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Q. If $A$ and $B$ are square matrices of order $3$ such that $ det\text{ }A=1 $ and $ det\text{ }B=-1 $ then $ det\text{ (-10}\,\text{AB)} $ is equal to

J & K CETJ & K CET 2010Determinants

Solution:

$ \det \,\,A=1, $
$ \det \,\,B=-1 $
$ \det \,(-10AB)={{(-10)}^{3}}\,\det \,(AB) $
$ [\because \,\,\det \,(kA)\,={{k}^{n}}\,\det \,(A)] $
(here, $ n=3 $ ) $ =-1000\,\det \,(A)\,\det \,(B) $
$ =1+\log \,abc $