First slide

Applications of determinant

Question

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

Moderate

A

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

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

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

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

Solution

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

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