To solve the linear system:\[\begin{align*}x_1 + x_2 + x_3 + x_4 &= 1, \\3x_1 + x_2 + x_3 - x_4 &= 5, \\5x_1 - x_2 + x_3 - 5x_4 &= 3.\end{align*}\]Describe the solution set geometrically.

Given system of linear equation are x1+x2+x3+x4=13x1+x2+x3−x4=55x1−x2+x3−5x4=3.......1 We have to…

Answered: Solve the linear system x1 + x2 + x3 +…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

To solve the linear system:

\[
\begin{align*}
x_1 + x_2 + x_3 + x_4 &= 1, \\
3x_1 + x_2 + x_3 - x_4 &= 5, \\
5x_1 - x_2 + x_3 - 5x_4 &= 3.
\end{align*}
\]

Describe the solution set geometrically.

Expert Solution

Step 1

Given system of linear equation are

$\begin{matrix} x_{1} + x_{2} + x_{3} + x_{4} = 1 \\ 3 x_{1} + x_{2} + x_{3} - x_{4} = 5 \\ 5 x_{1} - x_{2} + x_{3} - 5 x_{4} = 3 \end{matrix}\} . . . . . . . (1)$

We have to solve the given system of equation.

Write the system $(1)$ in augmented matrix form $(A | b)$ , which is given below

$(A | b) = [\begin{matrix} 1 & 1 & 1 & 1 & 1 \\ 3 & 1 & 1 & - 1 & 5 \\ 5 & - 1 & 1 & - 5 & 3 \end{matrix}]$

Now, we will convert the matrix $(A | b)$ into reduced row echelon form.

Swap the matrix row $R_{1} \leftrightarrow R_{3}$ , we get

$(A | b) = [\begin{matrix} 5 & - 1 & 1 & - 5 & 3 \\ 3 & 1 & 1 & - 1 & 5 \\ 1 & 1 & 1 & 1 & 1 \end{matrix}]$

Now, performing the row operation $R_{2} \to R_{2} - \frac{3}{5} R_{1}$ and $R_{3} \to R_{3} - \frac{1}{5} R_{1}$ , we get

$(A | b) = [\begin{matrix} 5 & - 1 & 1 & - 5 & 3 \\ 0 & \frac{8}{5} & \frac{2}{5} & 2 & \frac{16}{5} \\ 0 & \frac{6}{5} & \frac{4}{5} & 2 & \frac{2}{5} \end{matrix}]$

Now, performing the row operation $R_{3} \to R_{3} - \frac{3}{4} R_{2}$ , we get

$(A | b) = [\begin{matrix} 5 & - 1 & 1 & - 5 & 3 \\ 0 & \frac{8}{5} & \frac{2}{5} & 2 & \frac{16}{5} \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & - 2 \end{matrix}]$ .