# If C, D, E, and K is matrices, and CKD=E, is there any other way to determine matrix K beside expanding that equation with matrix multiplication?

If C is 3 x 2 matrix, D is 2 x 3 matrix, and E is 3 x 3 matrix, that K should be 2 x 2 matrix if CKD = E. With specific entries of each matrices, I could get the elements of K with expanding the equation with matrix multiplication. But, is there another way to determine K that briefer than to expanding that equation? Because matrix C and D can't be inverted due to the different numbers of each rows and columns.

• You could use the pseudoinverses $(C^+,D^+)$ of the matrices $(C,D)$ to write $$K=C^+ED^+$$Of course, then you'll need to figure out how to calculate those pseudoinverses. – greg Oct 21 '18 at 16:56

## 2 Answers

Delete the third rows of C and E, and the third columns of D and E.
Perhaps a different row and column instead if necessary.

• whoaa, it leads to the same answer if solving it with expanding the equation. But, why we have to delete the third rows of C and E and third columns of D and E? Is there any specific rules for doing this elimination? – Dziban N Oct 21 '18 at 9:40
• The third row is extra information that you don't need to solve the equation. If the first two rows of C were linearly dependent, you would delete one of those, and the corresponding row of E. – Empy2 Oct 21 '18 at 9:43
• how can we know that the third row of C and the third column of D are an extra information of the equation? – Dziban N Oct 21 '18 at 10:14

Multiply both sides by $$C'$$ (from the left), where the first row of $$C'$$ is orthogonal to the second column of $$C$$, and the second row of $$C'$$ is orthogonal to the first column of $$C$$. Such vectors exist and are easy to find.
Multiply both sides (from the right) by $$D'$$ whose first column is orthogonal to the second row of $$D$$, and vice versa.
We get $$C'CKDD'=C'ED',$$ where $$C'C$$ and $$DD'$$ are $$2×2$$ diagonal matrices. It is quick to finish.