Question

Let $A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$; $x, y, z \in I$ with the condition that $det(adj(adj A)) = 2^{12} \times 3^8 \times 5^4$, $S_1$ is set of all the matrices $A$, $S_2$ is set of all such matrices $A$ where trace of $A$ is divisible by 2, and $x, y, z \in N$ and $S_3$ is set of all such matrices $A$, where $(trace A^3) - 3 det A = 0$, and $x, y, z \in N$. Let the number of elements in set $S_1$, $S_2$ and $S_3$ be $2\alpha$, $\beta$ and $\gamma$ respectively.

Which of the following is/are correct?

• A. $\frac{\alpha}{\beta + \gamma} < 10$
• B. $\frac{\alpha}{\beta - \gamma} > 15$
• C. $\frac{\alpha}{\beta + \gamma} > 9$
• D. $\frac{\alpha}{\beta - \gamma} < 14$

Answer 1

Answer: (C), (D)

(3), (4)

Answer 2

Here's a breakdown of the solution:

1. Determine the condition on $x, y, z$:

Using the property $det(adj(M)) = (det(M))^{n-1}$. For matrix $A$, $det(adj(adj(A))) = (det(A))^4 = (xyz)^4$. Equating this to the given value, $(xyz)^4 = 2^{12} \times 3^8 \times 5^4$, which simplifies to $|xyz| = 2^3 \times 3^2 \times 5^1 = 360$. So, $xyz = \pm 360$.

2. Calculate $\alpha$ (for $S_1$):

$S_1$ contains matrices where $x, y, z \in I$ and $xyz = \pm 360$. The number of ways to factorize $360 = 2^3 \times 3^2 \times 5^1$ into three positive integer factors $(|x|, |y|, |z|)$ is calculated using stars and bars: $\binom{3+3-1}{3-1} \times \binom{2+3-1}{3-1} \times \binom{1+3-1}{3-1} = 10 \times 6 \times 3 = 180$. For each set of absolute values, there are 4 sign combinations that yield a positive product ($xyz = 360$) and 4 sign combinations that yield a negative product ($xyz = -360$). So, $2\alpha = 180 \times 4 + 180 \times 4 = 720 + 720 = 1440$. Thus, $\alpha = 720$.

3. Calculate $\beta$ (for $S_2$):

$S_2$ contains matrices where $x, y, z \in N$ (positive integers) and $x + y + z$ is even. Since $x, y, z \in N$, $xyz = 360$. For $x + y + z$ to be even, either all $x, y, z$ are even ($E + E + E = E$) or one is even and two are odd ($E + O + O = E$).

• If $x, y, z$ are all even: Let $x = 2x', y = 2y', z = 2z'$. Then $x'y'z' = 45 = 3^2 \times 5^1$. The number of solutions for $(x', y', z')$ is $\binom{2+3-1}{3-1} \times \binom{1+3-1}{3-1} = 6 \times 3 = 18$.

• If one is even and two are odd: Assume $x$ is even, $y, z$ are odd. This means $x$ must contain all factors of 2 ($2^3$), and $y, z$ contain only factors of 3 and 5. The number of ways to distribute $3^2$ and $5^1$ is $6 \times 3 = 18$. There are $\binom{3}{1} = 3$ ways to choose which variable is even. So, $3 \times 18 = 54$ solutions.

Total $\beta = 18 + 54 = 72$.

4. Calculate $\gamma$ (for $S_3$):

$S_3$ contains matrices where $x, y, z \in N$ and $trace(A^3) - 3det(A) = 0$. This means $x^3 + y^3 + z^3 - 3xyz = 0$. This identity is equivalent to $(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 0$, which simplifies to $(x + y + z) \frac{1}{2}((x - y)^2 + (y - z)^2 + (z - x)^2) = 0$. Since $x, y, z \in N$, $x + y + z \ne 0$. Thus, $(x - y)^2 + (y - z)^2 + (z - x)^2 = 0$, which implies $x = y = z$. From $xyz = 360$, we get $x^3 = 360$. Since 360 is not a perfect cube, there are no integer solutions for $x$. Therefore, $\gamma = 0$.

5. Check inequalities:

Substitute $\alpha = 720, \beta = 72, \gamma = 0$ into the given options.

(1) $\frac{720}{72 + 0} = 10$, so $10 < 10$ is false.

(2) $\frac{720}{72 - 0} = 10$, so $10 > 15$ is false.

(3) $\frac{720}{72 + 0} = 10$, so $10 > 9$ is true.

(4) $\frac{720}{72 - 0} = 10$, so $10 < 14$ is true.

The correct options are (3) and (4).