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If consider $$\left[ \begin{array}{ccc|c} 2&-1&3&b_1\\ a&1&0&b_2\\2&1&1&b_3 \end{array} \right]$$

a) If $$b_1,b_2,b_3$$ are not three zero. For what values of "a" would the system have an infinite solutions and which would have to satisfy b1,b2 and b3?

b)Suppose that $$b_1=b_2=b_3=0$$ and determine for which values of a the system has non-trivial solutions

I saw this problem on the page but i have a question, could this be resolved without using determinants? And what would be the complete process? I never taken the Linear Algebra class,but i'm interested in the subject; I only know some things about it.I think you could use RREF but i don't know to apply it.I could be like this or it's incorrect? $$\left[ \begin{array}{ccc|c} a&1&0&b_2\\ 0&\frac{a-2}{a}&1&\frac{ab_3-2b_2}{a}\\0&0&\frac{3a-2}{a-2}&\frac{b_1a-2b_1+ab_3+2b_3-4b_2}{a-2} \end{array} \right]$$

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  • $\begingroup$ The determinant is $4(a-1)$ and RREF yields $$\left( \begin{array}{cccc} 1 & 0 & 0 & \dfrac{\text{b1}+4 \text{b2}-3 \text{b3}}{4 (a-1)} \\ 0 & 1 & 0 & \dfrac{-a \text{b1}-4 \text{b2}+3 a \text{b3}}{4 (a-1)} \\ 0 & 0 & 1 & \dfrac{a \text{b1}-2 \text{b1}-4 \text{b2}+a \text{b3}+2 \text{b3}}{4 (a-1)} \\ \end{array} \right)$$ $\endgroup$
    – Moo
    Nov 4 '17 at 12:32
  • $\begingroup$ @Moo which operation you used to get there? $\endgroup$ Nov 4 '17 at 15:46
  • $\begingroup$ Gaussian Elimination. $\endgroup$
    – Moo
    Nov 4 '17 at 15:49
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The $b$'s are irrelevant to the question of infinite number of solutions. The matrix reduces to $$\begin{pmatrix} 1&0&1\\ 0&1&-1\\ 0&0&1-a\end{pmatrix}$$

So only $a=1$ gives an infinity of solutions.

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If we just reduce this matrix a bit to something easier to work with we have $$\left[ \begin{array}{ccc|c} 2&-1&3&b_1\\ 0&-4(a-1)&0&ab_1+4b_2-3ab_3\\0&2&-2&b_3-b_1 \end{array} \right]$$

It's clear to see that $-4(a-1)=0 \implies a=1$. The rest is irrelevant since you only asked for what values of $a$. But if you wanted to see for what values of $b$ it has infinite solution we have $b_1 = 3b_3 -4b_2$

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  • $\begingroup$ Then the values of b's will be 0 right? $\endgroup$ Nov 4 '17 at 12:45
  • $\begingroup$ @HectorJavierTorres Well if you need to know for what values of $b$ we have infinite solutions, you would get $b_1+4(1)b_2 - 3b_3=0\implies b_3=(b_1+4b_2)/3$. So the $b$'s don't need to be zero, but the expression above must hold. $\endgroup$ Nov 4 '17 at 12:51
  • $\begingroup$ how did you get the row two?What operation did you use? If it's not too much trouble because that's the part that does give me doubt $\endgroup$ Nov 4 '17 at 13:48
  • $\begingroup$ @HectorJavierTorres Sorry I made a mistake, $R_2 \to 2R_2 - aR_1$ is the initial transformation. Then $R_2 \to 2R_2-3aR_3$, I made the changes. $\endgroup$ Nov 4 '17 at 14:14
  • $\begingroup$ No problem, thank you very much; I understand right now $\endgroup$ Nov 4 '17 at 15:47

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