On Wikipedia, I saw the following formula

$$\det\begin{bmatrix}A & B\\ C & D\end{bmatrix} = \det(AD-BC)$$

if $C$ and $D$ commute. Is this always true?

Or is there a good counter example for each $2 \times 2$ block matrices?

  • $\begingroup$ Perhaps you could try proving it by considering them as linear transformations . . . $\endgroup$ – Shaun Nov 1 '13 at 23:28

Yes, it's always true, but note that there is a premise to fulfill: $C$ and $D$ have to commute, that is, we need $CD=DC$. If this condition is violated, the formula may fail to hold.

More generally, when the entries of $A,B,C,D$ be matices over a field (this includes, but isn't limited to, the cases where $A,B,C,D$ are real or complex matrices), we have $$ \det \pmatrix{A&B\\ C&D}= \begin{cases} \det(AD-BC) & \text{ if } CD=DC,\\ \det(DA-CB) & \text{ if } AB=BA,\\ \det(DA-BC) & \text{ if } BD=DB,\\ \det(AD-CB) & \text{ if } AC=CA. \end{cases} $$ A proof was given by John Silvester (see his paper). Essentially, suppose $AC=CA$ (this is the last case in the above). Then $$ \pmatrix{I&0\\ -C&A+xI}\pmatrix{A+xI&B\\ C&D}=\pmatrix{A+xI&B\\ AC-CA&(A+xI)D-CB}. $$ By assumption, $AC-CA=0$, so the RHS is block-triangular and $$\det(A+xI)\,\det\pmatrix{A+xI&B\\ C&D}=\det(A+xI)\,\det\left((A+xI)D-CB\right).$$ As $\det(A+xI)$ is a nonzero polynomial in $x$, we can divide both sides by it and obtain $\det\pmatrix{A+xI&B\\ C&D}=\det\left((A+xI)D-CB\right)$. Put $x=0$, the assertion follows. The proofs for the other three cases are similar.


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