Let A = left[ {begin{array}{*{20}{c}}5&{5α }&α 0&α &{5α }0&0&5end{array}

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3 and 4 .Determinants and Matrices

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Let $A = \left[ {\begin{array}{*{20}{c}}5&{5\alpha }&\alpha \\0&\alpha &{5\alpha }\\0&0&5\end{array}} \right]$, If ${\left| A \right|^2} = 25$, then $\left| \alpha \right|$ equals

A

$\frac{1}{5}$

B

$5$

C

${5^2}$

D

$1$

(AIEEE-2007)

Solution

$\left|A^{2}\right|=25 \Rightarrow|A|^{2}=25$

$\Rightarrow(25 \alpha)^{2}=25 \Rightarrow|\alpha|=\frac{1}{5}$

Standard 12

Mathematics

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