### Problem 4#### (i) Linear SystemConsider the linear system:\[\begin{align*}x_1 - 2x_3 &= 1 \\-x_1 + x_2 &= 3 \\x_2 - 2x_3 &= 0\end{align*}\]Determine if this system is consistent or not. If it is consistent, provide either its single unique solution or its general solution.#### (ii) Homogeneous SystemNow consider the corresponding homogeneous system:\[\begin{align*}x_1 - 2x_3 &= 0 \\-x_1 + x_2 &= 0 \\x_2 - 2x_3 &= 0\end{align*}\]Determine if this system is consistent or not. If it is consistent, provide either its single unique solution or its general solution.

4i Given system of equation is- x1−2x3=1−x1+x2=3x2−2x3=0........1 Now, write the augmented matrix…

Answered: 4) (i) Consider the linear system: x₁ −…

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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### Problem 4

#### (i) Linear System

Consider the linear system:

\[
\begin{align*}
x_1 - 2x_3 &= 1 \\
-x_1 + x_2 &= 3 \\
x_2 - 2x_3 &= 0
\end{align*}
\]

Determine if this system is consistent or not. If it is consistent, provide either its single unique solution or its general solution.

#### (ii) Homogeneous System

Now consider the corresponding homogeneous system:

\[
\begin{align*}
x_1 - 2x_3 &= 0 \\
-x_1 + x_2 &= 0 \\
x_2 - 2x_3 &= 0
\end{align*}
\]

Determine if this system is consistent or not. If it is consistent, provide either its single unique solution or its general solution.

Expert Solution

Step 1

$4 (i)$ Given system of equation is-

$\begin{matrix} x_{1} - 2 x_{3} = 1 \\ - x_{1} + x_{2} = 3 \\ x_{2} - 2 x_{3} = 0 \end{matrix}\} ........ (1)$

Now, write the augmented matrix form $(A | b)$ , which is given below

$(A | b) = (\begin{matrix} 1 & 0 & - 2 & 1 \\ - 1 & 1 & 0 & 3 \\ 0 & 1 & - 2 & 0 \end{matrix})$

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

$(A | b) ~ (\begin{matrix} 1 & 0 & - 2 & 1 \\ 0 & 1 & - 2 & 4 \\ 0 & 1 & - 2 & 0 \end{matrix}) ~ (\begin{matrix} 1 & 0 & - 2 & 1 \\ 0 & 1 & - 2 & 4 \\ 0 & 0 & 0 & 4 \end{matrix}) ~ (\begin{matrix} 1 & 0 & - 2 & 1 \\ 0 & 1 & - 2 & 4 \\ 0 & 0 & 0 & 1 \end{matrix}) ~ (\begin{matrix} 1 & 0 & - 2 & 0 \\ 0 & 1 & - 2 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) \Rightarrow (A | b) ~ (\begin{matrix} 1 & 0 & - 2 & 0 \\ 0 & 1 & - 2 & 0 \\ 0 & 0 & 0 & 1 \end{matrix})$