# Thread: Matrix in partitioned form.

1. ## Matrix in partitioned form.

Let $B$ be an $m * m$ matrix with inverse $B^-1$, $c$ an $m$-vector and $0$ the $m$ dimensional null vector. $A$ is given in the following partitioned form...

$A = \left[ \begin{array}{cccc} B & 0 \\ c^T & 1 \end{array} \right]$

Determine $A^{-1}$ symbolically in a partition form.

For which I get the following...

$A^{-1} = \left[ \begin{array}{cccc} d & e \\ f & g \end{array} \right]$

$\left[ \begin{array}{cccc} B & 0 \\ c^T & 1 \end{array} \right] \left[ \begin{array}{cccc} d & e \\ f & g \end{array} \right] = \left[ \begin{array}{cccc} 1 & 0 \\ 0 & 1 \end{array} \right]$

$Bd = 1$

$d = B^{-1}$

$Be = 0$

Since B is non-singular, B does not equal 0...

$e = 0$

$c^Td + f = 0$

$c^TB^{-1} + f = 0$

$f = -c^TB^{-1}$

$c^Te + g = 1$

$g = 1$

$A^{-1} = \left[ \begin{array}{cccc} B^{-1} & 0 \\ -c^TB^{-1} & 1 \end{array} \right]$

Is this correct so far?

The question then asks the dimension of $A$ and the dimensions of the submatrices in $A^{-1}$.

Am I right in thinking the dimension of A is m+1, the dimension of $B^{-1}$ is m and the dimension of $-c^TB^{-1}$ is 1?

Thanks for the help.

2. Originally Posted by sean.1986
Let $B$ be an $m * m$ matrix with inverse $B^-1$, $c$ an $m$-vector and $0$ the $m$ dimensional null vector. $A$ is given in the following partitioned form...

$A = \left[ \begin{array}{cccc} B & 0 \\ c^T & 1 \end{array} \right]$

Determine $A^{-1}$ symbolically in a partition form.

For which I get the following...

... [SNIP] ...

$A^{-1} = \left[ \begin{array}{cccc} B^{-1} & 0 \\ -c^TB^{-1} & 1 \end{array} \right]$

Is this correct so far? Yes.

The question then asks the dimension of $A$ and the dimensions of the submatrices in $A^{-1}$.

Am I right in thinking the dimension of A is m+1, the dimension of $B^{-1}$ is m and the dimension of $-c^TB^{-1}$ is 1?
A matrix is a two-dimensional structure. So the dimensions of A are that it is an $(m+1)\mathord\times(m+1)$ matrix; $B^{-1}$ is m×m; and $-c^TB^{-1}$ is 1×m (or the transpose of an m-vector). Also, the 0 is an m×1 matrix (or an m-vector), and the 1 is a 1&#215;1 matrix.

3. Ah that makes sense, thank you.