Let A = [ 1 0 0 0 0 ] and | 2 A | 3 = 2 21 where. Then a value of is [2024]

Question

Let $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$ and $| 2 A |^{3} = 2^{21}$ where $α, β \in Z .$ Then a value of $α$ is [2024]

Smart Achievers · Accepted Answer

(1)

Given, $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$

$\Rightarrow | A | = α^{2} - β^{2}$ ...(i)

Also, $| 2 A |^{3} = 2^{21}$

$\Rightarrow {(2^{3} | A |)}^{3} = 2^{21} \Rightarrow | A |^{3} = 2^{12} \Rightarrow | A | = 2^{4}$

From (i), we get, $α^{2} - β^{2} = 2^{4}$

$\Rightarrow α^{2} - β^{2} = 16 \Rightarrow (α + β) (α - β) = 8 \times 2$

So, $α + β = 8 and α - β = 2$

Solving, we get, $α = 5$ and $β = 3$

Solution

Q.
Let $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$ and $| 2 A |^{3} = 2^{21}$ where $α, β \in Z .$ Then a value of $α$ is [2024]

1 5

2 9

3 17

4 3

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Solution

Q. Let A=[1000αβ0βα] and |2A|3=221 where α,β∈Z. Then a value of α is [2024]

1 5

2 9

3 17

4 3

Ans. (1) Given, A=[1000αβ0βα] ⇒|A|=α2-β2 ...(i) Also, |2A|3=221 ⇒(23|A|)3=221⇒|A|3=212⇒|A|=24 From (i), we get, α2-β2=24 ⇒α2-β2=16⇒(α+β)(α-β)=8×2 So, α+β=8 and α-β=2 Solving, we get, α=5 and β=3

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Q.
Let $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & α & β \\ 0 & β & α \end{matrix}]$ and $| 2 A |^{3} = 2^{21}$ where $α, β \in Z .$ Then a value of $α$ is [2024]