Solve the following system of homogenous linear equations :
`2x+3y+4z=0`, `x+y+z=0`, `2x-y+3z=0`

Answer

To solve the given system of homogeneous linear equations:

1. **Write the equations:**
\[ \begin{align*} 1. & \quad 2x + 3y + 4z = 0 \\ 2. & \quad x + y + z = 0 \\ 3. & \quad 2x - y + 3z = 0 \end{align*} \]

2. **Form the coefficient matrix \( A \):**
The coefficient matrix \( A \) can be formed from the coefficients of \( x, y, z \) in the equations:
\[ A = \begin{pmatrix} 2 & 3 & 4 \\ 1 & 1 & 1 \\ 2 & -1 & 3 \end{pmatrix} \]

3. **Calculate the determinant of matrix \( A \):**
To determine if the system has a unique solution or infinite solutions, we need to calculate the determinant of \( A \):
\[ \text{det}(A) = 2 \begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} - 3 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} + 4 \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} \]

Now calculate each of the 2x2 determinants:
\[ \begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} = (1)(3) - (1)(-1) = 3 + 1 = 4 \]
\[ \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1 \]
\[ \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (1)(2) = -1 - 2 = -3 \]

Substitute these values back into the determinant calculation:
\[ \text{det}(A) = 2(4) - 3(1) + 4(-3) = 8 - 3 - 12 = -7 \]

4. **Analyze the determinant:**
Since \( \text{det}(A) = -7 \) which is not equal to zero, the system has a unique solution.

5. **Find the unique solution:**
For a homogeneous system \( Ax = 0 \), if the determinant is non-zero, the only solution is the trivial solution:
\[ x = 0, \quad y = 0, \quad z = 0 \]

Thus, the solution to the system of equations is:
\[ (x, y, z) = (0, 0, 0) \]

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