Let A be a 2 × 2 matrix with real entries such that A ' = A + I , where R- { - 1 , 1 } . If det ( A 2 - A ) = 4 , then the sum of all possible values of is equal to [2023]

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Solution

Q.
Let A be a $2 \times 2$ matrix with real entries such that $A' = α A + I,$ where $α \in R - {- 1, 1}$ . If $\det (A^{2} - A) = 4$ , then the sum of all possible values of $α$ is equal to [2023]

1 0

2 $\frac{3}{2}$

3 $\frac{5}{2}$

4 2

Ans.
(3)

We have, $A^{T} = α A + I$

$∴$ $A = α A^{T} + I \Rightarrow A = α (α A + I) + I$

$\Rightarrow A = α^{2} A + (α + 1) I \Rightarrow A (1 - α^{2}) = (α + 1) I$

$\Rightarrow A = \frac{I}{1 - α},$ Now, $| A |$ $= \frac{1}{{(1 - α)}^{2}} ...(i)$

$A - I = \frac{I}{1 - α} - I = \frac{α}{1 - α} I$

We know that

$| A^{2} - A |$ $= | A | | A - I |$ ...(ii)

$∴$ $| A - I |$ $= {(\frac{α}{1 - α})}^{2} ...(iii)$

Now, $| A^{2} - A |$ $= 4$ [ $∵$ Given]

$\Rightarrow \frac{1}{{(1 - α)}^{2}} \times \frac{α^{2}}{{(1 - α)}^{2}} = 4$ [Using (i), (ii) and (iii)]

$\Rightarrow \frac{α}{{(1 - α)}^{2}} = \pm 2 \Rightarrow 2 {(1 - α)}^{2} = \pm α$

Case I: $2 {(1 - α)}^{2} = α$ $\Rightarrow 2 α^{2} - 5 α + 2 = 0$

$∴$ $α_{1} + α_{2} = \frac{5}{2}$

Case II: $2 {(1 - α)}^{2} = - α$

$\Rightarrow 2 α^{2} - 3 α + 2 = 0 \Rightarrow α \notin R$

$∴$ $Sum of values of α = \frac{5}{2}$

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