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Let A= $\begin{bmatrix}B & C\\0 & D\end{bmatrix}$ be block triangular matrix (here $B$, $C$, $D$, $0$ are matrices, not scalars). Show that:

a) $A$ is invertible iff $B$ and $D$ are invertible.

b)Find $A^{-1}$ provided $B^{-1}$ and $D^{-1}$ are known.

c)Make use of b) to invert the matrix $\begin{bmatrix}-2 & 0 &-1 &-1\\-3& 1 &-3 &-3 \\ 0 &0&0&2\\ 0& 0& -1 & -3\end{bmatrix}$.

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Hint

$$\begin{bmatrix}B & C\\0 & D\end{bmatrix}=\begin{bmatrix}I & 0\\0 & D\end{bmatrix}\begin{bmatrix}B & C\\0 & I\end{bmatrix}$$ and $$A^{-1}=\begin{bmatrix}B^{-1} & ?\\0 & D^{-1}\end{bmatrix}$$

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Using block matrix determinant, you would get: $$\det{A} = \det{B}\det{D}$$ and the conclusion follows.

Alternatively, you can find explicitly the inverse matrix of $A$ following Romdhane and mookid's hints. You will see that $B^{-1}$ and $D^{-1}$ would be blocks, so they need to be invertible.

By doing so, you would solve item b.

The third item is just a matter of computations...

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  • $\begingroup$ if you could help me with another way it would be really helpful because determinant is a term which i have not taken yet @Victor $\endgroup$ – abcdef Oct 17 '14 at 21:36
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Hint: look for an inverse matrix in the form $\begin{bmatrix}B' & C'\\0 & D'\end{bmatrix}$.

This is justified because the left bottom part of the matrix is invariant in the process of the Gauss algorithm if it is already 0 at the start.


solution: $$\begin{bmatrix}B' & C'\\0 & D'\end{bmatrix} \begin{bmatrix}B & C\\0 & D\end{bmatrix} = \begin{bmatrix}I & 0\\0 & I\end{bmatrix} \iff BB' = I, DD' = I, B'C + C'D = 0 $$ then there is a solution iff $D$ and $B$ are invertible.

In this case the solution is $$ B' = B^{-1},\\ D' = D^{-1},\\ C' = - B^{-1}CD^{-1} $$

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  • $\begingroup$ @idAl: look at my edit for the solution. $\endgroup$ – mookid Oct 17 '14 at 21:47
  • $\begingroup$ Could you help me with part b) i found it but i am not quite sure @mookid... Is it [ B^-1 - B^-1* C* D^-1 ; 0 D^-1]?? $\endgroup$ – abcdef Oct 18 '14 at 23:49
  • $\begingroup$ please, it is in my answer. btw you did not either accept it or upvote it, did you at least read it? $\endgroup$ – mookid Oct 18 '14 at 23:51
  • $\begingroup$ Yes i am sorry i didn't notice it... I was trying to use it to solve part C but it was getting more complicated... How do i use it to find part c @mookid $\endgroup$ – abcdef Oct 18 '14 at 23:58
  • $\begingroup$ Can't you inverse the 2x2 matrices and made the product? $\endgroup$ – mookid Oct 18 '14 at 23:59

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