0
$\begingroup$

I have this 4x4 matrix A:

a, b, c, 0,
e, f, g, 0,
i, j, k, 0,
0, 0, 0, 1

And I want to convert it to this matrix B:

a, c, b, 0,
i, k, j, 0,
e, g, f, 0,
0, 0, 0, 1

What matrix do I multiply A to get B? And what is the general approach I could use to "divide" B by A?

$\endgroup$
2
$\begingroup$

Are you familiar with the notion of the matrix inverse?

That is, corresponding to "most" (but not all) square matrices $A$ there is some unique matrix $M$ (which we name $A^{-1}$) such that $A A^{-1} = A^{-1} A = I$ whher $I$ is the identity matrix (ones on the diagonal, zeros elsewhere). The exception is that if the determinant of $A$ is zero, then it will not have an inverse.

It is straightforward but messy to find the inverse of a matrix such as your given $A$. Look up Gaussian elimination for one easily understood technique. The answer for your case will be something like $$ A^{-1} = \frac1{\det A} \pmatrix{fk-gj & gi-ek & ce - ag &0 \\ cj-bk & af-eb & ce-ag & 0 \\ cj-eb & ja-ib & ak-ic & 0 \\ 0&0&0&1 } $$

Now use the fact that $(BA^{-1}) A = B$ so that $(BA^{-1})$ is the matrix you are looking for if you wand to multiply from the left. From the right it would be $(A^{-1}B)$ and in general those are not the same.

$\endgroup$
  • $\begingroup$ Thanks this is helpful! In my particular case the values are just swapped around. The second and the third row and columns are swapped. I thought that I could multiply A by a given matrix with 0 and 1 values (identity matrix with swapped values) to get the matrix B. Is this not the case? $\endgroup$ – Lenny White Aug 14 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.