Matrices – Adjoint and Determinant – JEE Main 23 Jan 2026 Shift 1

Solution:

We are given $|A| = 6$ and order $n=3$.
Let’s simplify the expression inside the adjoint step-by-step.
Recall the property $adj(kA) = k^{n-1} adj(A)$. For $n=3$, $adj(kA) = k^2 adj(A)$.
Also $adj(AB) = adj(B) adj(A)$.

First, simplify the innermost term $B = A^2 \cdot adj(2A)$.
$$ adj(2A) = 2^{3-1} adj(A) = 4 adj(A) $$
$$ B = A^2 \cdot (4 adj(A)) = 4 A (A \cdot adj(A)) $$
Using $A \cdot adj(A) = |A| I$:
$$ B = 4 A (|A| I) = 4 |A| A $$
Since $|A| = 6$:
$$ B = 24 A $$

Now we substitute $B$ back into the main expression:
$$ X = |adj(3 adj(B))| = |adj(3 adj(24A))| $$
Let $C = 3 adj(24A)$. We need $|adj(C)|$.
Formula: $|adj(C)| = |C|^{n-1} = |C|^2$.

Now find $|C|$:
$$ |C| = |3 adj(24A)| $$
Using $|kA| = k^n |A|$:
$$ |C| = 3^3 |adj(24A)| = 27 |adj(24A)| $$
Using $|adj(M)| = |M|^{n-1}$:
$$ |C| = 27 |24A|^2 $$

Now substitute $|C|$ into the expression for $X$:
$$ X = |C|^2 = (27 |24A|^2)^2 = 27^2 |24A|^4 $$
$$ X = (3^3)^2 |24A|^4 = 3^6 |24A|^4 $$

Now find $|24A|$:
$$ |24A| = 24^3 |A| = 24^3 \cdot 6 $$
Substitute this back:
$$ X = 3^6 (24^3 \cdot 6)^4 = 3^6 \cdot 24^{12} \cdot 6^4 $$

Prime factorization:
$24 = 2^3 \cdot 3$
$6 = 2 \cdot 3$
$$ X = 3^6 \cdot (2^3 \cdot 3)^{12} \cdot (2 \cdot 3)^4 $$
$$ X = 3^6 \cdot 2^{36} \cdot 3^{12} \cdot 2^4 \cdot 3^4 $$
$$ X = 2^{36+4} \cdot 3^{6+12+4} $$
$$ X = 2^{40} \cdot 3^{22} $$

Given $X = 2^m \cdot 3^n$.
So $m = 40$ and $n = 22$.
We need $m + n$:
$$ m + n = 40 + 22 = 62 $$

Ans. 62