# Math (matrices)

posted by
**James** on
.

No one answered my matrix question. Let me rephrase:

Let A, B, and 0 be 2x2 matrices. Assuming that A is invertible and 0 is all zeroes, what is the inverse of the matrix

[A|0]

[B|A]

(that is a 4x4 matrix represented as 4 2x2 matrices)

The answer is in the form:

[A^-1|0 ]

[C |A^-1]

where A^-1 is the inverse of A. But I have to solve for C.

I thought I could solve solve this problem, but my answer comes out wrong. I was hoping to see someone else try it.

C = -A^(-1)B A^(-1)

That's the right answer. How how did you do it?

the determinant is A^2

So C should be -B/det = -B/A^2

Or C = -BA^(-1)A^(-1)

How did you get the different multiplication order?

Put

X =

[A|0]

[B|A]

And

Y =

[A^-1|0 ]

[C |A^-1]

X Y must be the unit matrix. You can multiply the matrices X and Y as if these are two by two matrices. I.e. you treat the A, A^(-1) etc. as elements of the matrices. But you must take care of the multiplication order. I can prove this property for you later, if you want.

The element at the second row and first column of the product X Y must be the zero matrix. If you multiply X with Y you see that it is:

B A^(-1) + A C

Equating this to zero gives:

B A^(-1) + A C = 0

Multiplying by A^(-1) on the left gives the result:

C = -A^(-1)B A^(-1)

This is true in general, not just for 4 by 4 matrices. The matrix could have been a 100 by 100 matrix with A , B etc. 50 by 50 matrices.

Ah, that makes perfect sense.

I was using the standard formula for calculating the inverse of a 2x2 matrix.

[a|b]^-1

[c|d]

=

[d/det|-b/det]

[-c/det|a/det]

But in hindsight, I see that that formula doesn't apply for partioned matrices.

Problem Solved! Thank you so much Count Iblis!

Thanks!

B.t.w., if you google for "Dodgson condensation" you'll find an interesting method to calculate determinants!

help me with this it is really hard hank you <3