If A=15λ10,A−1=αA+βI and α+β=−2 then 4α2+β2+λ2 is equal to

Given A−1=αA+BI⇒110−5λ10−5−λ1=α15λ10+β1 00 1⇒1010−5λ=α+β ⇒1010−5λ=−2⇒λ=3And −510−5λ=5α⇒α=15 also 110−5λ=10α+β⇒β=−115Now 4α2+β2+λ2 =4125+12125+9=14

If A=15λ10,A−1=αA+βI and α+β=−2 then 4α2+β2+λ2 is equal to

Given A−1=αA+BI⇒110−5λ10−5−λ1=α15λ10+β1 00 1⇒1010−5λ=α+β ⇒1010−5λ=−2⇒λ=3And −510−5λ=5α⇒α=15 also 110−5λ=10α+β⇒β=−115Now 4α2+β2+λ2 =4125+12125+9=14

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If $A = [\begin{array}{cc} 1 & 5 \\ λ & 10 \end{array}], A^{- 1} = αA + βI a nd α + β = - 2$ then $4 α^{2} + β^{2} + λ^{2}$ is equal to

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Detailed Solution

Given $A^{- 1} = αA + BI \Rightarrow \frac{1}{10 - 5 λ} [\begin{array}{cc} 10 & - 5 \\ - λ & 1 \end{array}] = α [\begin{array}{cc} 1 & 5 \\ λ & 10 \end{array}] + β [\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$
$\Rightarrow \frac{10}{10 - 5 λ} = α + β \Rightarrow \frac{10}{10 - 5 λ} = - 2 \Rightarrow λ = 3$
And $\frac{- 5}{10 - 5 λ} = 5 α \Rightarrow α = \frac{1}{5} also \frac{1}{10 - 5 λ} = 10 α + β \Rightarrow β = \frac{- 11}{5}$
Now $4 α^{2} + β^{2} + λ^{2} = 4 (\frac{1}{25}) + \frac{121}{25} + 9 = 14$