Solving for Values in an Inverse Partitioned Matrix

[BraedenP]

Homework Statement

Given partitioned matrices:
A=\begin{bmatrix}<br /> I &amp; 0 &amp; 0\\ <br /> C &amp; 0 &amp; 0\\ <br /> A &amp; B &amp; I<br /> \end{bmatrix}<br /> ,A^{-1}=\begin{bmatrix}<br /> I &amp; 0 &amp; 0\\<br /> Z &amp; I &amp; 0\\<br /> X &amp; Y &amp; I<br /> \end{bmatrix}

Solve for matrices Z, X, and Y

Homework Equations

N/A

The Attempt at a Solution

I started by equating the product of those partitioned matrices to a partitioned identity matrix:

\begin{bmatrix}<br /> I &amp; 0 &amp; 0\\ <br /> C &amp; 0 &amp; 0\\ <br /> A &amp; B &amp; I<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> I &amp; 0 &amp; 0\\<br /> Z &amp; I &amp; 0\\<br /> X &amp; Y &amp; I<br /> \end{bmatrix}<br /> = \begin{bmatrix}<br /> I &amp; 0 &amp; 0\\<br /> 0 &amp; I &amp; 0\\<br /> 0 &amp; 0 &amp; I\end{bmatrix}

Then My multiplying out the entries and comparing them to the expected corresponding entries in the identity matrix, I generated some equations:

C+Z=0 \to Z=-C

A+BZ+X=0 \to X=-B(-C)-A \to X=BC-A

B+Y=0 \to B=-Y

Therefore:
Z=-C, X=BC-A, B=-Y

Am I right in my method, or am I out to lunch? Hopefully I broke no rules in my multiplication of the sub-matrices..

Any advice would be appreciated.

 

[fzero]

Looks good, but you probably want to express Y = -B in case you have a pedantic grader that might take points off. Also you have a typo in the center element of the original matrix (center element is I instead of 0), but you seem to have used the correct matrix for your calculation.

 

[BraedenP]

Looks good, but you probably want to express Y = -B in case you have a pedantic grader that might take points off. Also you have a typo in the center element of the original matrix (center element is I instead of 0), but you seem to have used the correct matrix for your calculation.

Oops! Yeah, you're right -- it is a typo. I used I in my calculation.

And yep, you're correct. He'll want it as Y=-B.. Not sure why I changed it around.

Thanks a heap!