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ia

How do I solve this problem? Ideally, I think $$a^2-4=1$$ is the best option to solve for a I set the entire expression equal to 1. Then to solve for b I just put $b=1$. Is this right? I feel like that's really wrong.

So update I got this enter image description here

I suppose my answer is just b does not equal 0 and a does not equal +- sqrt 4

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  • $\begingroup$ Do you know cramer's rule, Matrix inversion? $\endgroup$ Sep 22, 2020 at 19:17
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    $\begingroup$ Cramer's Rule would be folly. It usually is. $\endgroup$ Sep 22, 2020 at 19:18
  • $\begingroup$ @AnindyaPrithvi never heard of it. is that like the determinant method or something? $\endgroup$
    – Si Random
    Sep 22, 2020 at 19:26
  • $\begingroup$ Cramer's rule is an obscure way to solve very special cases of matrices, you're best to avoid it. $\endgroup$
    – o's1234
    Sep 22, 2020 at 19:29
  • $\begingroup$ This is Cramer's rule it is fast if you can compute determinants fast. This should work well in this case because 3/4 of all determinants will be $2 \times 2$ $\endgroup$ Sep 23, 2020 at 3:05

3 Answers 3

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There's a fundamental result for this kind of problems:

A non-homogenous linear system $AX=B$ has solutions if & only if the matrix $A$ and the augmented matrix $A|B$ have the same rank. When this condition is satisfied, the common rank is the codimension of the affine subspace of solutions.

Therefore, if $(a^2-4)b\ne 0$, the matrix $A$ has rank $3$, which is also the maximum rank of the augmented matrix, and there is a unique solution.

If $a=\pm 2$, $A$ has rank $2$, but the augmented matrix may have rank $3$. If it has rank $2$, the set of solutions has dimension $1$.

To determine the solutions, write the augmented matrix in reduced row echelon form. The set of solutions is the last column.

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This question is asking for ALL pairs of a and b for which the system is consistent, not just one.

You are likely better off trying to consider what makes the system inconsistent. Try doing Gaussian elimination on the matrix, and take good notes along the way. Any time you are dividing by a term containing a or b, determine what value of a or b would make that term zero (thus dividing by zero). These are values where the system is inconsistent.

Hope that helps.

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  • $\begingroup$ ah ok i gotchu. thanks for the stuff. i feel like idk how to do gaussian elimantion like this ill it'll be super complicated fractions so I wasn't sure if that was possible or not or if there was an easier way $\endgroup$
    – Si Random
    Sep 22, 2020 at 19:25
  • $\begingroup$ Awesome, yeah it's a pretty complex problem. Let me know if this works out for you, and mark mine as an answer if so :) $\endgroup$
    – o's1234
    Sep 22, 2020 at 19:28
  • $\begingroup$ will do ill try my best $\endgroup$
    – Si Random
    Sep 22, 2020 at 19:32
  • $\begingroup$ Do not worry if this material is rough at first, it took years for me to really get to the point where I felt comfortable with it. Just try to stick to the principles you already know from elementary math and you will always have that on your side. $\endgroup$
    – o's1234
    Sep 22, 2020 at 19:33
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    $\begingroup$ thank you for the encouragement. please see my updated question and if that answer is right or not. if so, how would I find the second part if it's unique or not? $\endgroup$
    – Si Random
    Sep 22, 2020 at 19:43
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Sorry for the late reply.

The system can be written as $$x_{1}+6x_{2}-9x_{3}=9$$ $$(a^2-4)x_{2}+x_{3}=-7$$ $$x_{3}b=7.$$

Clearly the last equation is only consistent if $b\neq0.$ So this will give $x_{3}=\frac{7}{b}.$ Now we substitute this in the second equation. This gives $$(a^2-4)x_{2}=-7-x_{3}=-7-\frac{7}{b}=-\frac{7b+7}{b} (\star)$$

Two cases to consider:

  1. If $a\neq2,-2$ then we have $x_{2}=-\frac{7b+7}{b(a^2-4)}$ and then we can use the first equation to obtain $x_{1}$ and this will give a unique solution.
  2. What happens if $a=2,-2$? Well in this case we have from $(\star)$ that $$0=-\frac{7b+7}{b}$$

and this forces $b=-1$ so the numerator vanishes. So in this case both equations (2) and (3) are satisfied and substituting into equation (1) we get that $x_{1}=-6x_{2}+9x_{3}+9=-6x_{2}+9(x_{3}+1)=-6x_{2}+9(-7+1)=-6x_{2}-54$ for $x_{3}=\frac{7}{b}=\frac{7}{-1}=-7$ and there are infinitely many solutions (since any $x_{2}$ satisfies the second equation because it vanishes for $a=\pm 2$).

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    $\begingroup$ wow this is amazing! thank you so much ares ur amazing as always! $\endgroup$
    – Si Random
    Sep 22, 2020 at 19:58

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