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I have tried solving the augmented matrix but I haven't been able to get any answers from that. Can someone explain how to do a question like this? Find a value of a such that this system is inconsistent

Thank you!

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  • $\begingroup$ Your question should be viewable right here for anyone who wants to see it, we should not be required to click over to an external source to see what you are asking. You can very easily type that question into your post using mathjax. $\endgroup$ – Lee Mosher Dec 7 '19 at 16:08
  • $\begingroup$ Have a look through the handy list of related problem at right. There are several examples there of this sort of problem. Study them and their related problems. $\endgroup$ – amd Dec 7 '19 at 20:17
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Basically, you have to find $a$ such that no $(x, y, z)$ could possibly work.

If you sum rows 1 and 3, and compare the result to row 2, you obtain potentially mutually exclusive conditions, ie:

a) $3x + 5y + 4z = a$

and

b) $3x + 5y + (3 + a^2)z = 0$

which implies

b - a) $(3 + a^2 - 4)z = -a$

Expressed this way, the problem is simpler: you have to find $a$ such that no $z$ could possibly work.

$a = -1$ and $a = 1$ respectively lead to $0z = 1$ and $0z = -1$, both of which are unsolvable conditions.

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  • $\begingroup$ Thank you! Now that you have solved it, it seems much simpler. But how do I build the intuition to do something like that? Like I knew I had to prove an a such that no x,y,z would work but it never occurred to me to do the row operations you just did. $\endgroup$ – fresh42juice Dec 7 '19 at 16:08
  • $\begingroup$ Well, first look where the $a$ are, vertically. If you can remove columns where the $a$s aren't, you'll simplify the problem (this is what I did with the $x$ and $y$ columns). The 'noticing that (1) + (3) gives idential $x$ and $y$ coeffs to (2)' comes with quick mental math. Since they're the same, subtraction sends their result to 0, and they disappear. Once the problem was simplified, I looked for the "classical impossible task" which is $0*t = b$ with $b \neq 0$. (Sometimes you'll have $t^2 < 0$ with $t \in R$ or other such impossible things.) Finally I tested the values. There you have it $\endgroup$ – Tristan Duquesne Dec 7 '19 at 16:30

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