• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q49E

Expert-verified Found in: Page 22 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # a. Find all solutions${{\mathbit{x}}}_{{\mathbf{1}}}{\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{2}}}{\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{3}}}{\mathbf{,}}{{\mathbit{x}}}_{{\mathbf{4}}}$ of the system .${\mathbit{x}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{\right)}\mathbf{,}{\mathbit{x}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{4}}\mathbf{\right)}$b. In partrole="math" localid="1659677484607" ${\mathbf{\left(}}{\mathbit{a}}{\mathbf{\right)}}$ , is there a solution with ${{\mathbit{x}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{1}}$and${{\mathbit{x}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{13}}$ ?

a. For arbitrary ${x}_{3}$ and ${x}_{4}$ , ${x}_{1}=3{x}_{3}-2{x}_{4},{x}_{2}=2{x}_{3}-{x}_{4}$

b. Yes, ${x}_{1}=1,{x}_{2}=5,{x}_{3}=9,{x}_{4}=13$

See the step by step solution

## (a)Step 1: Consider the equations to find other equations

Find the expression for, ${x}_{1}$.

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

Find the expression for, ${x}_{2}$ .

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

## Step 2: Express the variable in terms of other variables

Re-write ${x}_{1}$in terms of ${x}_{3}$and${x}_{4}$ .

$\begin{array}{c}{x}_{1}=2{x}_{2}-{x}_{3}\\ {x}_{1}=2\left(2{x}_{3}-{x}_{4}\right)-{x}_{3}\\ =3{x}_{3}-2{x}_{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(3\right)\end{array}$

Therefore, ${x}_{1}=3{x}_{3}-2{x}_{4},{x}_{2}=2{x}_{3}-{x}_{4}$

## (b)Step 3: Using the above equations, find the values of the variables

Consider the given values of the variables.

${x}_{1}=1$ and ${x}_{4}=13$

Substitute these values in the above equations to find the values of the remaining variables.

Consider the equation (3).

$\begin{array}{c}{x}_{1}=3{x}_{3}-2{x}_{4}\\ 1=3{x}_{3}-2×13\\ {x}_{3}=\frac{27}{3}\\ =9\end{array}$

Consider the equation (2).

$\begin{array}{c}{x}_{2}=2{x}_{3}-{x}_{4}\\ =2×9-13\\ =18-13\\ =5\end{array}$

Therefore, ${x}_{1}=1,{x}_{2}=5,{x}_{3}=9,{x}_{4}=13$.

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades. 