Question

If \(A=\left[\begin{array}{lll}a & x & y \\ x & b & z \\ y & z & c\end{array}\right], a, b, c, x, y\) \(z \in\\{1,2,3,4,5,6\\}, a, b, c, x, y, z\) are distinct then number of matrices \(\mathrm{A}\) with trace equal to 10 are Note: The trace of matrix is the sum of its diagonal entries.] (1) \(3 .(3 !)^{2}\) (2) \(2 .(3 !)^{2}\) (3) \((3 !)^{2}\) (4) \((3 !)^{3}\)

Short answer

108

Step-by-step solution

Understand the problem

The goal is to find the number of distinct matrices, where all elements in the matrix are allowed to take values from the set \(\{1,2,3,4,5,6\}\) and the diagonal elements of the matrix \(A\) sum up to 10.

Identify possible values for diagonal elements

The diagonal elements are \(a, b,\) and \(c\). They must be distinct and selected from \(\{1,2,3,4,5,6\}\), and must satisfy the equation: \(a + b + c = 10\).

List distinct combinations summing to 10

List the sets of three distinct numbers from \(\{1,2,3,4,5,6\}\) that sum to 10: \[\{1, 3, 6\}, \{1, 4, 5\}, \{2, 3, 5\}\].

Permute the diagonal combinations

Calculate the number of permutations for each combination of \(a, b, c\). Since all three elements in each combination must be distinct, each has \(3! = 6\) permutations. We have three such combinations.

Assign values to non-diagonal elements

The non-diagonal elements \(x, y, z\) can be any of the remaining three numbers from the set. Since these numbers must be distinct and placed in three different spots in the matrix, there are \(3! = 6\) permutations for these placements.

Calculate the total number of matrices

Multiply the number of diagonal permutations (\(3!\)) by the number of non-diagonal permutations (\(3!\)) and then by the number of valid diagonal combinations (3). Therefore, the total number is \[(3!\times 3!)\times 3 = 6 \times 6 \times 3 = 108\].

Key concepts

matrix diagonal elements

A matrix's diagonal elements are key to understanding its structure and properties. These elements are the entries from the top left to the bottom right. In our example, the matrix is structured as follows: Suppose you have a matrix \begin{bmatrix} a & x & y \ x & b & z \ y & z & c \end{bmatrix}The entries 'a', 'b', and 'c' form the diagonal elements.
These elements are crucial as they directly affect the trace of the matrix.
The trace of a matrix is the sum of all its diagonal elements.
In this context for our matrix, the trace is calculated as: \(a + b + c\).
In this problem, we need these diagonal elements to sum up to 10 and these elements must be distinct.

distinct combinations

When dealing with problems involving the selection of items, especially where they must be distinct, it's important to think about combinations.
Understand Distinct Combinations:
Here, distinct means each element is different. Given the set \{1, 2, 3, 4, 5, 6\}, we need to find all possible sets of three unique numbers that add up to 10.
The sets that meet these criteria are:

• \(\{1, 3, 6\}\)
• \(\{1, 4, 5\}\)
• \(\{2, 3, 5\}\)

Each set is distinct as no set contains any repeated numbers, and all numbers belong to the initial set. These sets are vital as they form the possible diagonal combinations for our matrix.

permutations

Once we have the distinct combinations for the diagonal elements, we need to consider permutations. Permutations are different ways of arranging items.
For instance, if we take the combination \(\{1, 3, 6\}\), we can arrange these three elements in \(3! = 6\) different ways.
This gives us each of the following permutations:

• \((1, 3, 6)\)
• \((1, 6, 3)\)
• \((3, 1, 6)\)
• \((3, 6, 1)\)
• \((6, 1, 3)\)
• \((6, 3, 1)\)

Each combination of diagonal elements, hence, offers 6 permutations.
In our case, there are 3 combinations, each providing 6 permutations, multiplying to give a total of \(3 \times 6 = 18\) possible permutations for the diagonal alone.
For the non-diagonal elements, 'x', 'y', and 'z', they can be arranged in \(3! = 6\) ways as well.
Combining these permutations with those of the diagonal elements, the total number of matrices is calculated by:\(3 \times 6 \times 6\), which results in 108 distinct matrices.