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Is $a_1, a_2, a_3$ a linear combination of $b$?

$a_1 = (1, -2, 0), a_2 = (0, 1, 2), a_3 = (5, -6, 8), b = (2, -1, 6)$

I used Gaussian elimination to get to. $$ \left[ \begin{array}{@{}ccc|c@{}} 1&0&5 & 2 \\ 0&1&4 & 3 \\ 0&0&0 & 0 \\ \end{array} \right] $$ so now $x_3$, is a free variable and I can't tell if it is a linear has a linear combination? Can someone tell me if it does and why?

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Are you asking whether $b$ is a linear combination of $a_1$, $a_2$ and $a_3$? If you are, then you already answered your question, that is $b = 2a_1 + 3a_2$. –  Eric Feb 1 '13 at 22:47
    
YES, THIS IS IT. Thank you. The book was wrong. –  Mateusz Pejko Feb 2 '13 at 17:43
    
Hey, we must have the same book; I had the exact same question. –  Azmisov Sep 12 '13 at 21:48

1 Answer 1

up vote 4 down vote accepted

Assuming you did the reduction properly, you just solve the remaining system:

$$x_1 + 5x_3 = 2$$

$$x_2 + 4x_3 = 3$$

Solving this, you get $x_2 = 3 -4 x_3$ and $x_1 = 2 - 5x_3$.

So, yes, you are free to choose $x_3$.

Update

Given any system of equations there are exactly three possibilities for the solution:

  • (1) There will not be a solution.

  • (2) There will be exactly one solution.

  • (3) There will be infinitely many solutions.

When you do RREF and you get a row of zeros, what does that mean? It means that you have infinitely many solutions (hence, free variables).

How can you tell the remaining two cases from the augmented system? Can you relate this back to other methods, with say, the determinant? It is very good to ask these questions, just as you are doing.

Update 2

$$ \left[ \begin{array}{@{}ccc|c@{}} 1 & -2 & 0 & 2 \\ 0 & 1 & 2 & -2 \\ 5 & -6 & 8 & 6 \\ \end{array} \right] $$

If you RREF, using the following:

  • -5 R1 + R3 $\rightarrow$ R3

  • R4/4 $\rightarrow$ R4

  • -R2 + R3 $\rightarrow$ R3 and then Negate and Divide by 2 $\rightarrow$ R3

  • 2R2 + R1 $\rightarrow$ R1, we end up with:

$$ \left[ \begin{array}{@{}ccc|c@{}} 1 & 0 & 4 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right] $$

So, I am confused if you have written the problem incorrectly or if you calculated the RREF incorrectly. Which is it?

In this solution, the last equation is showing $0 = 1$, which is nonsense, so there is no solution.

Of course, if we took:

$$\text{det} \left[ \begin{array}{@{}ccc|c@{}} 1 & -2 & 0 \\ 0 & 1 & 2 \\ 5 & -6 & 8 \\ \end{array} \right] = 0 $$

we already knew there would be no solutions, assuming you wrote your vectors correctly.

Regards

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Although that does give me a general solution, that doesn't answer the question. The answer at the back of the book states that the vectors does not have a linear combination, the question I'm asking is why as it does have a general solution as you stated. –  Mateusz Pejko Feb 1 '13 at 23:12
    
@MateuszPejko: See the update, is this what what you were looking for? Regards –  Amzoti Feb 1 '13 at 23:33
    
@MateuszPejko: Are you sure you got the RREF correct? Regards –  Amzoti Feb 1 '13 at 23:48
    
+1 Regards $\;$$ –  Babak S. Feb 2 '13 at 6:36
    
Good work...and supplemented, as needed...! $+ 1\;\;\land\;\; \large\checkmark\;\;\land \;\;\ddot\smile\;\;$ –  amWhy May 4 '13 at 2:20

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