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Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 6: Q32E (page 290)

Use Exercise 31 to find
$d e t [\begin{matrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 4 & 9 & 16 & 25 \\ 1 & 8 & 27 & 64 & 125 \\ 1 & 16 & 81 & 256 & 625 \end{matrix}]$
Do not use technology.

Short Answer

Expert verified

Therefore, the determinant of the given matrix is given by,

$d e t A = 288$ .

Step by step solution

01

By using Vandermonde’s Formula.

For $n = 1$ , we have a $2 脳 2$ matrix

$A = [\begin{matrix} 1 & 1 \\ a_{0} & a_{1} \end{matrix}]$

Using Vandermonde's formula, we have

$\underset{}{\underset{i > j}{鈭�} (a_{i} - a_{j}) = a_{1} - a_{0} = d e t A}$ .

02

To find the determinant.

Let $a_{0} = 1, a_{1} = 2, a_{2} = 3, a_{3} = 4, a_{4} = 5$ .

Now we can use Vandermonde's formula to find this determinant.

We compute,

$\begin{array}{l} d e t A = \underset{\frac{i, j - 1}{i > j}}{鈭�} (a_{i} - a_{j}) \\ d e t A = (2 - 1) (3 - 1) (4 - 1) (5 - 1) (3 - 2) (4 - 2) (5 - 2) (4 - 3) (5 - 3) (5 - 4) \\ d e t A = 288 \end{array}$

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Most popular questions from this chapter

Use Cramer's rule to solve the systems in Exercises 22 through 24.

22. $| \begin{matrix} 3 x + & 7 y = & 1 \\ 4 x + & 11 y = & 3 \end{matrix} |$

$d e t (A + B) = d e t A + d e t B$ for all $5 脳 5$ matricesA andB.

Demonstrate Theorem 6.3.6 for linearly dependent vector ${\overset{鈬赌}{v}}_{1}, . . . ., {\overset{鈬赌}{v}}_{m}$ .

$d e t (4 A) = 4 d e t A$ for all $4 脳 4$ matricesA.

Find the determinants of the linear transformations in Exercises 17 through 28.

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