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Chapter 11: Problem 19

Repeat Prob. 11.18 , but for the case of a six-dimensional Vandermonde matrix (see Prob. 10.14 ) where \(x_{1}=4, x_{2}=2, x_{3}=\) \(7, x_{4}=10, x_{5}=3,\) and \(x_{6}=5\)

Short Answer

Expert verified
The determinant of the six-dimensional Vandermonde matrix with given values \(x_1 = 4, x_2 = 2, x_3 = 7, x_4 = 10, x_5 = 3,\) and \(x_6 = 5\) can be calculated using the formula: \(\text{det}(V) = \prod_{1 \le j < i \le 6} (x_i - x_j)\). After substituting the given values and computing the differences and product, the determinant is found to be \(72576\).

Step by step solution

01

Write down the Vandermonde matrix with given values

First, we need to write down the six-dimensional Vandermonde matrix \(V\) using the given \(x\) values: \[ V = \begin{pmatrix} 1 & x_1 & x_1^2 & x_1^3 & x_1^4 & x_1^5\\ 1 & x_2 & x_2^2 & x_2^3 & x_2^4 & x_2^5\\ 1 & x_3 & x_3^2 & x_3^3 & x_3^4 & x_3^5\\ 1 & x_4 & x_4^2 & x_4^3 & x_4^4 & x_4^5\\ 1 & x_5 & x_5^2 & x_5^3 & x_5^4 & x_5^5\\ 1 & x_6 & x_6^2 & x_6^3 & x_6^4 & x_6^5 \end{pmatrix} \] Now substitute the given \(x\) values: \[ V = \begin{pmatrix} 1 & 4 & 16 & 64 & 256 & 1024\\ 1 & 2 & 4 & 8 & 16 & 32\\ 1 & 7 & 49 & 343 & 2401 & 16807\\ 1 & 10 & 100 & 1000 & 10000 & 100000\\ 1 & 3 & 9 & 27 & 81 & 243\\ 1 & 5 & 25 & 125 & 625 & 3125 \end{pmatrix} \]
02

Compute the determinant using the formula

Now we need to calculate the determinant of the Vandermonde matrix \(V\) using the formula: \[ \text{det}(V) = \prod_{1 \le j < i \le 6} (x_i - x_j) \] We can expand this product: \[ \text{det}(V) = (x_2-x_1)(x_3-x_1)(x_3-x_2)(x_4-x_1)(x_4-x_2)(x_4-x_3)(x_5-x_1)(x_5-x_2)(x_5-x_3)(x_5-x_4)(x_6-x_1)(x_6-x_2)(x_6-x_3)(x_6-x_4)(x_6-x_5) \] Now substitute the given \(x\) values: \[ \text{det}(V) = (2-4)(7-4)(7-2)(10-4)(10-2)(10-7)(3-4)(3-2)(3-7)(3-10)(5-4)(5-2)(5-7)(5-10)(5-3) \] Compute the differences inside the brackets: \[ \text{det}(V) = (-2)(3)(5)(6)(8)(3)(-1)(1)(-4)(-7)(1)(3)(-2)(-5)(2) \]
03

Calculate the final result

Calculate the product of all these numbers to find the determinant of the Vandermonde matrix: \[ \text{det}(V) = -2 \cdot 3 \cdot 5 \cdot 6 \cdot 8 \cdot 3 \cdot (-1) \cdot 1 \cdot (-4) \cdot (-7) \cdot 1 \cdot 3 \cdot (-2) \cdot (-5) \cdot 2 = 72576 \] Hence, the determinant of the Vandermonde matrix using given \(x\) values is \(72576\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant Calculation
The determinant of a matrix, in its most abstract sense, is a scalar value that is a unique property of the matrix, offering insight into properties such as the volume factor by which it scales geometrical figures and the existence of an inverse. Calculating the determinant of a Vandermonde matrix is a topic that combines the beauty of algebra with practical applications.

For a Vandermonde matrix, the determinant can be calculated using a special property of the matrix that relates to the differences between its elements. The general formula for the determinant of a Vandermonde matrix is expressed as a product of differences between the matrix's elements, which in our case involves repeating the difference operation on the provided values of x.

Understanding the Vandermonde Determinant

Creating a Vandermonde matrix involves powers of the given variables laid out across its rows. The simplicity of its determinant calculation, through the product of differences, belies the intricate interrelationship between its elements. By systematically calculating the differences between each term, as shown in the step by step solution, we leverage the unique structure of the Vandermonde matrix to efficiently arrive at the desired outcome, which in this instance is 72576.
Multidimensional Matrices
A multidimensional matrix is a mathematical object that can represent data or equations in more than two dimensions. The six-dimensional Vandermonde matrix within the exercise is a perfect example of such a matrix. Each dimension in the matrix corresponds to a new degree of freedom, showcasing how variables can interact across multiple dimensions. In higher-dimensional matrices, the concepts of rows and columns extend and the indexing becomes more complex, observing the order progression in powers for each successively higher dimension.

The Power of Higher Dimensions

Dealing with a matrix like the Vandermonde involves understanding that each additional dimension reflects an increase in the complexity and the potential of the mathematical relationships it can describe. In polynomial interpolation, for instance, a six-dimensional matrix could correspond to interpolating a polynomial of the fifth degree. Multiple dimensions in matrices allow for the portrayal of rich interactions in various fields such as physics, economics, and engineering.
Polynomial Interpolation
Polynomial interpolation is the process of determining a polynomial that passes through a set of given points. The Vandermonde matrix plays a pivotal role in polynomial interpolation, as it forms the coefficient matrix when solving the linear system to find the interpolating polynomial.

Linking Vandermonde and Interpolation

The particular structure of Vandermonde matrices, where each row is based on increasing powers of a given value, fits perfectly with polynomial terms of certain degrees. Interpolation in a six-dimensional space means constructing a polynomial of the fifth degree, and the calculated determinant gives us certainty about the existence and uniqueness of this polynomial. If the determinant were zero, this would imply linear dependence between the columns, signaling that no such unique polynomial exists. Fortunately, in the exercise provided, the nonzero determinant of the Vandermonde matrix guarantees a unique solution for the interpolation of the given points.

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Most popular questions from this chapter

The following tridiagonal system must be solved as part of a larger algorithm (Crank-Nicolson) for solving partial differential equations: $$\left[\begin{array}{cccc} 2.01475 & -0.020875 & & \\ -0.020875 & 2.01475 & -0.020875 & \\ & -0.020875 & 2.01475 & -0.020875 \\ & & -0.020875 & 2.01475 \end{array}\right]$$ $$\times\left\\{\begin{array}{l} T_{1} \\ T_{2} \\ T_{3} \\ T_{4} \end{array}\right\\}=\left\\{\begin{array}{c} 4.175 \\ 0 \\ 0 \\ 2.0875 \end{array}\right\\}$$ Use the Thomas algorithm to obtain a solution.

Perform the same calculations as in Example 11.1 , but for the tridiagonal system, $$\left[\begin{array}{ccc} 0.8 & -0.4 & \\ -0.4 & 0.8 & -0.4 \\ & -0.4 & 0.8 \end{array}\right]\left\\{\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right\\}=\left\\{\begin{array}{c} 41 \\ 25 \\ 105 \end{array}\right\\}$$

Given the pair of nonlinear simultaneous equations: $$\begin{aligned} &f(x, y)=4-y-2 x^{2}\\\ &g(x, y)=8-y^{2}-4 x \end{aligned}$$ (a) Use the Excel Solver to determine the two pairs of values of \(x\) and \(y\) that satisfy these equations. (b) Using a range of initial guesses \((x=-6 \text { to } 6 \text { and } y=-6 \text { to } 6\) ). determine which initial guesses yield each of the solutions.

Use the Gauss-Seidel method (a) without relaxation and (b) with relaxation \((\lambda=1.2)\) to solve the following system to a tolerance of \(\varepsilon_{s}=5 \%\). If necessary, rearrange the equations to achieve convergence. $$\begin{aligned} 2 x_{1}-6 x_{2}-x_{3} &=-38 \\ -3 x_{1}-x_{2}+7 x_{3} &=-34 \\ -8 x_{1}+x_{2}-2 x_{3} &=-20 \end{aligned}$$

Use the Gauss-Seidel method to solve the following system until the percent relative error falls below $\epsilon_{\mathrm{s}}=5 \%$$$\begin{aligned} 10 x_{1}+2 x_{2}-x_{3} &=27 \\ -3 x_{1}-6 x_{2}+2 x_{3} &=-61.5 \\ x_{1}+x_{2}+5 x_{3} &=-21.5 \end{aligned}$$
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