Solve the system of equations
` x+3y -2z =0, 2x-y+4z =0 ,x-11y+14z =0`

Answer

To solve the system of equations given by:

1. \( x + 3y - 2z = 0 \)
2. \( 2x - y + 4z = 0 \)
3. \( x - 11y + 14z = 0 \)

we will use matrix methods.

### Step 1: Write the system in matrix form

We can express the system of equations in the matrix form \( AX = 0 \), where:

\[ A = \begin{bmatrix} 1 & 3 & -2 \\ 2 & -1 & 4 \\ 1 & -11 & 14 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Find the determinant of matrix A

To check if the system has a unique solution, we need to find the determinant of matrix \( A \):

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -1 & 4 \\ -11 & 14 \end{vmatrix} - 3 \cdot \begin{vmatrix} 2 & 4 \\ 1 & 14 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & -1 \\ 1 & -11 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 4 \\ -11 & 14 \end{vmatrix} = (-1)(14) - (4)(-11) = -14 + 44 = 30 \)

2. \( \begin{vmatrix} 2 & 4 \\ 1 & 14 \end{vmatrix} = (2)(14) - (4)(1) = 28 - 4 = 24 \)

3. \( \begin{vmatrix} 2 & -1 \\ 1 & -11 \end{vmatrix} = (2)(-11) - (-1)(1) = -22 + 1 = -21 \)

Now substituting these values back into the determinant calculation:

\[ \text{det}(A) = 1 \cdot 30 - 3 \cdot 24 - 2 \cdot (-21) = 30 - 72 + 42 = 0 \]

### Step 3: Analyze the determinant

Since \( \text{det}(A) = 0 \), the matrix \( A \) is singular, which means that the system of equations does not have a unique solution. This suggests that either there are infinitely many solutions or no solution at all.

### Step 4: Find the rank of matrix A

To determine the nature of the solutions, we can find the rank of the matrix \( A \) and the augmented matrix \( [A|B] \).

Performing row reduction on \( A \):

\[ \begin{bmatrix} 1 & 3 & -2 \\ 2 & -1 & 4 \\ 1 & -11 & 14 \end{bmatrix} \]

Using row operations to simplify:

1. \( R_2 \leftarrow R_2 - 2R_1 \)
2. \( R_3 \leftarrow R_3 - R_1 \)

This gives:

\[ \begin{bmatrix} 1 & 3 & -2 \\ 0 & -7 & 8 \\ 0 & -14 & 16 \end{bmatrix} \]

Next, simplify \( R_3 \):

3. \( R_3 \leftarrow R_3 - 2R_2 \)

This results in:

\[ \begin{bmatrix} 1 & 3 & -2 \\ 0 & -7 & 8 \\ 0 & 0 & 0 \end{bmatrix} \]

The rank of matrix \( A \) is 2 (since there are 2 non-zero rows).

### Step 5: Find the rank of the augmented matrix

The augmented matrix \( [A|B] \) is:

\[ \begin{bmatrix} 1 & 3 & -2 & | & 0 \\ 2 & -1 & 4 & | & 0 \\ 1 & -11 & 14 & | & 0 \end{bmatrix} \]

Performing the same row operations, we find that the rank remains 2.

### Conclusion

Since the rank of \( A \) (2) is equal to the rank of the augmented matrix \( [A|B] \) (2) and is less than the number of variables (3), the system has infinitely many solutions.

The solutions can be expressed in terms of a free variable. Let’s set \( z = t \). Then we can express \( x \) and \( y \) in terms of \( t \).

From the second equation, substituting \( z = t \):

\[ 2x - y + 4t = 0 \implies y = 2x + 4t \]

Substituting \( y \) into the first equation:

\[ x + 3(2x + 4t) - 2t = 0 \implies x + 6x + 12t - 2t = 0 \implies 7x + 10t = 0 \implies x = -\frac{10}{7}t \]

Substituting \( x \) back into the equation for \( y \):

\[ y = 2\left(-\frac{10}{7}t\right) + 4t = -\frac{20}{7}t + 4t = -\frac{20}{7}t + \frac{28}{7}t = \frac{8}{7}t \]

Thus, the general solution is:

\[ x = -\frac{10}{7}t, \quad y = \frac{8}{7}t, \quad z = t \]

### Final Answer

The solution to the system of equations is:

\[ (x, y, z) = \left(-\frac{10}{7}t, \frac{8}{7}t, t\right) \quad \text{for any } t \in \mathbb{R} \]