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?


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.

  • $\begingroup$ 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

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