Solve the following systems of linear homogenous equations :
`3x+2y+7z=0`, `4x-3y-2z=0` and `5x+9y+23z=0`

Answer

To solve the system of linear homogeneous equations:

1. **Equations Given:**
\[ \begin{align*} 3x + 2y + 7z &= 0 \quad (1) \\ 4x - 3y - 2z &= 0 \quad (2) \\ 5x + 9y + 23z &= 0 \quad (3) \end{align*} \]

2. **Form the Augmented Matrix:**
The system can be represented in matrix form \(AX = 0\), where:
\[ A = \begin{bmatrix} 3 & 2 & 7 \\ 4 & -3 & -2 \\ 5 & 9 & 23 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad 0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

3. **Row Reduction:**
We will perform row operations to reduce the matrix to Row Echelon Form (REF).

Start with the matrix:
\[ \begin{bmatrix} 3 & 2 & 7 \\ 4 & -3 & -2 \\ 5 & 9 & 23 \end{bmatrix} \]

- **Step 1:** Make the leading coefficient of the first row a 1 by dividing the first row by 3:
\[ R_1 \rightarrow \frac{1}{3} R_1 \Rightarrow \begin{bmatrix} 1 & \frac{2}{3} & \frac{7}{3} \\ 4 & -3 & -2 \\ 5 & 9 & 23 \end{bmatrix} \]

- **Step 2:** Eliminate the first column entries below the first row:
\[ R_2 \rightarrow R_2 - 4R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - 5R_1 \]
This gives:
\[ R_2 \rightarrow \begin{bmatrix} 0 & -\frac{17}{3} & -\frac{34}{3} \end{bmatrix}, \quad R_3 \rightarrow \begin{bmatrix} 0 & \frac{7}{3} & \frac{14}{3} \end{bmatrix} \]

- **Step 3:** Now, the matrix looks like:
\[ \begin{bmatrix} 1 & \frac{2}{3} & \frac{7}{3} \\ 0 & -\frac{17}{3} & -\frac{34}{3} \\ 0 & \frac{7}{3} & \frac{14}{3} \end{bmatrix} \]

- **Step 4:** Multiply the second row by \(-\frac{3}{17}\):
\[ R_2 \rightarrow \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} \]

- **Step 5:** Now, eliminate the second column entry in the third row:
\[ R_3 \rightarrow R_3 - \frac{7}{3} R_2 \]
This gives:
\[ R_3 \rightarrow \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \]

- **Final Matrix:**
The final matrix is:
\[ \begin{bmatrix} 1 & \frac{2}{3} & \frac{7}{3} \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \]

4. **Determine the Rank:**
The rank of the matrix \(A\) is 2 (since there are 2 non-zero rows). The rank of the augmented matrix is also 2 (since the last row is all zeros).

5. **Solution:**
Since the rank of \(A\) is equal to the rank of the augmented matrix and is less than the number of variables (3), the system has infinitely many solutions.

Let \(z = k\) (a free variable), then from the second row:
\[ y + 2k = 0 \Rightarrow y = -2k \]
From the first row:
\[ x + \frac{2}{3}(-2k) + \frac{7}{3}k = 0 \Rightarrow x - \frac{4}{3}k + \frac{7}{3}k = 0 \Rightarrow x + k = 0 \Rightarrow x = -k \]

6. **Final Solution:**
The general solution can be expressed as:
\[ \begin{align*} x &= -k \\ y &= -2k \\ z &= k \end{align*} \]
where \(k\) is any real number.