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Find conditions on a such that the following linear system has no solution, exactly one solution, or infinitely many solutions. In the case when the system has a unique solution, find the solution in terms of a.

The system of equations given is as follows:

x + z = -1
4x + 4y + az = 0
-4x-ay-(a+4)z = -4

The solution I got was as follows:

i. The system has no solution when a = 0. ii. The system has infinite solutions when a = 8. iii. The system has a unique solution a is not equal to 8 or 0.

However, for the second part of the question I do not know how to express the unique solution in terms of a. Could I perhaps receive some help on this and also on checking through my answer?

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  • $\begingroup$ How did you solve the first part ? $\endgroup$
    – user65203
    Feb 1 '20 at 15:19
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Your system is$$\left\{\begin{array}{l}x+z=-1\\4x+4y+az=0\\-4x-ay-(a+4)z=-4.\end{array}\right.$$If you subtract from the second equation the first one times $4$ and if you add to the third equation the first one times $4$, you'll get$$\left\{\begin{array}{l}x+z=-1\\4y+(a-4)z=4\\-ay-az=-8.\end{array}\right.$$Since $a\neq0$, you can divide the third equation by $-a$ and the system becomes$$\left\{\begin{array}{l}x+z=-1\\4y+(a-4)z=4\\y+z=\frac8a.\end{array}\right.$$Now, if you subtract from the second equation the third one times $4$, you'll get$$\left\{\begin{array}{l}x+z=-1\\(a-8)z=4-\frac{32}a=\frac{4a-32}a\\y+z=\frac8a.\end{array}\right.$$Can you take it from here?

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  • $\begingroup$ Hello! Thanks for your help. I was able to take it from there. Just to confirm, the second equation should actually be 4y+(a−4)z=4 I think. $\endgroup$
    – XDXDXD
    Feb 1 '20 at 16:26
  • $\begingroup$ The answer I got was x = (-a-4)/a; y = 4/a and z = 4/a. Would that be correct? $\endgroup$
    – XDXDXD
    Feb 1 '20 at 16:26
  • $\begingroup$ Yes (in both cases). I've edited my answer. $\endgroup$ Feb 1 '20 at 16:33
  • $\begingroup$ Thanks a lot for your help! $\endgroup$
    – XDXDXD
    Feb 2 '20 at 4:00

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