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Q70E

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Found in: Page 39

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# 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 $\stackrel{\mathbf{\to }}{\mathbf{b}}$ in ${{\mathbf{\square }}}^{{\mathbf{n}}}$, solve the linear system ${\mathbit{A}}{\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{=}}\stackrel{\mathbf{⇀}}{\mathbf{b}}$ , expressing the components ${{\mathbit{x}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{n}}}$ of ${\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{x}}$ in terms of the components of $\stackrel{\mathbf{⇀}}{\mathbf{b}}$ . See Exercise 69 for the case n=3 .

The solution of the linear system ${A}{}\stackrel{\to }{\mathrm{x}}{=}\stackrel{⇀}{\mathrm{b}}$ is${x}_{i}=\frac{{b}_{1}+......+{b}_{n}}{n-1}-{b}_{i}, i=1,2,....,n$ .

See the step by step solution

## Step 1: Consider the system.

If A is an n x m matrix with row vectors $\stackrel{⇀}{{\omega }_{1}}{,}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}\stackrel{⇀}{{\omega }_{n}}$ and $\stackrel{⇀}{x}$ is a vector in Rm then, .

${\mathbit{A}}\stackrel{\mathbf{⇀}}{\mathbf{x}}{\mathbf{=}}\left[\begin{array}{c}-\stackrel{⇀}{{\omega }_{1}}-\\ .\\ .\\ .\\ -\stackrel{⇀}{{\omega }_{n}}-\end{array}\right]\stackrel{\mathbf{⇀}}{\mathbf{x}}{\mathbf{=}}\left[\begin{array}{c}-\stackrel{⇀}{{\omega }_{1}}.\stackrel{⇀}{x}-\\ .\\ .\\ .\\ -\stackrel{⇀}{{\omega }_{n}}.\stackrel{⇀}{x}-\end{array}\right]$

Consider the linear system.

$\left[\begin{array}{c}y+z=a\\ y+z=b\\ y+y=c\end{array}\right]$

The matrix form of the system is,

$\left[\begin{array}{ccccc}0& 1& 1& |& 1\\ 1& 0& 1& |& b\\ 1& 1& 0& |& c\end{array}\right]$

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

## Step 2: Compute the system

Consider the linear system, ${x}_{1},.......,{x}_{n}$of $\stackrel{⇀}{x}$$\stackrel{⇀}{x}$in terms of the components of$\stackrel{⇀}{b}$ .

$\left[\begin{array}{c}{x}_{1}+{x}_{2}+.......+{x}_{n}={b}_{1}\\ {x}_{1}+{x}_{2}+.......+{x}_{n}={b}_{1}\\ .\\ ;\\ {x}_{1}+{x}_{2}+.......+{x}_{n}={b}_{n-1}\\ {x}_{1}+{x}_{2}+.......+{x}_{n}={b}_{n}\end{array}\right]$

The solution, ${x}_{i}=\frac{{b}_{1}+......+{b}_{n}}{n-1}-{b}_{i}$ .

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

Hence, ${x}_{i}=\frac{{b}_{1}+......+{b}_{n}}{n-1}-{b}_{i}, i=1,2,....,n$ is the solution of the linear system $A\stackrel{\to }{x}=\stackrel{\to }{b}.$