Let a ∈ R and A be a matrix of order 3×3 such that det(A) = -4 and A + I = [ 1 a 1 \ 2 1 0 \ a 1 2 ], where I is the identity matrix of order 3×3. If det((a+1)adj((a-1)A)) is 2^m 3^n, m, n ∈ {0, 1, 2,..., 20}, then m+n is equal to :

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

Maths Q15 – JEE-MAIN 2025

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Home Maths MatricesQuestion #295

Let

a \in R

and A be a matrix of order

3 \times 3

such that

det (A) = - 4

and

A + I = [\begin{matrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{matrix}]

, where I is the identity matrix of order

3 \times 3

. If

det ((a + 1) adj ((a - 1) A))

2^{m} 3^{n}

m, n \in {0, 1, 2, . . ., 20}

, then

m + n

is equal to :

Clue 1 of 3: Determine the matrix A and the value of 'a'

First, subtract the identity matrix from the given sum to find A, then calculate its determinant and equate it to the given value to solve for $a$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine matrix A and the value of 'a'✦ Active

Given $A + I = [\begin{matrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{matrix}]$ , we find $A$ by subtracting the identity matrix $I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ :

A = [\begin{matrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{matrix}] - [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{matrix}]

Now, calculate $det (A)$ :

det (A) = 0 (0 - 0) - a (2 - 0) + 1 (2 - 0) = - 2 a + 2

Given $det (A) = - 4$ , we have:

- 2 a + 2 = - 4 ⟹ - 2 a = - 6 ⟹ a = 3

Step 2: Simplify the expression using determinant properties○ Expand

Substitute $a = 3$ into the expression $det ((a + 1) adj ((a - 1) A))$ :

det ((3 + 1) adj ((3 - 1) A)) = det (4 adj (2 A))

Using the property $det (k M) = k^{n} det (M)$ where $n = 3$ (order of matrix A):

det (4 adj (2 A)) = 4^{3} det (adj (2 A)) = 64 det (adj (2 A))

Using the property $det (adj (M)) = (det (M))^{n - 1}$ where $n = 3$ :

64 det (adj (2 A)) = 64 (det (2 A))^{3 - 1} = 64 (det (2 A))^{2}

Now, calculate $det (2 A)$ using $det (k M) = k^{n} det (M)$ :

det (2 A) = 2^{3} det (A) = 8 \cdot (- 4) = - 32

Substitute this back into the expression:

64 (det (2 A))^{2} = 64 (- 32)^{2} = 64 \cdot 1024

💡 Teacher's Secret Hint

Remember to apply the determinant properties for scalar multiplication and adjoints in the correct order.

Step 3: Express in powers of 2 and 3 and find m+n○ Expand

Express the result in terms of powers of 2 and 3:

64 \cdot 1024 = 2^{6} \cdot 2^{10} = 2^{16}

Given that the expression is $2^{m} 3^{n}$ , by comparing with $2^{16}$ :

m = 16

n = 0

Both $m = 16$ and $n = 0$ are within the set ${0, 1, 2, . . ., 20}$ . Finally, calculate $m + n$ :

m + n = 16 + 0 = 16

💡 Teacher's Secret Hint

Ensure all calculations are correct, especially powers of 2.

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Maths Question 15 – JEE-MAIN 2025

Clue 1 of 3: Determine the matrix A and the value of 'a'

🧠 Full Solution Path