Solve the following system of equations by matrix method.

(1) x - y + z = 0 x + 2y - z = 0 4x + y - 5z = 0

(ii) x + y + 2 = 0 x - 2y - 8z = 0 x + y - 3z = 0

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.

Solve the following system of equations by matrix method.

(i) x - y + z = 0, x + 2y - z = 0, 4x + y - 5z = 0

(ii) x + y + 2z = 0, x - 2y - 8z = 0, x + y - 3z = 0

Learn how to solve systems of linear equations using the matrix method. This tutorial covers key concepts such as determinants, matrix inverses, and the unique solution conditions for 3x3 systems. Suitable for high school and college students studying linear algebra.

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