Given, $$A= \left(\begin{matrix} a_{1,1}&a_{1,2} \\ a_{2,1}&a_{2,2} \\ \end{matrix}\right) $$

$$ B= \left(\begin{matrix} b_{1,1}&b_{1,2} \\ b_{2,1}&b_{2,2} \\ \end{matrix}\right) $$

Show $$ \det\left(\begin{matrix} A&0 \\ 0&B \\ \end{matrix}\right)=\det(A)\det(B) $$

I know how to solve the right side. But I wasn't sure about the left determinant. I tried to find out $\det(AB)$ where $AB$ is found by using matrix multiplication and I get

$$ \det\left(\begin{matrix} a_{1,1}b_{1,1}+a_{1,2}b_{2,1}&a_{1,1}b_{1,2}+a_{1,2}b_{2,2}\\ a_{2,1}b_{1,1}+a_{2,2}b_{2,1}&a_{2,1}b_{1,2}+a_{2,2}b_{2,2} \\ \end{matrix}\right) $$

and I don't think this is right as this doesn't equal to the right side


closed as off-topic by Jendrik Stelzner, iadvd, Leucippus, Parcly Taxel, user91500 Sep 14 '16 at 5:33

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    $\begingroup$ Expand on the first row. $\endgroup$ – vadim123 Sep 13 '16 at 16:12

Using the usual method for determinants by the first column we have:

\begin{align*} \det\begin{pmatrix}A & 0\\ 0 & B\end{pmatrix} &=\begin{vmatrix}a_{11} & a_{12} & 0 & 0\\ a_{21} & a_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22}\end{vmatrix}=a_{11}\begin{vmatrix}a_{22} & 0 & 0\\ 0 & b_{11} & b_{12}\\ 0 & b_{21} & b_{22}\end{vmatrix} - a_{21}\begin{vmatrix}a_{12} & 0 & 0\\ 0 & b_{11} & b_{12}\\ 0 & b_{21} & b_{22}\end{vmatrix}\\ &=a_{11}a_{22}\begin{vmatrix}b_{11} & b_{12}\\ b_{21} & b_{22} \end{vmatrix}-a_{21}a_{12}\begin{vmatrix}b_{11} & b_{12}\\ b_{21} & b_{22} \end{vmatrix}\\ &=\begin{vmatrix}b_{11} & b_{12}\\ b_{21} & b_{22} \end{vmatrix}\cdot (a_{11}a_{22}-a_{21}a_{12})\\ &=\det(B)\cdot \det(A). \end{align*}


You can take a standard approach and just expand both sides of $\det (A)\det (B)=\det (AB)$ to see if the equation holds true.

Starting with the left side we have: $$\begin{align}\det (A)\det (B)&=(a_{1,1}a_{2,2}-a_{2,1}a_{1,2})(b_{1,1}b_{2,2}-b_{2,1}b_{1,2})\\&=a_{1,1}a_{2,2}b_{1,1}b_{2,2}-a_{1,1}a_{2,2}b_{2,1}b_{1,2}-b_{1,1}b_{2,2}a_{2,1}a_{1,2}+a_{2,1}a_{1,2}b_{1,2}b_{2,1}\end{align}$$ For the right side we have: $$\begin{align}\det (AB)&=(a_{1,1}b_{1,1}+a_{1,2}b_{2,1})(a_{2,1}b_{1,2}+a_{2,2}b_{2,2})-(a_{2,1}b_{1,1}+a_{2,2}b_{2,1})(a_{1,1}b_{1,2}+a_{1,2}b_{2,2})\\&=a_{1,1}b_{1,1}a_{2,1}b_{1,2}+a_{1,1}b_{1,1}a_{2,2}b_{2,2}+a_{1,2}b_{2,1}a_{2,1}b_{1,2}+a_{1,2}b_{2,1}a_{2,2}b_{2,2}-a_{2,1}b_{1,1}a_{1,1}b_{1,2}-a_{2,1}b_{1,1}a_{1,2}b_{2,2}-a_{2,2}b_{2,1}a_{1,1}b_{1,2}-a_{2,2}b_{2,1}a_{1,2}b_{2,2}\\&=a_{1,1}b_{1,1}a_{2,2}b_{2,2}+a_{1,2}b_{2,1}a_{2,1}b_{1,2}-a_{2,1}b_{1,1}a_{1,2}b_{2,2}-a_{2,2}b_{2,1}a_{1,1}b_{1,2}\end{align}$$ We can now see that the two sides do equal (messy but it works).

  • $\begingroup$ That's correct, but you don't even have to expand the determinants. It's enough to notice that the matrix in the OP is actually $A B$. $\endgroup$ – dxiv Sep 13 '16 at 17:41
  • $\begingroup$ @dxiv I took it as he/she is trying to show that $\det (A)\det (B)=\det (AB)$ without already assuming that fact about determinants. $\endgroup$ – Dave Sep 13 '16 at 17:45
  • $\begingroup$ That is possible, of course. Given the title of the question, however, I don't know that the OP did in fact realize that the $matrix$ is actually $A B$. $\endgroup$ – dxiv Sep 13 '16 at 17:48
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    $\begingroup$ @dxiv That is a good point. I suppose if they are able to use that property of determinants the question does just boil down to recognition that $matrix=AB$. $\endgroup$ – Dave Sep 13 '16 at 17:51

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