If A=5412 and if (A+2I)−1=k1A+k2I then k1,k2=

We have A2−β1A+β2I=0β1=7;β2=6∴A2−7A+6I=0⇒(A+2I)(A−9I)+24I=0⇒(A+2I)(A−9I)−24=I⇒(A+2I)−1=(A−9I)−24⇒(A+2I)−1=−124A+38IClearly k1,k2=−124,38

If A=5412 and if (A+2I)−1=k1A+k2I then k1,k2=

We have A2−β1A+β2I=0β1=7;β2=6∴A2−7A+6I=0⇒(A+2I)(A−9I)+24I=0⇒(A+2I)(A−9I)−24=I⇒(A+2I)−1=(A−9I)−24⇒(A+2I)−1=−124A+38IClearly k1,k2=−124,38

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$If A = [\begin{array}{cc} 5 & 4 \\ 1 & 2 \end{array}] and if (A + 2 I)^{- 1} = k_{1} A + k_{2} I then (k_{1}, k_{2}) =$

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$(\frac{1}{24}, \frac{3}{8})$

$(\frac{1}{4}, \frac{2}{3})$

$(\frac{- 1}{24}, \frac{3}{8})$

$(\frac{- 1}{4}, \frac{2}{3})$

answer is C.

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Detailed Solution

$We have A^{2} - β_{1} A + β_{2} I = 0$

$\begin{array}{l} β_{1} = 7; β_{2} = 6 \\ ∴ A^{2} - 7 A + 6 I = 0 \\ \Rightarrow (A + 2 I) (A - 9 I) + 24 I = 0 \\ \Rightarrow (A + 2 I) \frac{(A - 9 I)}{- 24} = I \\ \Rightarrow (A + 2 I)^{- 1} = \frac{(A - 9 I)}{- 24} \\ \Rightarrow (A + 2 I)^{- 1} = (\frac{- 1}{24}) A + (\frac{3}{8}) I \end{array}$