JEE Main 2024 — 5 April, Shift 1 — Question 11

Given that $A$ and $B$ are two square matrices of order $n = 3$ with determinants:

$$|A| = 3 \quad \text{and} \quad |B| = 2$$

We need to evaluate the determinant of the following expression:

$$E = \left| A^{\mathrm{T}} A (\operatorname{adj}(2A))^{-1} (\operatorname{adj}(4B)) (\operatorname{adj}(AB))^{-1} A A^{\mathrm{T}} \right|$$

Key Properties of Determinants and Adjoints For any square matrices $X$ and $Y$ of order $n=3$, and a scalar $k$: $|X \cdot Y| = |X| \cdot |Y|$ $|X^{\mathrm{T}}| = |X|$ $|kX| = k^3 |X|$ $|\operatorname{adj}(X)| = |X|^{n-1} = |X|^2$ Expanding the Determinant Expression} Using the multiplicative property of determinants, we can break the main expression down into the product of individual determinants:

$$|E| = |A^{\mathrm{T}}| \cdot |A| \cdot \left| (\operatorname{adj}(2A))^{-1} \right| \cdot |\operatorname{adj}(4B)| \cdot \left| (\operatorname{adj}(AB))^{-1} \right| \cdot |A| \cdot |A^{\mathrm{T}}|$$

Since $|A^{\mathrm{T}}| = |A|$, we can group the terms involving $A$:

$$|E| = |A|^4 \cdot \frac{1}{|\operatorname{adj}(2A)|} \cdot |\operatorname{adj}(4B)| \cdot \frac{1}{|\operatorname{adj}(AB)|}$$

Simplifying Each Adjoint Term

1. For $|\operatorname{adj}(2A)|$:

$$|2A| = 2^3 |A| = 8|A|$$ $$|\operatorname{adj}(2A)| = |2A|^2 = (8|A|)^2 = 64|A|^2$$

2. For $|\operatorname{adj}(4B)|$:

$$|4B| = 4^3 |B| = 64|B|$$ $$|\operatorname{adj}(4B)| = |4B|^2 = (64|B|)^2 = 4096|B|^2$$

3. For $|\operatorname{adj}(AB)|$:

$$|AB| = |A| \cdot |B|$$ $$|\operatorname{adj}(AB)| = |AB|^2 = (|A||B|)^2 = |A|^2 |B|^2$$

Substitution and Final Evaluation Now, substitute these simplified expressions back into the main equation:

$$|E| = |A|^4 \cdot \frac{1}{64|A|^2} \cdot 4096|B|^2 \cdot \frac{1}{|A|^2 |B|^2}$$

Combining the constants and grouping the matrix variables:

$$|E| = \frac{4096}{64} \cdot \frac{|A|^4}{|A|^2 \cdot |A|^2} \cdot \frac{|B|^2}{|B|^2}$$ $$|E| = 64 \cdot \frac{|A|^4}{|A|^4} \cdot \frac{|B|^2}{|B|^2}$$

Since the determinant variables fully cancel out, we obtain:

$$|E| = 64 \cdot 1 \cdot 1 = 64$$

The value of the given determinant expression is $64$.