Partitioned Matrices

Imagine a matrix A, $A=\left[\begin{array}{cccc}1& 0& 0& 5\\ 0& 1& 0& 6\\ 0& 0& 1& 6\\ 7& 9& 1& 6\end{array}\right]$ Partitioned it could look like this,

A&= \begin{bmatrix} \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} &\begin{bmatrix} 5\\6\\6 \end{bmatrix} \\\begin{bmatrix} 7&9&1 \end{bmatrix} & \begin{bmatrix} 6 \end{bmatrix} \end{bmatrix} \\ &=\begin{bmatrix} I_3 & U \\ V & X \end{bmatrix} \end{aligned}

We can even perform matrix multiplication, $\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\end{array}\right]\left[\begin{array}{cc}2& -1\\ 0& -1\\ 0& 1\end{array}\right]$

$$\begin{array}{rl}& =\left[\begin{array}{cc}{I}_2& X\end{array}\right]\left[\begin{array}{c}U\\ V\end{array}\right]\\ & =I_{2}U+XY\\ & =\left[\begin{array}{cc}2& -1\\ 0& -1\end{array}\right]+\left[\begin{array}{c}1\\ 1\end{array}\right]\left[\begin{array}{cc}0& 1\end{array}\right]\\ & =\left[\begin{array}{cc}2& -1\\ 0& -1\end{array}\right]+\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]\\ & =\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right]\end{array}$$

Find the Inverse

Compute equations the inverse $\left[\begin{array}{cc}A& B\\ 0& C\end{array}\right]$.

$$\left[\begin{array}{cc}A& B\\ 0& C\end{array}\right]\left[\begin{array}{cc}W& X\\ Y& Z\end{array}\right]=I=\left[\begin{array}{cc}{I}_n& 0\\ 0& I_n\end{array}\right]$$ $$\begin{array}{c}\begin{array}{rl}0W+CY& =0\\ CY& =0\\ C^{-1}CY& =C^{-1}0\\ Y& =0\end{array}\end{array}$$ $$\begin{array}{c}\begin{array}{rl}0X+CZ& =I_n\\ C^{-1}CZ& =C^{-1}{I}_n\\ Z& =C^{-1}\end{array}\end{array}$$

$\text{We know }Y=0\text{ as it was calculated above}$

$$\begin{array}{c}\begin{array}{rl}AW+BY& =I_n\\ AW+B0& =I_n\\ A^{-1}AW& =A^{-1}{I}_n\\ W& =A^{-1}\end{array}\end{array}$$ $$\begin{array}{c}\begin{array}{rl}AX+BZ& =0\\ A^{-1}AX& =-A^{-1}BZ\\ X& =-A^{-1}BZ\\ X& =-A^{-1}B{C}^{-1}\end{array}\end{array}$$

So, putting this back into a matrix:

$${\left[\begin{array}{cc}A& B\\ 0& C\end{array}\right]}^{-1}=\boxed{\left[\begin{array}{cc}{A}^{-1}& -A^{-1}B{C}^{-1}\\ 0& C^{-1}\end{array}\right]}$$