# Blockwise if we split up a and b into multiple blocks

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• Blockwise: If we split up A and B into multiple blocks where the sizes would match, we can do regular multiplication using those blocks! If A and B were both split into 2 × 2 chunks, each block being a square. Then, C 11 = A 11 B 11 + A 12 B 21 ! 1.3 Finding Inverses with Gauss-Jordan Say we want to find an inverse of A . We have the following equation: A [ c 1 c 2 · · · c n ] = I n 2
where c i is the i -th column of A - 1 . Now, each column in I n are linear combinations of columns of A - namely, i th column of I n is Ac i . So each column in I n gives us a system of linear equations, that can be solved by Gauss elimination. The way of solving n linear systems at once is called the Gauss-Jordan method. We work with an augmented matrix of form [ A | I n ] and we eliminate A to be I n . We can say: Y E j [ A | I n ] = [ I n | ?] Say we found a set of E s that make the above equation hold. Then, we got: Y E j A = I n ⇐⇒ Y E j = A - 1 = Y E j I n the last equality telling us that the right half of the augmented matrix after the elimination is A - 1 , thus proving the validity of the algorithm. 1.3.1 Inverses of Products and Transposes What is ( AB ) - 1
Example: let A = " 2 1 8 7 # We can represent the elimination with a elimination matrix: " 1 0 - 4 1 # A = " 1 0 - 4 1 # " 2 1 8 7 # = " 2 1 0 3 # Thus 3
" 2 1 4 7 # | {z } A = " 1 0 - 4 1 # | {z } E - 1 " 2 1 0 3 # | {z } U We can factor out a diagonal matrix so U will only have ones on the diagonal: A = " 1 0 - 4 1 # " 2 0 0 3 # " 1 1 2 0 1 # 1.4.1 A case for 3 × 3 matrices Say n = 3 . Then we can represent the elimination is E 32 E 31 E 21 A = U Now, the following holds: A = E - 1 21 E - 1 31 E - 1 32 U = LU and we call the product of elimination matrices L . Question: how is L 21 . Since each elimination matrix subtracts an upper from from lower rows - so everything
is moving “downwards”. A nonzero number cannot move “up”. Now how do we calculate L