Question

Consider two matrices $A = \begin{bmatrix} x_1 & x_2 & x_3 \\ 0 & x_2 & x_1 \\ 0 & 0 & x_3 \end{bmatrix}$ and $B = \begin{bmatrix} y_1 & 0 & 0 \\ y_3 & y_2 & 0 \\ y_2 & y_1 & y_3 \end{bmatrix}$, where each of $x_i, y_j \in \{-1, 0, 1\} \forall i, j = 1, 2, 3$, and if N is the number of possible ordered pair of matrices A and B for which det A = det B. Then the value of $\frac{N}{131}$, is

Answer 1

3

Answer 2

The given matrices are $A = \begin{bmatrix} x_1 & x_2 & x_3 \\ 0 & x_2 & x_1 \\ 0 & 0 & x_3 \end{bmatrix}$ and $B = \begin{bmatrix} y_1 & 0 & 0 \\ y_3 & y_2 & 0 \\ y_2 & y_1 & y_3 \end{bmatrix}$, where each of $x_i, y_j \in \{-1, 0, 1\}$ for $i, j = 1, 2, 3$.

The determinant of matrix A is given by the product of its diagonal elements since it is an upper triangular matrix: det A = $x_1 \cdot x_2 \cdot x_3$.

The determinant of matrix B is given by the product of its diagonal elements since it is a lower triangular matrix: det B = $y_1 \cdot y_2 \cdot y_3$.

We are looking for the number of possible ordered pairs of matrices (A, B) for which det A = det B. This condition is $x_1 x_2 x_3 = y_1 y_2 y_3$.

The elements $x_i$ and $y_j$ can each take values from the set $S = \{-1, 0, 1\}$. There are $3^3 = 27$ possible combinations for the triplet $(x_1, x_2, x_3)$ and $3^3 = 27$ possible combinations for the triplet $(y_1, y_2, y_3)$. The total number of possible ordered pairs of matrices (A, B) is $27 \times 27 = 729$.

Let $P_x = x_1 x_2 x_3$ and $P_y = y_1 y_2 y_3$. Since $x_i, y_j \in \{-1, 0, 1\}$, the possible values for $P_x$ and $P_y$ are $\{-1, 0, 1\}$.

We need to count the number of triplets $(x_1, x_2, x_3)$ with elements from $\{-1, 0, 1\}$ that result in each possible product value $k \in \{-1, 0, 1\}$. Let $C(k)$ be this number.

Case 1: $x_1 x_2 x_3 = 1$. For the product to be 1, none of the $x_i$ can be 0. So $x_i \in \{-1, 1\}$. The product is 1 if there is an even number of -1's.

• Zero -1's: (1, 1, 1). There is $\binom{3}{0} = 1$ such triplet.
• Two -1's: (-1, -1, 1), (-1, 1, -1), (1, -1, -1). There are $\binom{3}{2} = 3$ such triplets. So, $C(1) = 1 + 3 = 4$.

Case 2: $x_1 x_2 x_3 = -1$. For the product to be -1, none of the $x_i$ can be 0. So $x_i \in \{-1, 1\}$. The product is -1 if there is an odd number of -1's.

• One -1: (-1, 1, 1), (1, -1, 1), (1, 1, -1). There are $\binom{3}{1} = 3$ such triplets.
• Three -1's: (-1, -1, -1). There is $\binom{3}{3} = 1$ such triplet. So, $C(-1) = 3 + 1 = 4$.

Case 3: $x_1 x_2 x_3 = 0$. The product is 0 if at least one $x_i$ is 0. The total number of triplets is $3^3 = 27$. The number of triplets with no 0 is $2^3 = 8$ (each element is -1 or 1). These 8 triplets give products 1 or -1. So, the number of triplets with at least one 0 is $27 - 8 = 19$. $C(0) = 19$.

Check: $C(1) + C(-1) + C(0) = 4 + 4 + 19 = 27$. This is correct.

We need to find the number of pairs of triplets $((x_1, x_2, x_3), (y_1, y_2, y_3))$ such that $x_1 x_2 x_3 = y_1 y_2 y_3$. This equality holds if the common product is 1, -1, or 0. Let $N$ be the number of such pairs. $N = (\text{Number of pairs where } P_x=P_y=1) + (\text{Number of pairs where } P_x=P_y=-1) + (\text{Number of pairs where } P_x=P_y=0)$.

Number of pairs where $P_x = P_y = 1$: The number of choices for $(x_1, x_2, x_3)$ is $C(1)=4$. The number of choices for $(y_1, y_2, y_3)$ is $C(1)=4$. The number of pairs is $C(1) \times C(1) = 4 \times 4 = 16$.

Number of pairs where $P_x = P_y = -1$: The number of choices for $(x_1, x_2, x_3)$ is $C(-1)=4$. The number of choices for $(y_1, y_2, y_3)$ is $C(-1)=4$. The number of pairs is $C(-1) \times C(-1) = 4 \times 4 = 16$.

Number of pairs where $P_x = P_y = 0$: The number of choices for $(x_1, x_2, x_3)$ is $C(0)=19$. The number of choices for $(y_1, y_2, y_3)$ is $C(0)=19$. The number of pairs is $C(0) \times C(0) = 19 \times 19 = 361$.

The total number of ordered pairs of matrices (A, B) for which det A = det B is the sum: $N = 16 + 16 + 361 = 32 + 361 = 393$.

The question asks for the value of $\frac{N}{131}$. $\frac{N}{131} = \frac{393}{131}$. Dividing 393 by 131: $393 = 3 \times 131$. So, $\frac{393}{131} = 3$.

The final answer is 3.

Explanation of the solution:

1. Calculate the determinants of matrices A and B. Since A is upper triangular and B is lower triangular, det A = $x_1 x_2 x_3$ and det B = $y_1 y_2 y_3$.
2. The condition det A = det B becomes $x_1 x_2 x_3 = y_1 y_2 y_3$.
3. The elements $x_i, y_j$ belong to $\{-1, 0, 1\}$. The possible values for the product of three elements from this set are $\{-1, 0, 1\}$.
4. Count the number of triplets $(x_1, x_2, x_3)$ from $\{-1, 0, 1\}^3$ whose product is 1, -1, and 0. Let these counts be $C(1), C(-1), C(0)$ respectively.
   • $C(1) = 4$ (triplets are (1,1,1) and permutations of (-1,-1,1))
   • $C(-1) = 4$ (triplets are (-1,-1,-1) and permutations of (-1,1,1))
   • $C(0) = 19$ (total triplets 27, non-zero products 8, so $27-8=19$)
5. The number of pairs of triplets $((x_1, x_2, x_3), (y_1, y_2, y_3))$ such that their products are equal is the sum of the number of pairs where both products are 1, both are -1, and both are 0.
   • Number of pairs with product 1 = $C(1) \times C(1) = 4 \times 4 = 16$.
   • Number of pairs with product -1 = $C(-1) \times C(-1) = 4 \times 4 = 16$.
   • Number of pairs with product 0 = $C(0) \times C(0) = 19 \times 19 = 361$.
6. The total number of ordered pairs of matrices (A, B) is $N = 16 + 16 + 361 = 393$.
7. Calculate the required value $\frac{N}{131} = \frac{393}{131} = 3$.

The final answer is 3.