Problem 15
Question
Let \(\left\\{\Delta_{1}, \Delta_{2}, \Delta_{3}, \ldots, \Delta_{k}\right\\}\) be the set of third order determinants that can be made with the distinct nonzero real numbers \(a_{1}, a_{2}, a_{3}, \ldots, a_{9}\). Then (A) \(k=9 !\) (B) \(\sum_{i=1}^{k} \Delta_{i}=0\) (C) at least one \(\Delta_{i}=0\) (D) None of these
Step-by-Step Solution
Verified Answer
The correct answer is B: \(\sum_{i=1}^{k} \Delta_{i}=0\).
1Step 1: Understand Third Order Determinants
A third order determinant is a 3x3 matrix composed of nine elements, where the value of the determinant is calculated using specific rules for expanding along rows or columns. Here, each determinant will be made from distinct nonzero real numbers given by the set \( \{a_1, a_2, a_3, \ldots, a_9\} \).
2Step 2: Calculate the Number of Third Order Determinants
To find the set of all possible third order determinants, we choose 9 distinct numbers, and we can form a matrix by choosing any 3 numbers for the first row, any 3 numbers for the second row (out of the remaining 6), and the remaining 3 for the third row. Therefore, the number of different determinants is the number of permutations of 9 distinct elements, i.e., \(9!\).
3Step 3: Evaluate the Sum of All Determinants
Each row arrangement provides another set of permutations with 3! possibilities per configuration. As signs change according to the permutation when calculating determinants, for a complete set of permutations, the positive and negative contributions cancel out. Therefore, the sum of all determinants, \(\sum_{i=1}^k \Delta_i\), is 0.
4Step 4: Check if Any Determinant Equals Zero
If all nine numbers are nonzero and arranged randomly into different permutations, there will not necessarily be a repeated element, leading to at least one zero determinant. However, due to the uniqueness and non-repetition of numbers, no elemental repetition guarantees zero unless numbers form collinear vectors (which isn’t generally true with random nonzero numbers).
5Step 5: Conclusion on the Satisfaction of Conditions
The calculation and evaluation show that \(B\) is valid since the determinants form a closed loop of permutations with all values cancelling out. \(C\) isn't always true since non-occurrence of repeated elements in permutations doesn't guarantee a zero determinant.
Key Concepts
Third Order DeterminantsMatrix PermutationsDeterminant Properties
Third Order Determinants
Third order determinants refer to the determinants of a 3x3 matrix. These determinants are calculated with matrices composed of nine elements arranged in three rows and three columns. Each element within the matrix is distinct, especially when discussing problems involving permutations. To calculate the determinant of a 3x3 matrix, you select an element from the first row, then multiply it by the determinant of the 2x2 matrix formed by deleting the row and column of that element. You continue this process for each element in the first row, applying an alternating sign for each calculation.Mathematically, if the matrix is:\[\begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \]The determinant, \(\Delta\), is calculated as:\(\Delta = a(ei - fh) - b(di - fg) + c(dh - eg)\)Understanding how each element and calculation fits into this pattern is key to solving problems involving third order determinants.
Matrix Permutations
Matrix permutations involve arranging elements in different orders within the matrix, affecting the calculated determinant. For a 3x3 matrix where each row is made up of numbers taken from a larger set, you can create multiple permutations by changing the position of these numbers. Given nine distinct numbers, as in our problem, permutations are about rearranging these numbers into potential combinations.To determine the number of different third order determinants, consider this: the first row can have any three of the nine elements. The second row can take any three of the remaining six, and the final three automatically fill the third row. As a result, you end up with a number of permutations equal to the factorial of nine, \(9!\), with each arrangement representing a different third order determinant.
Determinant Properties
Determinants have unique properties that make them intriguing in mathematical computations, particularly relevant in problems involving permutations.
- Determinants that result from complete sets of permutations can cancel each other out. This is due to their alternating sign characteristic when calculated through permutations. This explains why the sum of all possible third order determinants, in our example, equals zero.
- Another property is that a determinant may be zero if the matrix rows or columns are linearly dependent. This means that if any row or column can be represented as a combination of others, the determinant will be zero. However, with distinct non-zero numbers arranged randomly, assuring a zero determinant is not straightforward.
- A determinant's value does not change if rows or columns are swapped an even number of times, but if swapped an odd number of times, the determinant changes sign.
Other exercises in this chapter
Problem 13
If \(a_{0}, a_{1} a_{2}, a_{3}, a_{4}\) are in A.P with the common difference \(d\), the value of \(\left|\begin{array}{lll}a_{1} a_{2} & a_{1} & a_{0} \\\ a_{2
View solution Problem 14
If \(\alpha, \beta, \gamma\) are different from and are the roots of \(a x^{3}+\) \(b x^{2}+c x+d=0\) and \((\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)=\frac{25
View solution Problem 16
If \(f(x)=\left|\begin{array}{ccc}(1+x)^{a} & (1+2 x)^{b} & 1 \\ 1 & (1+x)^{a} & (1+2 x)^{b} \\ (1+2 x)^{b} & 1 & (1+x)^{a}\end{array}\right|, a, b\) being posi
View solution Problem 17
If \(P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\e
View solution