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Q50E

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Found in: Page 22

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# For an arbitrary positive integer${\mathbit{n}}{\mathbf{\ge }}{\mathbf{3}}$ , find all solutions ${\mathbit{x}}_{\mathbf{1}}\mathbf{,}{\mathbit{x}}_{\mathbf{2}}\mathbf{,}{\mathbit{x}}_{\mathbf{3}}\mathbf{,}\mathbf{.}\mathbf{.....}\mathbf{,}{\mathbit{x}}_{\mathbf{n}}$of the simultaneous equations ${{\mathbit{x}}}_{{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{\right)}{\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{3}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{4}}\mathbf{\right)}{\mathbf{,}}{\mathbf{.}}{\mathbf{....}}{\mathbf{,}}{{\mathbit{x}}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{n}\mathbf{-}\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{n}}\mathbf{\right)}$. Note that we are asked to solve the simultaneous equations ${{\mathbit{x}}}_{{\mathbf{k}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{k}\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{\right)}$, for ${\mathbit{k}}{\mathbf{=}}{\mathbf{2}}{\mathbf{,}}{\mathbf{3}}{\mathbf{,}}{\mathbf{.}}{\mathbf{....}}{\mathbf{,}}{\mathbit{n}}{\mathbf{-}}{\mathbf{1}}$ .

${x}_{n}=3{x}_{n+2}-2{x}_{n+3}\text{\hspace{0.17em}}$ is the expression using which the solutions ${x}_{1},{x}_{2},{x}_{3},......,{x}_{n}$of the simultaneous equations can be obtained.

See the step by step solution

## Step 1: Consider the given equations.

The expression for${x}_{2}$ is,

${x}_{2}=\frac{1}{2}\left({x}_{1}+{x}_{3}\right)$

Substitute the above equation in ${x}_{3}$

$\begin{array}{l}{x}_{3}=\frac{1}{2}\left({x}_{2}+{x}_{4}\right)\\ ⇒{x}_{3}=\frac{1}{2}\left(\left(\frac{1}{2}\left({x}_{1}+{x}_{3}\right)\right)+{x}_{4}\right)\\ ⇒{x}_{3}=\frac{1}{4}{x}_{1}+\frac{1}{2}{x}_{3}+\frac{1}{2}{x}_{4}\\ \therefore {x}_{1}=3{x}_{3}-2{x}_{4}\text{\hspace{0.17em}}\end{array}$

## Step 2: Compute the expressions for remaining variables.

Similarly,

${x}_{2}=3{x}_{4}-2{x}_{5}\text{\hspace{0.17em}}$

Continuing the same, the expression for the ${n}^{th}$ term will be,

${x}_{n}=3{x}_{n+2}-2{x}_{n+3}\text{\hspace{0.17em}}$

## Step 3: Compute the expressions for remaining variables.

Similarly,

${x}_{2}=3{x}_{4}-2{x}_{5}\text{\hspace{0.17em}}$

Continuing the same, the expression for the ${n}^{th}$ term will be,

${x}_{n}=3{x}_{n+2}-2{x}_{n+3}\text{\hspace{0.17em}}$

The solutions ${x}_{1},{x}_{2},{x}_{3},......,{x}_{n}$ of the simultaneous equations can be found using the expression,${x}_{n}=3{x}_{n+2}-2{x}_{n+3}\text{\hspace{0.17em}}$ .