To solve this system of linear equations, we’ll set it up as an augmented matrix and use row reduction (Gaussian elimination) to find the general solution. The matrix setup and operations will be as follows:

The matrix equation given is: $\begin{pmatrix} 0 & -1 & 2 \\ 1 & -2 & 1 \\ 1 & -4 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

1. Set up the augmented matrix: $\begin{pmatrix} 0 & -1 & 2 & | & 0 \\ 1 & -2 & 1 & | & 0 \\ 1 & -4 & 5 & | & 0 \end{pmatrix}$

2. Row Reduction Steps:

   • Start by making the entry in the first row and first column zero (it's already zero).

   • Swap row 1 and row 2 to get a leading 1 in the first row: $\begin{pmatrix} 1 & -2 & 1 & | & 0 \\ 0 & -1 & 2 & | & 0 \\ 1 & -4 & 5 & | & 0 \end{pmatrix}$

   • Subtract row 1 from row 3: $\begin{pmatrix} 1 & -2 & 1 & | & 0 \\ 0 & -1 & 2 & | & 0 \\ 0 & -2 & 4 & | & 0 \end{pmatrix}$

   • Divide row 2 by -1: $\begin{pmatrix} 1 & -2 & 1 & | & 0 \\ 0 & 1 & -2 & | & 0 \\ 0 & -2 & 4 & | & 0 \end{pmatrix}$

   • Add 2 times row 2 to row 3: $\begin{pmatrix} 1 & -2 & 1 & | & 0 \\ 0 & 1 & -2 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$

3. Interpret the reduced row echelon form: The matrix now represents the equations: $x - 2y + z = 0$ $y - 2z = 0$ The third row gives no new information (0 = 0), so we have two equations with three variables, implying there will be free variables.

4. Express variables in terms of the free variable $z$:

   • From the second equation, $y = 2z$.
   • Substitute $y = 2z$ into the first equation: $x - 2(2z) + z = 0 \Rightarrow x - 4z + z = 0 \Rightarrow x = 3z$.

   So, we can express $x$, $y$, and $z$ in terms of $z$ as: $x = 3z, \quad y = 2z, \quad z = z$

5. General Solution: Let $z = t$, where $t$ is a parameter. Then the solution can be written as: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3t \\ 2t \\ t \end{pmatrix} = t \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$

6. Basis for the Solution Space: The solution space is spanned by the vector: $\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$ So, a basis for the solution space is: $\left\{ \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \right\}$

Summary:

The general solution to the system is: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = t \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}, \quad t \in \mathbb{R}$ And the basis for the solution space is $\left\{ \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \right\}$.

Would you like further details on any step, or do you have questions?

Related Questions:

1. How can we verify if a vector is in the solution space of a given system?
2. What is the geometric interpretation of the solution space for this system?
3. How would the solution change if the system had a non-zero right-hand side?
4. What are the conditions for a system of equations to have infinitely many solutions?
5. How does the rank of the coefficient matrix relate to the solution space?

Tip:

When solving systems of linear equations, keep track of free variables carefully, as they indicate the number of parameters in the solution set.