JEE Mathematics 2025 Question with Solution

Given: $X^TAX=0$ for all nonzero vectors $X$, and

$$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix},\qquad A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}$$

Also,

$$\det(\operatorname{adj}(2(A+I)))=2^{\alpha}3^{\beta}5^{\gamma}$$

Find: $\alpha^2+\beta^2+\gamma^2$.

Since $X^TAX=0$ for all $X$, the matrix $A$ must be skew-symmetric, so $A^T=-A$. Hence let

$$A=\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}$$

Using

$$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}a+b\\-a+c\\-b-c\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix}$$

and

$$A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}2a+b\\-a+2c\\-2b-c\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}$$

we get the equations

$$a+b=1,\quad -a+c=4,\quad -b-c=-5$$ $$2a+b=0,\quad -a+2c=4,\quad -2b-c=-8$$

From $a+b=1$ and $2a+b=0$,

$$a=-1,\qquad b=2$$

Then from $-a+c=4$,

$$c=3$$

Therefore,

$$A=\begin{bmatrix} 0 & -1 & 2\\ 1 & 0 & 3\\ -2 & -3 & 0 \end{bmatrix}$$

Now,

$$A+I=\begin{bmatrix} 1 & -1 & 2\\ 1 & 1 & 3\\ -2 & -3 & 1 \end{bmatrix}$$

so

$$2(A+I)=\begin{bmatrix} 2 & -2 & 4\\ 2 & 2 & 6\\ -4 & -6 & 2 \end{bmatrix}$$

Its determinant is

$$\det(2(A+I))=120$$

For a $3\times 3$ matrix $M$,

$$\det(\operatorname{adj}(M))=(\det M)^{3-1}=(\det M)^2$$

Hence,

$$\det(\operatorname{adj}(2(A+I)))=(120)^2=14400$$

Now factorize:

$$14400=2^6\cdot 3^2\cdot 5^2$$

Therefore,

$$\alpha=6,\qquad \beta=2,\qquad \gamma=2$$

So,

$$\alpha^2+\beta^2+\gamma^2=6^2+2^2+2^2=36+4+4=44$$

Therefore, the required value is $44$.

Given: $\det(\operatorname{adj}(2(A+I)))=2^{\alpha}3^{\beta}5^{\gamma}$ after finding the matrix $A$ from the vector relations.

Find: $\alpha^2+\beta^2+\gamma^2$.

The key determinant identity is

$$\det(\operatorname{adj}(M))=(\det M)^{n-1}$$

for an $n\times n$ matrix $M$. Here $n=3$, so once $\det(2(A+I))$ is known, the rest follows directly.

From the given vector equations and skew-symmetry, the solution obtains

$$A=\begin{bmatrix} 0 & -1 & 2\\ 1 & 0 & 3\\ -2 & -3 & 0 \end{bmatrix}$$

Thus,

$$2(A+I)=\begin{bmatrix} 2 & -2 & 4\\ 2 & 2 & 6\\ -4 & -6 & 2 \end{bmatrix}$$

and

$$\det(2(A+I))=120$$

Now apply the adjugate determinant property:

$$\det(\operatorname{adj}(2(A+I)))=(120)^2=14400=2^6\cdot 3^2\cdot 5^2$$

Hence,

$$\alpha=6,\quad \beta=2,\quad \gamma=2$$

Therefore,

$$\alpha^2+\beta^2+\gamma^2=36+4+4=44$$

So the final answer is $44$.