Maths Q3 – JEE-MAIN 2025

Let A be a

3 \times 3

real matrix such that

A^{2} (A - 2 I) - 4 (A - I) = O

, where I and O are the identity and null matrices, respectively. If

A^{5} = α A^{2} + β A + γ I

, where

α, β

, and

γ

are real constants, then

α + β + γ

is equal to :

The given equation provides a polynomial relationship that the matrix A satisfies.

Step 1: Derive the fundamental matrix relation✦ Active

The given equation for matrix A is $A^{2} (A - 2 I) - 4 (A - I) = O$ . Expand this equation to find a relation for $A^{3}$ :

A^{3} - 2 A^{2} - 4 A + 4 I = O

A^{3} = 2 A^{2} + 4 A - 4 I (1)

Step 2: Calculate

A^{4}

and

A^{5}

○ Expand

Multiply equation (1) by A to find $A^{4}$ , then substitute $A^{3}$ again:

A^{4} = A \cdot A^{3} = A (2 A^{2} + 4 A - 4 I) = 2 A^{3} + 4 A^{2} - 4 A

A^{4} = 2 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 4 A

A^{4} = 4 A^{2} + 8 A - 8 I + 4 A^{2} - 4 A = 8 A^{2} + 4 A - 8 I (2)

Now, multiply equation (2) by A to find $A^{5}$ , and substitute $A^{3}$ once more:

A^{5} = A \cdot A^{4} = A (8 A^{2} + 4 A - 8 I) = 8 A^{3} + 4 A^{2} - 8 A

A^{5} = 8 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 8 A

A^{5} = 16 A^{2} + 32 A - 32 I + 4 A^{2} - 8 A

A^{5} = 20 A^{2} + 24 A - 32 I

💡 Teacher's Secret Hint

Ensure careful distribution and collection of terms at each step to avoid algebraic errors.

Step 3: Determine

α, β, γ

and their sum○ Expand

Comparing the expression for $A^{5}$ with the given form $A^{5} = α A^{2} + β A + γ I$ , we get:

α = 20

β = 24

γ = - 32

Finally, calculate the sum $α + β + γ$ :

α + β + γ = 20 + 24 + (- 32) = 44 - 32 = 12

💡 Teacher's Secret Hint

Double-check the signs, especially for the constant term $γ$ .

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Maths Question 3 – JEE-MAIN 2025