Solving for Values in an Inverse Partitioned Matrix

[BraedenP]

Homework Statement

Given partitioned matrices:
$$A=\begin{bmatrix} I & 0 & 0\\ C & 0 & 0\\ A & B & I \end{bmatrix} ,A^{-1}=\begin{bmatrix} I & 0 & 0\\ Z & I & 0\\ X & Y & I \end{bmatrix}$$

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:

$$\begin{bmatrix} I & 0 & 0\\ C & 0 & 0\\ A & B & I \end{bmatrix} \begin{bmatrix} I & 0 & 0\\ Z & I & 0\\ X & Y & I \end{bmatrix} = \begin{bmatrix} I & 0 & 0\\ 0 & I & 0\\ 0 & 0 & I\end{bmatrix}$$

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!