JEE Mathematics 2026 Question with Solution

Given: $A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$

Find: The determinant of $A^{2025} - 3A^{2024} + A^{2023}$.

Using the characteristic polynomial of $A$,

$$\lambda^2 - 7\lambda + 1 = 0$$

so by the Cayley--Hamilton theorem,

$$A^2 - 7A + I = 0$$

Hence,

$$A^2 = 7A - I$$

Now factor the given expression:

$$A^{2025} - 3A^{2024} + A^{2023} = A^{2023}(A^2 - 3A + I)$$

Substitute $A^2 = 7A - I$:

$$A^2 - 3A + I = (7A - I) - 3A + I = 4A$$

Therefore,

$$A^{2025} - 3A^{2024} + A^{2023} = 4A^{2024}$$

Taking determinant on both sides for a $2 \times 2$ matrix,

$$\det(4A^{2024}) = 4^2 \det(A^{2024})$$

Also,

$$\det(A) = 2 \cdot 5 - 3 \cdot 3 = 1$$

so

$$\det(A^{2024}) = (\det A)^{2024} = 1$$

Thus,

$$\det(4A^{2024}) = 16$$

the solution concludes with $24$, but this arithmetic step is inconsistent with the determinant property above. Among the given options, the mathematically correct value is $16$, which corresponds to option B.

Given: $A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$

Find: $\det\left(A^{2025} - 3A^{2024} + A^{2023}\right)$.

If $\lambda_1, \lambda_2$ are eigenvalues of $A$, then for the matrix

$$A^{2023}(A^2 - 3A + I)$$

the corresponding eigenvalues are

$$\lambda_i^{2023}(\lambda_i^2 - 3\lambda_i + 1)$$

for $i = 1, 2$.

From Cayley--Hamilton,

$$\lambda_i^2 - 7\lambda_i + 1 = 0$$

so

$$\lambda_i^2 - 3\lambda_i + 1 = 4\lambda_i$$

Therefore each transformed eigenvalue becomes

$$4\lambda_i^{2024}$$

Hence the determinant is the product:

$$(4\lambda_1^{2024})(4\lambda_2^{2024}) = 16(\lambda_1\lambda_2)^{2024}$$

But $\lambda_1\lambda_2 = \det(A) = 1$, so

$$16(\lambda_1\lambda_2)^{2024} = 16$$

Therefore, the determinant equals $16$, so the correct mathematical option is B.