QJEEM24Ma23ec: JEE Main 2024 Mathematics - Class 12 JEEM24Ma23ec Solution: If and , then is equal to :

The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant. Given $\mathrm{A}$ , $\mathrm{B}$ , and that $\mathrm{C} = \mathrm{ABA}^{\mathrm{T}}$ , and $\mathrm{X} = \mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}$ , we find $\operatorname{det} \mathrm{X}$ as follows:

First, we express $|\mathrm{C}|$ , the determinant of $\mathrm{C}$ , in terms of the determinants of $\mathrm{A}$ and $\mathrm{B}$ , using the property that $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ :

$\begin{aligned} |\mathrm{C}| &= |\mathrm{ABA}^{\mathrm{T}}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{A}^{\mathrm{T}}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{B}| \quad \text{since } |\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|. \end{aligned}$

Next, we find $|\mathrm{X}|$ , using the property $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ and substituting $|\mathrm{C}|$ :

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}| = |\mathrm{A}^{\mathrm{T}}| \cdot |\mathrm{C}|^2 \cdot |\mathrm{A}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{C}|^2. \end{aligned}$

Substituting the expression for $|\mathrm{C}|$ obtained earlier:

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}|^2 \cdot (|\mathrm{A}|^2 \cdot |\mathrm{B}|)^2 \\\\ &= (|\mathrm{A}|^3 \cdot |\mathrm{B}|)^2. \end{aligned}$

The determinants of $\mathrm{A}$ and $\mathrm{B}$ are calculated as:

Finally, substituting these values into our expression for $|\mathrm{X}|$ :

$\begin{aligned} |\mathrm{X}| &= (3^3 \cdot 1)^2 \\\\ &= 729. \end{aligned}$

Therefore, $\operatorname{det} \mathrm{X} = 729$ .