Let A = [alpha -1; 6 beta], alpha > 0, such that det(A)=0 and alpha + beta = 1. If I denotes 2x2 identity matrix, then the matrix (I+A)^8 is :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – JEE-MAIN 2025

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Home Maths MatricesQuestion #296

Let

A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}]

α > 0

, such that

det (A) = 0

and

α + β = 1

. If

I

denotes

2 \times 2

identity matrix, then the matrix

(I + A)^{8}

is :

Clue 1 of 3: Matrix Properties

Use the given conditions for the determinant and trace of matrix $A$ to determine its specific properties.

🥷

Ninja Strategy•Idempotent Matrix Property and Element Check

Recognize that $A$ is an idempotent matrix ( $A^{2} = A$ ) from its characteristic equation, which implies $(I + A)^{n} = I + (2^{n} - 1) A$ . Then, calculate only the top-right element of $(I + A)^{8}$ as $0 + (2^{8} - 1) \times (- 1) = - 255$ and eliminate options that do not match.

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Step 1: Determine elements of matrix A✦ Active

Given $A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}]$ , $α > 0$ , $det (A) = 0$ and $α + β = 1$ . The determinant is $α β - (- 1) (6) = α β + 6$ . Setting $det (A) = 0$ gives $α β + 6 = 0 ⟹ α β = - 6$ . We have a system of equations: $α + β = 1$ and $α β = - 6$ . These are the roots of the quadratic equation $x^{2} - (α + β) x + α β = 0$ , which is $x^{2} - x - 6 = 0$ . Factoring gives $(x - 3) (x + 2) = 0$ , so $x = 3$ or $x = - 2$ . Since $α > 0$ , we must have $α = 3$ and $β = - 2$ . Thus, the matrix $A$ is:

A = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}]

Step 2: Find the property of matrix A and formula for

(I + A)^{n}

○ Expand

The trace of $A$ is $Tr (A) = α + β = 1$ . The determinant is $det (A) = 0$ . The characteristic equation of $A$ is $λ^{2} - (Tr (A)) λ + det (A) = 0$ , which simplifies to $λ^{2} - λ = 0$ . By the Cayley-Hamilton theorem, $A^{2} - A = 0$ , so $A^{2} = A$ . This means $A$ is an idempotent matrix. For an idempotent matrix $A$ , the power $(I + A)^{n}$ follows the pattern $(I + A)^{n} = I + (2^{n} - 1) A$ . We need to calculate $(I + A)^{8}$ , so $n = 8$ :

(I + A)^{8} = I + (2^{8} - 1) A

💡 Teacher's Secret Hint

Remember that $A^{2} = A$ is a key property for simplifying matrix powers involving $(I + A)$ .

Step 3: Calculate

(I + A)^{8}

○ Expand

Substitute $2^{8} = 256$ into the formula: $2^{8} - 1 = 255$ . Now, substitute $A$ into the expression:

(I + A)^{8} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + 255 [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}]

Perform the scalar multiplication and matrix addition:

(I + A)^{8} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 255 \times 3 & 255 \times (- 1) \\ 255 \times 6 & 255 \times (- 2) \end{matrix}]

(I + A)^{8} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 765 & - 255 \\ 1530 & - 510 \end{matrix}]

(I + A)^{8} = [\begin{matrix} 1 + 765 & 0 - 255 \\ 0 + 1530 & 1 - 510 \end{matrix}] = [\begin{matrix} 766 & - 255 \\ 1530 & - 509 \end{matrix}]

This result matches option 3.

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Maths Question 18 – JEE-MAIN 2025

Clue 1 of 3: Matrix Properties

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