Matrices – Adjoint and Determinant – JEE Main 22 Jan 2026 Shift 1

Solution:

Given $A + A^T = O \Rightarrow A^T = -A$, so $A$ is a skew-symmetric matrix.
The diagonal elements of a skew-symmetric matrix are 0.
Let $A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}$.

Using the first condition:
$$ \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -a \\ -a \\ -b+c \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} $$
Comparing elements:
1. $-a = 3 \Rightarrow a = -3$
2. $-b + c = 2 \Rightarrow c – b = 2$ … (i)

Using the second condition $A^2 X = Y \Rightarrow A(AX) = Y$. Since $AX = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}$:
$$ A \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 19 \\ -24 \end{pmatrix} $$
$$ \begin{pmatrix} 0 & -3 & b \\ 3 & 0 & c \\ -b & -c & 0 \end{pmatrix} \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} -9+2b \\ 9+2c \\ -3b-3c \end{pmatrix} = \begin{pmatrix} -3 \\ 19 \\ -24 \end{pmatrix} $$

Comparing elements:
1. $-9 + 2b = -3 \Rightarrow 2b = 6 \Rightarrow b = 3$
2. $9 + 2c = 19 \Rightarrow 2c = 10 \Rightarrow c = 5$
Check with (i): $c – b = 5 – 3 = 2$ (Satisfied).

So, matrix $A = \begin{pmatrix} 0 & -3 & 3 \\ 3 & 0 & 5 \\ -3 & -5 & 0 \end{pmatrix}$.
We need to find $K = \det(\text{adj}(2\text{adj}(A+I)))$.
Let $B = A+I$.
$$ B = \begin{pmatrix} 1 & -3 & 3 \\ 3 & 1 & 5 \\ -3 & -5 & 1 \end{pmatrix} $$
$$ |B| = 1(1+25) – (-3)(3+15) + 3(-15+3) $$
$$ |B| = 26 + 54 – 36 = 44 = 2^2 \cdot 11 $$

Now evaluate the determinant expression.
Recall property $|\text{adj}(M)| = |M|^{n-1}$. Here $n=3$.
Let $C = 2\text{adj}(B)$.
$$ K = |\text{adj}(2\text{adj}(B))| = |\text{adj}(C)| = |C|^2 $$
$$ |C| = |2\text{adj}(B)| = 2^3 |\text{adj}(B)| = 8 |B|^2 $$
$$ K = (8|B|^2)^2 = 64 |B|^4 = 2^6 |B|^4 $$

Substitute $|B| = 44$:
$$ K = 2^6 (44)^4 = 2^6 (2^2 \cdot 11)^4 $$
$$ K = 2^6 \cdot 2^8 \cdot 11^4 = 2^{14} \cdot 11^4 $$

Given form $2^{\alpha} \cdot 3^{\beta} \cdot 11^{\gamma}$.
Comparing powers:
$\alpha = 14$
$\beta = 0$
$\gamma = 4$

Sum $\alpha + \beta + \gamma = 14 + 0 + 4 = 18$.

Ans. (18)