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I have learned one way to get $4\times 4$ determinant. That is, divide a matrix $A$ by 4 part where each part is $2\times 2$ matrix: $$A = \left(\begin{array}{cc} B & C \\ D & E \end{array}\right)$$ Then $$\det A = \det B \det E - \det C \det D.$$

But I cannot prove it. Please give me a help.

AugSB
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user331899
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    This is not true. Take the matrix with columns $e_1$, $e_3$, $e_2$ and $e_4$. This is invertible, but all 4 $2\times 2$-sub-matrices are not. – Andreas Cap Apr 25 '16 at 07:36
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    Goes to show how worthwhile it is to try to come up with a counter-example before wasting a lot of time fruitlessly trying to prove something that is patently pointless. –  Apr 25 '16 at 07:38
  • Hint: in the expansion of a $4\times4$ determinant, there are $4!=24$ terms, while your expression has $2!\cdot2!\cdot2!=8$ of them. –  Apr 25 '16 at 08:00
  • You might have a look at the paper Silvester: Determinants of block matrices – Martin Sleziak Apr 25 '16 at 09:07
  • Similar question: http://math.stackexchange.com/questions/1460407/is-this-problem-statement-for-proving-the-determinant-of-a-block-matrix-possibly – Martin Sleziak Apr 25 '16 at 09:11

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The expression for the determinant of a block matrix is $$A = \left(\begin{array}{cc} B & C \\ D & E \end{array}\right)=\mbox{det}(B)\mbox{det}(E-DB^{-1}C)$$ where the matrices $B,C,D,E$ can be the $2\times2$ matrices you mentioned. You can find it here https://en.wikipedia.org/wiki/Determinant#Block_matrices Be careful! this expresion is only working if $B$ is invertible ($\mbox{det}(B)\ne0$). If $C=0$ or $D=0$, this expression simplifies to $\mbox{det}(B)\mbox{det}(E)$. So, an intelligent strategy is the use of gaussian elimination to simplify the matrix to this form and to get a simpler problem.

Finally, if the matrix has many zeroes in a column or a row, it is convenient the use of minors to compute the determinant http://mathworld.wolfram.com/DeterminantExpansionbyMinors.html which involve the computation of only $3\times3$ determinants. I hope this will be helpful!

user3209698
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  • The formula is fine, but it only works if $B$ is invertible, which is not true in general. In fact, the way to obtain this formula essentially is a variant of Gauss elimination used to obtain a block-upper-triangular matrix. (You just multiply the given matrix by $\begin{pmatrix} \mathbb I & -B^{-1}C \ 0 & \mathbb I\end{pmatrix}$, which does not change the determinant.) – Andreas Cap Apr 25 '16 at 09:55
  • Thank you for the remark @AndreasCap, I have included a comment about the existence of the inverse of $B$ – user3209698 Apr 25 '16 at 10:07