Consider the following system of linear equations in variables $x_1, x_2, x_3,$ where $a,b$ are some fixed real numbers.

$$x_1+x_2-x_3 = 1$$

$$2x_1+x_3 = 1$$

$$x_1-ax_2 + 2x_3 = b$$

Find the values of $a,b$ for which the system has: (i) infinitely many solutions, (ii) exactly one solution, (iii) no solutions.

Does anyone have an, idea of how I should approach this exercise. I tried to reduce the matrix.

  • $\begingroup$ For the infinitely many solutions, a would be 1, and b would be 0. When you use elimination method by multiplying the second equation by $-1$, and add all the left sides together, and the right sides together, you are left with $(1-a)x_2=b$, which when you have $a=1, b=0$, produces result $0=0$, which leads to infinite solutions. $\endgroup$ – KKZiomek Nov 25 '16 at 5:14

From: $2x_1+x_3=1$

Subtract: $x_1+x_2-x_3=1$

Obtain: $x_1-x_2+2x_3=0$

From: $x_1-ax_2+2x_3=b$

Subtract: $x_1-x_2+2x_3=0$

Obtain: $(1-a)x_2=b$


  • $(a=1)\wedge(b=0)\implies0=0\implies\infty$ solutions
  • $(a=1)\wedge(b\neq0)\implies0\neq0\implies0$ solutions
  • $(a\neq1)\implies x_2=b/(1-a)\implies1$ solution
  • $\begingroup$ Hey barak manos, Thank you very much. If you please could show me what you mean by From, subtract and obtaion. And maybe give a litte words on theory so I can leran it. :) $\endgroup$ – Mohamed Rasul Nov 26 '16 at 5:46
  • $\begingroup$ @MohamedRasul: You're welcome. For example: Take the LHS of the first equation ($2x_1+x_3$), and subtract from it the LHS of the second equation ($x_1+x_2-x_3$). Take the RHS of the first equation ($1$), and subtract from it the RHS of the second equation ($1$). What you get is $x_1-x_2+2x_3=0$. $\endgroup$ – barak manos Nov 26 '16 at 8:09
  • $\begingroup$ @MohamedRasul: You're welcome. Feel free to accept the answer by clicking on the V next to it :) $\endgroup$ – barak manos Nov 29 '16 at 11:55

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