# Determining number of solutions for system of equations

I'm having difficulty determining the number of solutions of a system of equations.

Suppose I want to solve the following system of equations

$$\begin{bmatrix} 5x_1+3x_2+2x_3\\ 2x_1-8x_2+6x_3\\ 7x_1+4x_2+15x_3 \end{bmatrix} =\begin{bmatrix} 0\\0\\0 \end{bmatrix}$$

Solving by elimination

$$\begin{split} 2R_1&:\space 10x_1 &+ 6x_2 &+4x_3 &=0\\ 5R_2&:\space 10x_1 &-40x_2 &+30x_3 &=0\\ 2R_1-5R_2&:\space &+46x_2 &-26x_3 &=0 \end{split}$$

A solution to this would be $$x_2=\frac{13}{23}$$, $$x_3=1$$

From $$R_1$$, $$x_1=-\frac{17}{23}$$

Since these 3 values satisfy $$R_1$$ and $$R_2$$, but don't satisfy $$R_3$$, does this automatically mean that the system only has the trivial solution $$R_1=R_2=R_3=0$$?

• Do you mean $$5R_2$$? Commented Feb 25, 2019 at 15:49
• Oh yea. Thx for correcting Commented Feb 25, 2019 at 15:52

## 2 Answers

No, it means you picked the "wrong" solution of the reduced equation. Maybe there is a different value of $$x_2,x_3$$ which would force a different value for $$x_1$$...

UPDATE

To understand if the system will admit non-trivial solutions, eliminate the variables one by one from the equations in a systematic way. For example, you combined $$R_1$$ and $$R_2$$ to get $$46x_2 -26x_3 = 0$$ and in a similar way, compute $$2R_3 - 7R_2$$ to eliminate $$x_1$$ as well, getting the equation $$64x_2 -12x_3 = 0$$

Now note your new system (once you divide top equation by 2 and bottom one by 4 to simplify) is $$\begin{split} 23x_2 &- 13x_3 &= 0\\ 16x_2 &- 3x_3 &= 0 \end{split}$$ It is now easy to eliminate $$x_2$$. Note from the last equation, $$x_2 = 3/16 x_3$$ and plugging that into the top equation: $$0 = 23x_2 - 13x_3 = 23 \times \frac{3}{16} x_3 - 13 x_3 = x_3\left(\frac{23 \cdot 3}{16} - 13\right)$$ which means $$x_3=0$$ is the only possible value. Hence, $$x_2 = \frac{3}{16} x_3 = 0$$ and you can plug into any of the original equations to prove $$x_1=0$$ as well, and the system only admits the trivial solution.

• How would you whether or not the system has infinite solutions or only the trivial one? Can you tell by manipulating the equations? (I haven't learned matrices yet) Commented Feb 25, 2019 at 19:27
• @AnsonPang see the update Commented Feb 25, 2019 at 19:57

Ok,dividing the second equation by $$2$$ we get $$x_1-4x_2+3x_3=0$$ Multiplying this equation by $$-5$$ and adding to the first we get $$23x_2-13x_3=0$$ and analogously $$16x_2-3x_3=0$$ and from $$x_3=\frac{16}{3}x_2$$ and the equation above we get $$x_1=x_2=x_3=0$$

• My book says this has only the trivial solution Commented Feb 25, 2019 at 17:45
• Ok, I will check it Commented Feb 25, 2019 at 19:21