Question

Let $A = \begin{bmatrix} x & 3 & 5 \\ 2 & x & 7 \\ x & 3 & x \end{bmatrix}$ be a matrix where x is an integer. If det(adj(adjA)) = $6^4$ then number of possible values of x is

• A. 0
• B. 1
• C. 2
• D. more than 3

Answer 1

Answer: (C)

2

Answer 2

The given matrix is $A = \begin{bmatrix} x & 3 & 5 \\ 2 & x & 7 \\ x & 3 & x \end{bmatrix}$. The order of the matrix is $n=3$. We are given that $\det(\text{adj}(\text{adj}A)) = 6^4$.

For a square matrix $A$ of order $n$, the determinant of the adjugate of the adjugate of $A$ is given by the property: $\det(\text{adj}(\text{adj}A)) = (\det A)^{(n-1)^2}$

In this case, $n=3$, so the formula becomes: $\det(\text{adj}(\text{adj}A)) = (\det A)^{(3-1)^2} = (\det A)^{2^2} = (\det A)^4$.

We are given $\det(\text{adj}(\text{adj}A)) = 6^4$. So, we have $(\det A)^4 = 6^4$. This equation implies $\det A = \pm 6$.

Now, we need to calculate the determinant of the matrix $A$ in terms of $x$. $\det A = \det \begin{bmatrix} x & 3 & 5 \\ 2 & x & 7 \\ x & 3 & x \end{bmatrix}$ Expanding along the first row: $\det A = x \begin{vmatrix} x & 7 \\ 3 & x \end{vmatrix} - 3 \begin{vmatrix} 2 & 7 \\ x & x \end{vmatrix} + 5 \begin{vmatrix} 2 & x \\ x & 3 \end{vmatrix}$ $\det A = x(x \cdot x - 7 \cdot 3) - 3(2 \cdot x - 7 \cdot x) + 5(2 \cdot 3 - x \cdot x)$ $\det A = x(x^2 - 21) - 3(2x - 7x) + 5(6 - x^2)$ $\det A = x^3 - 21x - 3(-5x) + 30 - 5x^2$ $\det A = x^3 - 21x + 15x + 30 - 5x^2$ $\det A = x^3 - 5x^2 - 6x + 30$

We have two possible cases for $\det A$:

Case 1: $\det A = 6$ $x^3 - 5x^2 - 6x + 30 = 6$ $x^3 - 5x^2 - 6x + 24 = 0$ We are looking for integer solutions for $x$. We can test integer divisors of 24 as potential roots. Let $P(x) = x^3 - 5x^2 - 6x + 24$. $P(2) = 2^3 - 5(2^2) - 6(2) + 24 = 8 - 5(4) - 12 + 24 = 8 - 20 - 12 + 24 = 0$. So, $x=2$ is an integer root. We can factor the polynomial using synthetic division or polynomial division. Dividing $x^3 - 5x^2 - 6x + 24$ by $(x-2)$:

2 | 1  -5  -6   24
  |    2  -6  -24
  ----------------
    1  -3  -12   0

The quotient is $x^2 - 3x - 12$. So, the equation is $(x-2)(x^2 - 3x - 12) = 0$. The roots are $x=2$ or $x^2 - 3x - 12 = 0$. For the quadratic equation $x^2 - 3x - 12 = 0$, the discriminant is $\Delta = (-3)^2 - 4(1)(-12) = 9 + 48 = 57$. Since 57 is not a perfect square, the roots $x = \frac{3 \pm \sqrt{57}}{2}$ are not integers. Thus, the only integer solution from Case 1 is $x=2$.

Case 2: $\det A = -6$ $x^3 - 5x^2 - 6x + 30 = -6$ $x^3 - 5x^2 - 6x + 36 = 0$ We are looking for integer solutions for $x$. We can test integer divisors of 36 as potential roots. Let $Q(x) = x^3 - 5x^2 - 6x + 36$. $Q(3) = 3^3 - 5(3^2) - 6(3) + 36 = 27 - 5(9) - 18 + 36 = 27 - 45 - 18 + 36 = 63 - 63 = 0$. So, $x=3$ is an integer root. We can factor the polynomial using synthetic division or polynomial division. Dividing $x^3 - 5x^2 - 6x + 36$ by $(x-3)$:

3 | 1  -5  -6   36
  |    3  -6  -36
  ----------------
    1  -2  -12   0

The quotient is $x^2 - 2x - 12$. So, the equation is $(x-3)(x^2 - 2x - 12) = 0$. The roots are $x=3$ or $x^2 - 2x - 12 = 0$. For the quadratic equation $x^2 - 2x - 12 = 0$, the discriminant is $\Delta = (-2)^2 - 4(1)(-12) = 4 + 48 = 52$. Since 52 is not a perfect square, the roots $x = \frac{2 \pm \sqrt{52}}{2} = \frac{2 \pm 2\sqrt{13}}{2} = 1 \pm \sqrt{13}$ are not integers. Thus, the only integer solution from Case 2 is $x=3$.

The possible integer values of $x$ for which $\det(\text{adj}(\text{adj}A)) = 6^4$ are $x=2$ and $x=3$. The number of possible values of $x$ is 2.

The final answer is $\boxed{2}$.

Explanation of the solution:

1. Use the property $\det(\text{adj}(\text{adj}A)) = (\det A)^{(n-1)^2}$ for an $n \times n$ matrix $A$.
2. For the given $3 \times 3$ matrix ($n=3$), this simplifies to $\det(\text{adj}(\text{adj}A)) = (\det A)^4$.
3. Given $\det(\text{adj}(\text{adj}A)) = 6^4$, we get $(\det A)^4 = 6^4$, which implies $\det A = \pm 6$.
4. Calculate the determinant of the matrix $A$ in terms of $x$: $\det A = x^3 - 5x^2 - 6x + 30$.
5. Set $\det A = 6$ and solve the cubic equation $x^3 - 5x^2 - 6x + 24 = 0$ for integer values of $x$. The only integer solution is $x=2$.
6. Set $\det A = -6$ and solve the cubic equation $x^3 - 5x^2 - 6x + 36 = 0$ for integer values of $x$. The only integer solution is $x=3$.
7. The possible integer values of $x$ are 2 and 3. The number of such values is 2.

The final answer is $\boxed{2}$.