JEE Main 2024 — 29 January, Shift 1 — Question 14

Given the matrix:

$$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{pmatrix}$$

And the equation:

$$|2A|^3 = 2^{21}$$

where $\alpha, \beta \in \mathbb{Z}$.

Step 1: Calculate the determinant of matrix $A$, $|A|$. The determinant of $A$ can be calculated by expanding along the first row:

$$|A| = 1 \cdot \begin{vmatrix} \alpha & \beta \\ \beta & \alpha \end{vmatrix} - 0 \cdot \text{cofactor} + 0 \cdot \text{cofactor}$$ $$|A| = 1 \cdot (\alpha \cdot \alpha - \beta \cdot \beta)$$ $$|A| = \alpha^2 - \beta^2$$

Step 2: Use the property of determinants for scalar multiplication. For an $n \times n$ matrix $A$ and a scalar $c$, $|cA| = c^n|A|$. In this case, $n=3$ (since $A$ is a $3 \times 3$ matrix) and $c=2$. So, we have:

$$|2A| = 2^3 |A|$$ $$|2A| = 8 (\alpha^2 - \beta^2)$$

Step 3: Substitute $|2A|$ into the given equation. We are given $|2A|^3 = 2^{21}$. Substituting the expression for $|2A|$:

$$(8 (\alpha^2 - \beta^2))^3 = 2^{21}$$ $$(2^3 (\alpha^2 - \beta^2))^3 = 2^{21}$$

Applying the power rule $(xy)^k = x^k y^k$:

$$(2^3)^3 (\alpha^2 - \beta^2)^3 = 2^{21}$$ $$2^9 (\alpha^2 - \beta^2)^3 = 2^{21}$$

Step 4: Solve for $\alpha^2 - \beta^2$. Divide both sides by $2^9$:

$$(\alpha^2 - \beta^2)^3 = \frac{2^{21}}{2^9}$$ $$(\alpha^2 - \beta^2)^3 = 2^{21-9}$$ $$(\alpha^2 - \beta^2)^3 = 2^{12}$$

Take the cube root of both sides:

$$\alpha^2 - \beta^2 = (2^{12})^{1/3}$$ $$\alpha^2 - \beta^2 = 2^{12/3}$$ $$\alpha^2 - \beta^2 = 2^4$$ $$\alpha^2 - \beta^2 = 16$$

Step 5: Find integer values for $\alpha$ that satisfy the equation. We can factor the left side as a difference of squares:

$$(\alpha - \beta)(\alpha + \beta) = 16$$

Since $\alpha$ and $\beta$ are integers, $(\alpha - \beta)$ and $(\alpha + \beta)$ must be integer factors of 16. Let $x = \alpha - \beta$ and $y = \alpha + \beta$. Then $y - x = (\alpha + \beta) - (\alpha - \beta) = 2\beta$, which is an even integer. Also, $x + y = (\alpha - \beta) + (\alpha + \beta) = 2\alpha$, which is an even integer. This implies that $x$ and $y$ must have the same parity. Since their product $xy = 16$ is even, both $x$ and $y$ must be even.

Possible pairs of even integer factors $(x, y)$ for 16, where $xy=16$: If $x=2$ and $y=8$: $\alpha - \beta = 2$ $\alpha + \beta = 8$ Adding the two equations: $2\alpha = 10 \Rightarrow \alpha = 5$. Subtracting the first from the second: $2\beta = 6 \Rightarrow \beta = 3$. Check: $5^2 - 3^2 = 25 - 9 = 16$. This is a valid pair of integer solutions for $\alpha$ and $\beta$. So, $\alpha = 5$ is a possible value.

If $x=-8$ and $y=-2$: $\alpha - \beta = -8$ $\alpha + \beta = -2$ Adding the two equations: $2\alpha = -10 \Rightarrow \alpha = -5$. Subtracting the first from the second: $2\beta = 6 \Rightarrow \beta = 3$. Check: $(-5)^2 - 3^2 = 25 - 9 = 16$. This is also a valid pair of integer solutions. So, $\alpha = -5$ is a possible value.

If $x=4$ and $y=4$: $\alpha - \beta = 4$ $\alpha + \beta = 4$ Adding the two equations: $2\alpha = 8 \Rightarrow \alpha = 4$. Subtracting the first from the second: $2\beta = 0 \Rightarrow \beta = 0$. Check: $4^2 - 0^2 = 16 - 0 = 16$. This is a valid pair of integer solutions. So, $\alpha = 4$ is a possible value.

If $x=-4$ and $y=-4$: $\alpha - \beta = -4$ $\alpha + \beta = -4$ Adding the two equations: $2\alpha = -8 \Rightarrow \alpha = -4$. Subtracting the first from the second: $2\beta = 0 \Rightarrow \beta = 0$. Check: $(-4)^2 - 0^2 = 16 - 0 = 16$. This is a valid pair of integer solutions. So, $\alpha = -4$ is a possible value. The question asks for "a value of $\alpha$". Any of $5, -5, 4, -4$ would be a correct answer.

The final answer is $\boxed{5}$.