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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.

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    $\begingroup$ 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. $\endgroup$ – Andreas Cap Apr 25 '16 at 7:36
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    $\begingroup$ 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. $\endgroup$ – user328032 Apr 25 '16 at 7:38
  • $\begingroup$ Hint: in the expansion of a $4\times4$ determinant, there are $4!=24$ terms, while your expression has $2!\cdot2!\cdot2!=8$ of them. $\endgroup$ – Yves Daoust Apr 25 '16 at 8:00
  • $\begingroup$ You might have a look at the paper Silvester: Determinants of block matrices $\endgroup$ – Martin Sleziak Apr 25 '16 at 9:07
  • $\begingroup$ Similar question: math.stackexchange.com/questions/1460407/… $\endgroup$ – Martin Sleziak Apr 25 '16 at 9:11
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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!

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  • $\begingroup$ 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.) $\endgroup$ – Andreas Cap Apr 25 '16 at 9:55
  • $\begingroup$ Thank you for the remark @AndreasCap, I have included a comment about the existence of the inverse of $B$ $\endgroup$ – seoanes Apr 25 '16 at 10:07

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