A=55αα0α5α005, if |A2|=25 then |α|=

GIven,A=55αα0α5α005 detA=5[5α−0]−0−0 detA=25α W.K.T |An|=|A|n∀n∈I+ If n=2 |A2|=52 |A2|=25 |A|2=25 (25α)2=25 625α2=25 α2=25625 |α|=15

A=55αα0α5α005, if |A2|=25 then |α|=

GIven,A=55αα0α5α005 detA=5[5α−0]−0−0 detA=25α W.K.T |An|=|A|n∀n∈I+ If n=2 |A2|=52 |A2|=25 |A|2=25 (25α)2=25 625α2=25 α2=25625 |α|=15

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$A = [\begin{matrix} 5 & 5 α & α \\ 0 & α & 5 α \\ 0 & 0 & 5 \end{matrix}]$ , if $| A^{2} | = 25$ then $| α | =$

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$5^{2}$

$\frac{1}{5}$

answer is D.

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Detailed Solution

GIven,

$A = [\begin{matrix} 5 & 5 α & α \\ 0 & α & 5 α \\ 0 & 0 & 5 \end{matrix}] \det A = 5 [5 α - 0] - 0 - 0 \det A = 25 α W . K . T | A^{n} | = {| A |}^{n} \forall n \in I^{+} I f n = 2 | A^{2} | = 5^{2} | A^{2} | = 25 {| A |}^{2} = 25 {(25 α)}^{2} = 25 625 α^{2} = 25 α^{2} = \frac{25}{625} | α | = \frac{1}{5}$