, {-5, 166}, {1, -43}}
\[ -{\bf W}\,{\bf Q}\,{\bf S}^{-1} = \begin{bmatrix} -\frac{61}{5} & \frac{2059}{5} \\ -5 & 166 \\ 1 & -43 \end{bmatrix} , \]
-Inverse[S].R.W
{{1001/5, -(178/5), -(1557/5)}, {1357/5, -(241/5), -(2109/5)}}
\[ - {\bf S}^{-1} {\bf R}\, {\bf W} = \begin{bmatrix} \frac{1001}{5} &- \frac{178}{5} & - \frac{1557}{5} \\ \frac{1357}{5} & -\frac{241}{5} & -\frac{2109}{5} \end{bmatrix} , \]
Inverse[S] + Inverse[S].R.W.Q.Inverse[S]
{{59/5, -(1956/5)}, {78/5, -(2652/5)}}
\[ {\bf S}^{-1} + {\bf S}^{-1} {\bf R}\,{\bf W}\,{\bf Q}\,{\bf S}^{-1} = \begin{bmatrix} \frac{59}{5} & - \frac{1956}{5} \\ \frac{78}{5} & -\frac{2652}{5} \end{bmatrix} . \] Substituting these expressions into formula (4), we obtain the inverse matrix.
End of Example 16
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