JEE Mathematics 2024 Question with Solution

Given: $A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix}$, $B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $C = ABA^T$ and $X = A^T C^2 A$.

Find: $\det(X)$.

First, compute

$$\det(A) = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 3$$

and

$$\det(B) = 1$$

Now, using $C = ABA^T$,

$$\det(C) = \det(A)\det(B)\det(A^T)$$

Since $\det(A^T) = \det(A)$,

$$\det(C) = (\det(A))^2 \det(B) = 3^2 \cdot 1 = 9$$

Next, for $X = A^T C^2 A$,

$$\det(X) = \det(A^T)\det(C^2)\det(A)$$

Also, $\det(A^T) = \det(A)$ and $\det(C^2) = (\det(C))^2$. Therefore,

$$\det(X) = \det(A) \cdot (\det(C))^2 \cdot \det(A)$$ $$= 3 \cdot 9^2 \cdot 3 = 729$$

Therefore, the correct option is B and $\det(X) = 729$.

Given: $A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix}$, $B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $C = ABA^T$ and $X = A^T C^2 A$.

Find: $\det(X)$.

Use the determinant properties:

• $\det(AB) = \det(A)\det(B)$
• $\det(A^T) = \det(A)$
• $\det(A^n) = (\det(A))^n$

First,

$$\det(A) = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3$$

For matrix $B$,

$$\det(B) = (1)(1) - (0)(0) = 1$$

Now,

$$C = ABA^T \Rightarrow \det(C) = \det(A)\det(B)\det(A^T)$$

So,

$$\det(C) = 3 \cdot 1 \cdot 3 = 9$$

Since $X = A^T C^2 A$,

$$\det(X) = \det(A^T)\det(C^2)\det(A)$$ $$= \det(A) \cdot (\det(C))^2 \cdot \det(A)$$ $$= 3 \cdot 9^2 \cdot 3$$ $$= 3 \cdot 81 \cdot 3 = 729$$

Hence, $\det(X) = 729$. The correct option is B.