Q70E Question:聽Let 聽A be the n x n ... [FREE SOLUTION]

91影视

Linear Algebra and its Applications

David C. Lay, Steven R. Lay and Judi J. McDonald

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 483 pages

ISBN: 978-03219822384

Chapter 1: Q70E (page 39)

Question:Let A be the n x n matrix with 0^'s on the main diagonal, and 1's everywhere else. For an arbitrary vector $\overset{鈫�}{b}$ in ${鈻�}^{n}$ , solve the linear system $A \overset{鈫�}{x} = \overset{鈬赌}{b}$ , expressing the components $x_{1}, . . . . . . ., x_{n}$ of $\overset{鈫�}{x}$ in terms of the components of $\overset{鈬赌}{b}$ . See Exercise 69 for the case n=3 .

Short Answer

Expert verified

The solution of the linear system $A \overset{鈫�}{x} = \overset{鈬赌}{b}$ is $x_{i} = \frac{b_{1} + . . . . . . + b_{n}}{n - 1} - b_{i}, 鈥� i = 1, 2, . . . ., n$ .

Step by step solution

Consider the system.

IfA is an n x m matrix with row vectors $\overset{鈬赌}{蝇_{1}}, . . . . . . . . . . \overset{鈬赌}{蝇_{n}}$ and $\overset{鈬赌}{x}$ is a vector in R^m then, .

$A \overset{鈬赌}{x} = [\begin{matrix} - \overset{鈬赌}{蝇_{1}} - \\ . \\ . \\ . \\ - \overset{鈬赌}{蝇_{n}} - \end{matrix}] \overset{鈬赌}{x} = [\begin{matrix} - \overset{鈬赌}{蝇_{1}} . \overset{鈬赌}{x} - \\ . \\ . \\ . \\ - \overset{鈬赌}{蝇_{n}} . \overset{鈬赌}{x} - \end{matrix}]$

Consider the linear system.

$[\begin{matrix} y + z = a \\ y + z = b \\ y + y = c \end{matrix}]$

The matrix form of the system is,

$[\begin{matrix} 0 & 1 & 1 & | & 1 \\ 1 & 0 & 1 & | & b \\ 1 & 1 & 0 & | & c \end{matrix}]$

The solution is, $x = \frac{b + c - a}{2}, y = \frac{a + c - b}{2}, z = \frac{a + b - c}{2}$ .

Compute the system

Consider the linear system, $x_{1}, . . . . . . ., x_{n}$ of $\overset{鈬赌}{x}$ $\overset{鈬赌}{x}$ in terms of the components of $\overset{鈬赌}{b}$ .

$[\begin{matrix} x_{1} + x_{2} + . . . . . . . + x_{n} = b_{1} \\ x_{1} + x_{2} + . . . . . . . + x_{n} = b_{1} \\ . \\ ; \\ x_{1} + x_{2} + . . . . . . . + x_{n} = b_{n - 1} \\ x_{1} + x_{2} + . . . . . . . + x_{n} = b_{n} \end{matrix}]$

The solution, $x_{i} = \frac{b_{1} + . . . . . . + b_{n}}{n - 1} - b_{i}$ .

Where, $i = 1, 2, . . . ., n$

Hence, $x_{i} = \frac{b_{1} + . . . . . . + b_{n}}{n - 1} - b_{i}, 鈥� i = 1, 2, . . . ., n$ is the solution of the linear system $A \overset{鈫�}{x} = \overset{鈫�}{b} .$