Matrices – Adjoint and Determinant – JEE Main 2 Apr 2025 Shift 1

Solution:

First, find matrix $A$:
$$ A = (A+I) – I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} – \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix} $$

Calculate $\det(A)$:
$$ |A| = -2(a(1) – 1(1)) = -2(a-1) $$
Given $|A| = -4$, so:
$$ -2(a-1) = -4 \Rightarrow a-1 = 2 \Rightarrow a = 3 $$

We need to find $D = \det((a+1)\text{adj}((a-1)A))$.
Substitute $a=3$:
$$ D = \det(4 \text{adj}(2A)) $$

For a $3 \times 3$ matrix, $\det(kM) = k^3 \det(M)$ and $\det(\text{adj}(M)) = (\det(M))^2$.
$$ D = 4^3 \det(\text{adj}(2A)) = 4^3 (\det(2A))^2 $$

Now, $\det(2A) = 2^3 \det(A) = 8(-4) = -32$.
$$ D = 64 (-32)^2 = 2^6 (2^5)^2 = 2^6 \cdot 2^{10} = 2^{16} $$

Given $D = 2^m 3^n$, we have $2^{16} \cdot 3^0$.
Thus, $m = 16$ and $n = 0$.
$$ m + n = 16 + 0 = 16 $$

Ans. (4)