-2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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: $(1-a)x_2=b$


Conclude:

  • $(a=1)\wedge(b=0)\implies0=0\implies\infty$ solutions
  • $(a=1)\wedge(b\neq0)\implies0\neq0\implies0$ solutions
  • $(a\neq1)\implies x_2=b/(1-a)\implies1$ solution
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.