Let \alpha \in(0, \infty) and A=[\beginarraylll1 & 2 & \alpha 1 & 0 & 1 0 & 1 & 2\endarray] . If \operatornamedet(\operatornameadj(2 A-A^T) \cdot \operatornameadj(A-2 A^T))=2^8 , then (\operatornamedet(A))^2 is equal to:

Matrices and Determinants PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q33 with Solution

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Correct answer: A

For any $3\times 3$ matrix $M$ , $\det(\operatorname{adj} M)=\big(\det M\big)^{3-1}=\big(\det M\big)^2.$

Also, $\det(XY)=\det X\,\det Y.$

Hence,

=\det\big(\operatorname{adj}(2A-A^T)\big)\det\big(\operatorname{adj}(A-2A^T)\big).$$ So, $$=\big(\det(2A-A^T)\big)^2\big(\det(A-2A^T)\big)^2.$$ Given this equals $2^8$, we get $$\big(\det(2A-A^T)\det(A-2A^T)\big)^2=2^8.$$ Thus, $$\left|\det(2A-A^T)\det(A-2A^T)\right|=2^4=16.$$ --- 2. **Relate $A-2A^T$ and $2A-A^T$** Observe that $$A-2A^T=-(2A^T-A)=-(2A-A^T)^T.$$ For a $3\times 3$ matrix, $$\det(-M)=(-1)^3\det(M)=-\det(M),$$ and $$\det(M^T)=\det(M).$$ Therefore, $$\det(A-2A^T)=\det\big(-(2A-A^T)^T\big)=-\det(2A-A^T).$$ Hence, $$\det(2A-A^T)\det(A-2A^T)=-\big(\det(2A-A^T)\big)^2.$$ So its absolute value is $$\big(\det(2A-A^T)\big)^2=16.$$ Thus, $$\det(2A-A^T)=\pm 4.$$ --- 3. **Compute $2A-A^T$** Given $$A=\begin{bmatrix}1&2&\alpha\\1&0&1\\0&1&2\end{bmatrix}, \qquad A^T=\begin{bmatrix}1&1&0\\2&0&1\\\alpha&1&2\end{bmatrix}.$$ So, $$2A=\begin{bmatrix}2&4&2\alpha\\2&0&2\\0&2&4\end{bmatrix},$$ therefore $$2A-A^T=\begin{bmatrix} 2-1 & 4-1 & 2\alpha-0\\ 2-2 & 0-0 & 2-1\\ 0-\alpha & 2-1 & 4-2 \end{bmatrix} = \begin{bmatrix} 1&3&2\alpha\\ 0&0&1\\ -\alpha&1&2 \end{bmatrix}.$$ --- 4. **Find its determinant** Expand along the second row: $$\det(2A-A^T)= 0-0+1\cdot \begin{vmatrix} 1&3\\ -\alpha&1 \end{vmatrix}.

Thus, $\det(2A-A^T)=1\cdot(1\cdot 1-3(-\alpha))=1+3\alpha.$

So, $1+3\alpha=\pm 4.$

Since $\alpha\in(0,\infty)$ , we must have $1+3\alpha>1$ , hence only $1+3\alpha=4$ is possible. Therefore, $\alpha=1.$

Then $A=\begin{bmatrix}1&2&1\\1&0&1\\0&1&2\end{bmatrix}.$

Compute determinant by first row expansion:

-2\begin{vmatrix}1&1\\0&2\end{vmatrix} +1\begin{vmatrix}1&0\\0&1\end{vmatrix}.$$ So, $$\det(A)=1(0\cdot 2-1\cdot 1)-2(1\cdot 2-1\cdot 0)+1(1\cdot 1-0\cdot 0).$$ $$\det(A)=-1-4+1=-4.$$ Hence, $$(\det A)^2=(-4)^2=16.$$ --- 6. **Option check** - A: $16$ ✅ - B: $36$ ❌ - C: $49$ ❌ - D: $1$ ❌ Therefore, the correct option is **A**.