# 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 $$5R_2$$? – Dr. Sonnhard Graubner Feb 25 '19 at 15:49
• Oh yea. Thx for correcting – Anson Pang Feb 25 '19 at 15:52

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 – Anson Pang Feb 25 '19 at 17:45
• Ok, I will check it – Dr. Sonnhard Graubner Feb 25 '19 at 19:21

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) – Anson Pang Feb 25 '19 at 19:27
• @AnsonPang see the update – gt6989b Feb 25 '19 at 19:57