Let A- 1 2- -1 4 - If A-1- I- A- - - - I is 2- 2 identity matrix- then 4- - is

Given: A=[ 1 amp; 2; nbsp; 1 amp; 4 ] |A|=4+2=6 ∴ A^ 1=16[ 4 amp; 2; nbsp;1 amp; 1 ]=[ 2/3 amp; 1/3; nbsp; 1/6 amp; 1/6 ] Now, α I+β A= ...

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Standard XII

Mathematics

Equality of Matrices

Let A=[ 1 2; ...

Question

Let $A = [\begin{matrix} 1 & 2 - 1 & 4 \end{matrix}] .$ If $A^{- 1} = α I + β A, α, β \in R, I$ is $2 \times 2$ identity matrix, then $4 (α - β)$ is

\frac{8}{3}

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Solution

The correct option is C $4$
Given: $A = [\begin{matrix} 1 & 2 - 1 & 4 \end{matrix}]$
$| A | = 4 + 2 = 6$
$∴ A^{- 1} = \frac{1}{6} [\begin{matrix} 4 & - 2 1 & 1 \end{matrix}] = [\begin{matrix} 2 / 3 & - 1 / 3 1 / 6 & 1 / 6 \end{matrix}]$
Now, $α I + β A = [\begin{matrix} α + β & 2 β - β & α + 4 β \end{matrix}]$
Since $A^{- 1} = α I + β A$
$∴ [\begin{matrix} 2 / 3 & - 1 / 3 1 / 6 & 1 / 6 \end{matrix}] = [\begin{matrix} α + β & 2 β - β & α + 4 β \end{matrix}]$
$∴ β = - \frac{1}{6}$ and $α = \frac{5}{6}$
$4 (α - β) = 4 (\frac{5}{6} + \frac{1}{6}) = 4$

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L e t M = [\begin{matrix} {sin}^{4} θ & - 1 - {sin}^{2} θ 1 + {cos}^{2} θ & {cos}^{4} θ \end{matrix}] = α I + β M^{- 1},

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α = α (θ) a n d β = β (θ)

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α^{*}

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Let A=[12−14]. If A−1=αI+βA, α, β∈R, I is 2×2 identity matrix, then 4(α−β) is

The correct option is C 4Given: A=[12−14] |A|=4+2=6 ∴A−1=16[4−211]=[2/3−1/31/61/6] Now, αI+βA=[α+β2β−βα+4β] Since A−1=αI+βA ∴[2/3−1/31/61/6]=[α+β2β−βα+4β] ∴β=−16 and α=56 4(α−β)=4(56+16)=4

Let $A = [\begin{matrix} 1 & 2 - 1 & 4 \end{matrix}] .$ If $A^{- 1} = α I + β A, α, β \in R, I$ is $2 \times 2$ identity matrix, then $4 (α - β)$ is