Question

Use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\{$$\begin{array}{r}x+y+z=0\\ 2x+3y+z=0\\ 3x+5y+z=0\end{array}$$\right. $$

Short answer

The system of equations has infinitely many solutions, with $x=0$, $y=z=s$, where $s$ is any real number.

Step-by-step solution

Set Up the Augmented Matrix

First, set up the augmented matrix which corresponds to the system of linear equations. An augmented matrix includes the coefficients of the variables and constants from the system of equations without the variables. In this case, it becomes:$$\text{Missing end{bmatrix}}$$

Apply Gauss-Jordan Elimination

Perform row operations to get the matrix in row-reduced echelon form. Swap first and third row, then subtract the first row multiplied by 2 from the second row and the first row multiplied by 3 from the new third row. We get:$$\text{Missing end{bmatrix}}$$Then, add the third row to the second row:$$\text{Missing end{bmatrix}}$$Then, swap second and third row, and multiply the new second row by -1:$$\text{Missing end{bmatrix}}$$And finally, divide the first row by 3:$$\text{Missing end{bmatrix}}$$And subtract the first row multipled by 5/3 from the second row:$$\text{Missing end{bmatrix}}$$

Find the Solution

From the final matrix, write the corresponding system of linear equations, which gives $$\begin{array}{rlrl}x& =0\text{y}-z& =0\text{0}& =0\end{array}$$The third equation suggests that z is a free variable (can take any real number as its value). So, if we let $z=s$ (with $s$ being any real number), then $y=z=s$, and $x=0$. So the system has infinitely many solutions.