Question

Let \( A = \{1, 4, 7\} \), \( B = \{2, 3, 8\} \). Let \( R \) be a relation defined as \[ \{((a_1, b_1), (a_2, b_2)) \in (A \times B) \times (A \times B) : (a_2 + b_1) \text{ divides } (a_1 + b_2)\} \], then find the number of such relations.

Hint:

To count relations efficiently, group the sums \( a+b \) from the sets. Possible sums from \( A \) and \( B \) are \( \{3, 4, 9, 6, 7, 12, 9, 10, 15\} \).

Answer 1

Correct Answer: 31

Step 1: Understanding the Concept:
A matrix \( M \) multiplied by the standard basis vectors gives its columns. Given the outputs for specific input vectors, we can find the individual columns of \( M \). Once the columns are identified, the matrix equation \( M \vec{X} = \vec{B} \) becomes a system of linear equations in terms of the variables \( x, y, z \).
Step 2: Key Formula or Approach:
Let \( \vec{c_1}, \vec{c_2}, \vec{c_3} \) be the columns of matrix \( M \).
From \( M \begin{pmatrix} 1
0
0 \end{pmatrix} = \begin{pmatrix} 1
2
3 \end{pmatrix} \), we have \( \vec{c_1} = \begin{pmatrix} 1
2
3 \end{pmatrix} \).
From \( M \begin{pmatrix} 0
0
1 \end{pmatrix} = \begin{pmatrix} -1
1
2 \end{pmatrix} \), we have \( \vec{c_3} = \begin{pmatrix} -1
1
2 \end{pmatrix} \).
From \( M \begin{pmatrix} 1
1
0 \end{pmatrix} = \begin{pmatrix} 1
3
5 \end{pmatrix} \), we have \( \vec{c_1} + \vec{c_2} = \begin{pmatrix} 1
3
5 \end{pmatrix} \).
Subtracting \( \vec{c_1} \): \[ \vec{c_2} = \begin{pmatrix} 1
3
5 \end{pmatrix} - \begin{pmatrix} 1
2
3 \end{pmatrix} = \begin{pmatrix} 0
1
2 \end{pmatrix} \]
Step 3: Detailed Explanation:
Now, the equation \( M \begin{pmatrix} x
y
z \end{pmatrix} = \begin{pmatrix} 2
4
7 \end{pmatrix} \) can be written as: \[ x \begin{pmatrix} 1
2
3 \end{pmatrix} + y \begin{pmatrix} 0
1
2 \end{pmatrix} + z \begin{pmatrix} -1
1
2 \end{pmatrix} = \begin{pmatrix} 2
4
7 \end{pmatrix} \]
This yields the following system of linear equations:
1) \( x - z = 2 \)
2) \( 2x + y + z = 4 \)
3) \( 3x + 2y + 2z = 7 \)
Multiply equation (2) by 2: \[ 4x + 2y + 2z = 8 \quad \text{(4)} \]
Subtract equation (3) from equation (4): \[ (4x + 2y + 2z) - (3x + 2y + 2z) = 8 - 7 \] \[ \Rightarrow x = 1 \]
Substitute \( x = 1 \) into equation (1): \[ 1 - z = 2 \Rightarrow z = -1 \]
Substitute \( x = 1 \) and \( z = -1 \) into equation (2): \[ 2(1) + y + (-1) = 4 \Rightarrow y + 1 = 4 \Rightarrow y = 3 \]
Step 4: Final Answer:
The values are \( x = 1 \), \( y = 3 \), and \( z = -1 \).
\[ x + y + z = 1 + 3 + (-1) = 3 \]
Thus, the sum is 3.