To solve the given systems of equations using the matrix method, let's address each system separately.

System (i)

The equations are:

1. $x - y + z = 0$
2. $x + 2y - z = 0$
3. $4x + y - 5z = 0$

Step 1: Set up the matrix equation

The system can be written in matrix form as: $\begin{bmatrix} 1 & -1 & 1 \\ 1 & 2 & -1 \\ 4 & 1 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Let: $A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 2 & -1 \\ 4 & 1 & -5 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

We now have the matrix equation $AX = B$.

Step 2: Check if there is a unique solution by finding the determinant of $A$

If $\det(A) \neq 0$, the system has a unique solution. If $\det(A) = 0$, then the system has either no solution or infinitely many solutions.

Calculate $\det(A)$: $\det(A) = 1 \begin{vmatrix} 2 & -1 \\ 1 & -5 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -1 \\ 4 & -5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix}$ $= 1(2 \cdot -5 - (-1) \cdot 1) + 1(1 \cdot -5 - (-1) \cdot 4) + 1(1 \cdot 1 - 2 \cdot 4)$ $= 1(-10 + 1) + 1(-5 + 4) + 1(1 - 8) = -9 - 1 - 7 = -17$ Since $\det(A) = -17 \neq 0$, the system has a unique solution.

Step 3: Solve for $X$ using $X = A^{-1}B$

Since $B = 0$, this system has a unique solution at $X = 0$, meaning $x = 0$, $y = 0$, and $z = 0$. This is the only solution for this system.

---

System (ii)

The equations are:

1. $x + y + 2z = 0$
2. $x - 2y - 8z = 0$
3. $x + y - 3z = 0$

Step 1: Set up the matrix equation

The system can be written in matrix form as: $\begin{bmatrix} 1 & 1 & 2 \\ 1 & -2 & -8 \\ 1 & 1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Let: $A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & -2 & -8 \\ 1 & 1 & -3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Step 2: Check if there is a unique solution by finding the determinant of $A$

Calculate $\det(A)$: $\det(A) = 1 \begin{vmatrix} -2 & -8 \\ 1 & -3 \end{vmatrix} - 1 \begin{vmatrix} 1 & -8 \\ 1 & -3 \end{vmatrix} + 2 \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix}$ $= 1(-2 \cdot -3 - (-8) \cdot 1) - 1(1 \cdot -3 - (-8) \cdot 1) + 2(1 \cdot 1 - (-2) \cdot 1)$ $= 1(6 - 8) - 1(-3 + 8) + 2(1 + 2) = -2 + 5 + 6 = 9$ Since $\det(A) = 9 \neq 0$, the system has a unique solution.

Step 3: Solve for $X$

Since $B = 0$, the only solution to this system is $x = 0$, $y = 0$, and $z = 0$.

---

Final Answer

For both systems (i) and (ii), the unique solution is: $x = 0, \quad y = 0, \quad z = 0$

Would you like further details on any step? Here are some related questions for practice:

1. How would you approach solving a similar system where $\det(A) = 0$?
2. Can you solve a 3x3 system of equations using Gaussian elimination?
3. What is the role of matrix inverses in solving systems of linear equations?
4. How do you interpret a solution of $x = 0, y = 0, z = 0$ in the context of linear systems?
5. How would you determine if a system has no solution or infinitely many solutions?

Tip: For systems with zero determinants, consider checking for consistency and looking for parametric solutions.