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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: Q8E (page 308)

The equation $d e t (- A) = d e t A$ holds for all $6 脳 6$ matrices.

Short Answer

Expert verified

Therefore, the given condition is true.

Step by step solution

01

Matrix Definition.

Matrix is a setof numbers arranged in rows and columns so as to form a rectangular array.

The numbers are called the elements, or entries, of the matrix.

If there are m rows and n columns, the matrix is said to be an 鈥�m by n鈥� matrix, written 鈥� $m 脳 n$ .鈥�

02

To check whether the given condition is true or false.

If A is a $6 脳 6$ matrix,

Then

$\det (- A) = (- 1)^{6} d e t A$

$d e t (- A) = d e t A$ .

Therefore, the given condition is satisfied.

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Most popular questions from this chapter

Vandermonde determinants (introduced by Alexandre-Th茅ophile Vandermonde). Consider distinct real numbers $a_{0}, a_{1}, . . . . ., a_{n .}$ . We define $(n + 1) 脳 (n + 1)$ the matrix

$A = [\begin{matrix} 1 & 1 & . . . . 1 \\ a_{0} & a_{1} & . . . . a_{n} \\ a_{0}^{2} & a_{1}^{2} & . . . . a_{1}^{2} \\ a_{0}^{n} & a_{1}^{n} & . . . . a_{n}^{n} \end{matrix}]$

Vandermonde showed that

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

the product of all differences $(a_{i} - a_{j})$ , where exceeds j.
a. Verify this formula in the case of $n = 1$ .
b. Suppose the Vandermonde formula holds for $n = 1$ . You are asked to demonstrate it for n. Consider the function

$f (t) = \det [\begin{matrix} 1 & 1 & . . . & 1 & 1 \\ a_{0} & a_{1} & . . . & a_{n - 1} & t \\ a_{0}^{2} & a_{1}^{2} & . . . & a_{n - 1} & t^{2} \\ 鈰� & 鈰� & . . . & 鈰� & 鈰� \\ a_{0}^{n} & a_{1}^{n} & . . . & a_{n - 1}^{n} & t^{n} \end{matrix}]$

Explain why f(t) is a polynomial of $n^{t h}$ degree. Find the coefficient k of $t^{n}$ using Vandermonde's formula for $a_{0}, . . ., a_{n - 1}$ . Explain why

role="math" localid="1659522435181" $f (a_{0}) = f (a_{1}) = . . . = f (a_{n - 1}) = 0$

Conclude that

$f (t) = k (t - a_{0}) (t - a_{1}) . . . (t - a_{n - 1})$

for the scalar k you found above. Substitute $t = a_{n}$ to demonstrate Vandermonde's formula.

Consider a $4 脳 4$ matrix A with rows ${\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}, {\overset{鈫�}{v}}_{4}$ . If, find the determinants in Exercises 11 through16.

12. localid="1659509477853" $d e t [\begin{matrix} {\overset{鈫�}{v}}_{1} \\ {\overset{鈫�}{v}}_{2} \\ {\overset{鈫�}{v}}_{3} \\ {\overset{鈫�}{v}}_{4} \end{matrix}]$

Find all $2 脳 2$ matrices Asuch that $adj (A) = A^{T}$ .

Show that $A (adjA) = 0 = (adjA) A$ for all noninvertible $n 脳 n$ matrices A. See Exercise 42.

There exist invertible $2 脳 2$ matrices A andBsuch that $d e t (A + B) = d e t A + d e t B$ .

See all solutions

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