If P = left[ {begin{array}{*{20}{c}}1&α &31&3&32&4&4end{array}} right] is a

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3 and 4 .Determinants and Matrices

medium

If $P = \left[ {\begin{array}{*{20}{c}}1&\alpha &3\\1&3&3\\2&4&4\end{array}} \right]$ is the adjoint of $3×3 $ matrix $A$ and $ |A|=4$ then $\alpha $ is equal to

A

$4$

B

$11$

C

$5$

D

$0$

(JEE MAIN-2013)

Solution

$|P|=1(12-12)-\alpha(4-6)+3(4-6)$

$=2 \alpha-6$

$|P|=|A|^{2}=16$

$2 \alpha-6=16$

$\alpha=11.$

Standard 12

Mathematics

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