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Show that the equation $Ax=b$ is not consistent for all possible $b$, and describe the set of all $b$ for which the equation is consistent, both algebraically and geometrically.

$$A = \begin{bmatrix} 1& 3& -4\\ -2& 1& 2\\ 3 &2& -6 \end{bmatrix}$$

$$b = \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}$$

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1 Answer 1

HINT: Set up the augmented matrix,

$$\left[\begin{array}{rrr|r} 1&3&-4&b_1\\ -2&1&2&b_2\\ 3&2&-6&b_3 \end{array}\right]\;,$$

and do a row reduction. The numbers have been chosen to make it very easy.

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I have done row reduction to get: 1 0 14 = b1-3(b2+2b1), 0 1 -6 = b2+2b1, 0, 0, 0 = b3-3b1+b2+2b1. (Sorry, still learning MathJax). It seems that the third column contains free variables? $ ( \begin{array}{ccc} 1 & 0 & 14 \\ 0 & 1 & -6 \\ 0 & 0 & 0 \end{array}) $ –  Faeynrir Sep 5 '12 at 8:27
    
@Faeynrir: It does, but that’s not the real point here. I get $$\begin{bmatrix}1&3&-4&b_1\\0&7&-6&b_2+2b_1\\0&0&0&b_3+b_2-b_1\end{bmatrix}$$ In order for this to be consistent, $b_3+b_2-b_1$ has to be $0$, since that last line corresponds to the equation $0x_1+0x_2+0x_3=b_3+b_2-b_1$. –  Brian M. Scott Sep 5 '12 at 8:35
    
I see, that makes sense. So, the algebraic description of the solution set where A is consistent is simply where $0x_1+0x_2+0x_3 = b_3+b_2-1b_1$; I'll have to think a bit about the geometric description. –  Faeynrir Sep 5 '12 at 8:59
    
@Faeynrir: Here’s a hint: what does the graph of $z+y-x=0$ look like? –  Brian M. Scott Sep 5 '12 at 9:00
    
It's a plane with part of the plane following / covering x=0. So, geometrically, the solution set here is similar, but following $b_3+b_2-b_1=0$? –  Faeynrir Sep 5 '12 at 9:04

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