-1

2x2 Matrix M [a b]
__________[c d]

M^-1 = 1/(ad-bc) [d -b]
____________[-c a]

If a and b commute, M^-1 = I(ad-bc)^-1 [d -b]
____________________________[-c a]

I is the 2x2 identity matrix

Could you please explain how that works?

edit:
Where a, b, c and d are 2x2 matrices

Francis TP
  • 47
  • 2
    What do you mean by "a and b commute"? Are a and b numbers or matrices? – Ben Grossmann Dec 06 '16 at 20:46
  • they are matrices – Francis TP Dec 06 '16 at 21:11
  • If $a,b,c,d$ are matrices, then how are we supposed to interpret $$ (ad - bc)^{-1} \pmatrix{d&-b\-c&a} $$ the matrices being multiplied here are nonconformable – Ben Grossmann Dec 06 '16 at 21:25
  • Also, please see this tutorial on how to format math on this site. – Ben Grossmann Dec 06 '16 at 21:26
  • In any case, the formula for the inverse of a $2\times 2$ block matrix can be derived using the Schur complement – Ben Grossmann Dec 06 '16 at 21:32
  • what are u talking about? all matrices are 2x2. I still don't understand what do you want to do. $M^{-1}=(AD-BC)^{-1} \begin{pmatrix} D & -B \ -C & A \ \end{pmatrix}$ works as far as $AD-BC$ can even be inverted (so $det(AD-BC) \neq 0$) – Kazz Dec 06 '16 at 23:55
  • and what's with that commutation? so A and B commute, what about C and D ? – Kazz Dec 06 '16 at 23:56
  • oh i know what don't see! if $A,B,C$ are $n x n$ matrices then for example $A\begin{pmatrix} B & 1 \ 0 & C \ \end{pmatrix}=\begin{pmatrix} AB & A \ 0 & AC \ \end{pmatrix}$. $\begin{pmatrix} 1 & 2 & 3 & 4 \ 2 & 3 & 4 & 5 \ 3 & 4 & 5 & 6 \ 4 & 5 & 6 & 7 \ \end{pmatrix}$is not the same as$\begin{pmatrix} \begin{pmatrix} 1 & 2 \ 2 & 3 \ \end{pmatrix} & \begin{pmatrix} 3 & 4 \ 4 & 5 \ \end{pmatrix} \ \begin{pmatrix} 3 & 4 \ 4 & 5 \ \end{pmatrix} & \begin{pmatrix} 5 & 6 \ 6 & 7 \ \end{pmatrix} \ \end{pmatrix}$ – Kazz Dec 07 '16 at 00:05
  • This in general does not work. Your $M^{-1}$ is neither a left inverse nor a right inverse of $M$. The problem is that multiplication of "scalars" in your case is not commutative. However, if all the "scalars" $a,b,c,d$ here are by themselves matrices over a field, and they all commute with each other (not just among $a$ and $b$) and $ad-bc$ is invertible , then $a,b,c,d$ will also commute with the "scalar" $(ad-bc)^{-1}$ and your $M^{-1}$ is indeed a (left and right) inverse of $M$. – user1551 Dec 07 '16 at 01:40

1 Answers1

0

I suppose $a,b,c,d\in \mathcal M_2(F)$ for some field $F$ and $M\in \mathcal M_2(\mathcal M_2(F))$.

What you've described does not work. Your $M^{-1}$ is neither a left inverse nor a right inverse of $M$. The problem is that multiplication of "scalars" in your case is not commutative. In particular, $(ad-bc)^{-1}$ does not necessarily commute with $a,b,c$ or $d$.

However, if all the "scalars" $a,b,c,d$ here are by themselves matrices over a field, and they all commute with each other (not just among $a$ and $b$) and $ad-bc$ is invertible , then $a,b,c,d$ will automatically commute with the "scalar" $(ad-bc)^{-1}$ and your $M^{-1}$ is indeed a (left and right) inverse of $M$.

That said, concerning determinants, we do have the following result. Let $$M'=\left[\begin{array}{c|c}a&b\\ \hline c&d\end{array}\right]\in \mathcal M_4(F)$$ (which is a $4\times4$ matrix over the field $F$) be the block matrix whose four sub-blocks are $a,b,c,d\in \mathcal M_2(F)$. The difference between $M$ and $M'$ is that $M$ is a $2\times2$ matrix whose entries are taken from $\mathcal M_2(F)$, while $M'$ is a $4\times4$ matrix whose entries are taken directly from $F$. Provided that $ab=ba$, we have $$ \det M'=\det(da-cb) $$ (where the left hand side is the determinant of a $4\times4$ matrix over $F$ and the right hand side is the determinant of a $2\times2$ matrix over $F$).

Note that in the above, we only need $ab=ba$; we don't need all four matrices $a,b,c,d$ to commute. In general, however, if $M'$ is a block matrix of larger size (say, it has 3 blocks by 3 blocks), then to guarantee that we can do something similar, we do need all sub-blocks to commute with each other. This is a result by John Silvester and this has been discussed a couple of times on this site. See my answer to a previous question, for instance.

Community
  • 1
user1551
  • 139,064

  • However, if all the "scalars" a,b,c,da,b,c,d here are by themselves matrices over a field, and they all commute with each other (not just among aa and bb) and ad−bcad−bc is invertible , then a,b,c,da,b,c,d will automatically commute with the "scalar" (ad−bc)−1(ad−bc)−1 and your M−1M−1 is indeed a (left and right) inverse of MM.

    Why?

     – Francis TP Dec 07 '16 at 21:37
  • @FrancisTP If $a,b,c,d$ commute, they commute with $ad-bc$ too. Yet $(ad-bc)^{-1}$ is a polynomial in $ad-bc$ (Cayley-Hamilton theorem). Hence $a,b,c,d$ commute with $(ad-bc)^{-1}$. – user1551 Dec 08 '16 at 04:19