Let A=left[begin{array}{lll}1 & 0 & 0 0 & α & β 0 & β & αend{array}right

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3 and 4 .Determinants and Matrices

hard

Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha\end{array}\right]$ and $|2 A|^3=2^{21}$ where $\alpha, \beta \in Z$, Then a value of $\alpha $ is

A

$3$

B

$5$

C

$17$

D

$9$

(JEE MAIN-2024)

Solution

$ |A|=\alpha^2-\beta^2 $

$ |2 A|^3=2^{21} \Rightarrow|A|=2^4 $

$ \alpha^2-\beta^2=16 $

$ (\alpha+\beta)(\alpha-\beta)=16 \Rightarrow \alpha=4 \text { or } 5$

Standard 12

Mathematics

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