JEE Main 2024 — 4 April, Shift 1 — Question 20

Given the matrix:

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

First, let us find the determinant of $A$, denoted as $|A|$:

$$|A| = 1(0 - 1) - 2(2 - 0) + \alpha(1 - 0)$$ $$|A| = -1 - 4 + \alpha = \alpha - 5$$

Establishing the Relation Between $2A - A^{\mathrm{T}}$ and $A - 2A^{\mathrm{T}}$}

Let $P = 2A - A^{\mathrm{T}}$ and $Q = A - 2A^{\mathrm{T}}$. Let us find the transpose of matrix $P$:

$$P^{\mathrm{T}} = (2A - A^{\mathrm{T}})^{\mathrm{T}} = 2A^{\mathrm{T}} - A = -(A - 2A^{\mathrm{T}}) = -Q$$

Taking the determinant on both sides:

$$|P^{\mathrm{T}}| = |-Q|$$

Since $A$ is a $3 \times 3$ matrix, $Q$ is also of order $3$. Using the property $|kM| = k^n |M|$ for a matrix of order $n$:

$$|P^{\mathrm{T}}| = (-1)^3 |Q| = -|Q|$$

Since the determinant of a matrix equals the determinant of its transpose ($|P^{\mathrm{T}}| = |P|$), we have:

$$|P| = -|Q| \implies |Q| = -|P|$$

Simplifying the Given Determinant Equation

We are given:

$$\det(\operatorname{adj}(2A - A^{\mathrm{T}}) \cdot \operatorname{adj}(A - 2A^{\mathrm{T}})) = 2^8$$ $$\det(\operatorname{adj}(P) \cdot \operatorname{adj}(Q)) = 2^8$$

Using the property $\det(XY) = \det(X)\det(Y)$:

$$|\operatorname{adj}(P)| \cdot |\operatorname{adj}(Q)| = 2^8$$

Recall the property for the determinant of an adjoint matrix of order $n=3$:

$$|\operatorname{adj}(M)| = |M|^{n-1} = |M|^2$$

Substituting this into the equation:

$$|P|^2 \cdot |Q|^2 = 2^8$$

Since $|Q| = -|P|$, then $|Q|^2 = (-|P|)^2 = |P|^2$:

$$|P|^2 \cdot |P|^2 = 2^8$$ $$|P|^4 = 2^8 \implies |P| = \pm 2^2 = \pm 4$$

Calculating $|P|$ Explicitly

Let us compute matrix $P = 2A - A^{\mathrm{T}}$:

$$A^{\mathrm{T}} = \begin{bmatrix} 1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2 \end{bmatrix}$$ $$P = 2\begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2 \end{bmatrix}$$ $$P = \begin{bmatrix} 2-1 & 4-1 & 2\alpha-0 \\ 2-2 & 0-0 & 2-1 \\ 0-\alpha & 2-1 & 4-2 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 2\alpha \\ 0 & 0 & 1 \\ -\alpha & 1 & 2 \end{bmatrix}$$

Expanding along the second row to find $|P|$:

$$|P| = -1 \cdot \det \begin{bmatrix} 1 & 3 \\ -\alpha & 1 \end{bmatrix}$$ $$|P| = -1 \cdot (1 - (-3\alpha)) = -(1 + 3\alpha)$$

Finding the Value of $(\det(A))^2$}

From Step 2, we have $|P| = \pm 4$.

Case 1: If $|P| = -4$

$$-(1 + 3\alpha) = -4 \implies 1 + 3\alpha = 4 \implies 3\alpha = 3 \implies \alpha = 1$$

Substituting $\alpha = 1$ into $|A| = \alpha - 5$:

$$|A| = 1 - 5 = -4$$ $$(|A|)^2 = (-4)^2 = 16$$

Case 2: If $|P| = 4$

$$-(1 + 3\alpha) = 4 \implies 1 + 3\alpha = -4 \implies 3\alpha = -5 \implies \alpha = -\frac{5}{3}$$

Substituting $\alpha = -\frac{5}{3}$ into $|A| = \alpha - 5$:

$$|A| = -\frac{5}{3} - 5 = -\frac{20}{3}$$ $$(|A|)^2 = \left(-\frac{20}{3}\right)^2 = \frac{400}{9}$$

For standard structural formats where $\alpha$ is restricted to integer configurations, the primary evaluated value is $16.$