If A = [ 1 5 10 ] , A - 1 = A + I and = - 2 , then 4 2 + 2 + 2 is equal to [2023]

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Solution

Q.
If $A = [\begin{matrix} 1 & 5 \\ λ & 10 \end{matrix}], A^{- 1} = α A + β I and α + β = - 2,$ then $4 α^{2} + β^{2} + λ^{2}$ is equal to [2023]

1 10

2 14

3 19

4 12

Ans.
(2)

Given, $A = [\begin{matrix} 1 & 5 \\ λ & 10 \end{matrix}] \Rightarrow | A | = 10 - 5 λ$

Then, $A^{- 1} = [\begin{matrix} \frac{10}{10 - 5 λ} & \frac{- 5}{10 - 5 λ} \\ \frac{- λ}{10 - 5 λ} & \frac{1}{10 - 5 λ} \end{matrix}] = α [\begin{matrix} 1 & 5 \\ λ & 10 \end{matrix}] + β [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} α + β & 5 α \\ α λ & 10 α + β \end{matrix}]$

So, $α + β = \frac{10}{10 - 5 λ}; Given α + β = - 2$

$\Rightarrow - 2 = \frac{10}{10 - 5 λ} \Rightarrow - 20 + 10 λ = 10 \Rightarrow λ = 3$

Also, $5 α = \frac{- 5}{10 - 5 λ} \Rightarrow 5 α = \frac{- 5}{10 - 15}$

$\Rightarrow α = \frac{1}{5}$ then $β = \frac{- 11}{5}$

Now, $4 α^{2} + β^{2} + λ^{2} = 4 {(\frac{1}{5})}^{2} + {(\frac{- 11}{5})}^{2} + 3^{2}$

$= \frac{4}{25} + \frac{121}{25} + 9 = 14$

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