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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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124,38

14,23

−124,38

−14,23

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detailed solution

Correct option is C

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

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