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Show that the system of equations $Ax=b$ is not consistent for all $b$ in $\mathbb{R}^3$ $$A = \begin{bmatrix} 1& 2& -1\\ -2& 0& 2\\ -1 &1& -2 \end{bmatrix}$$

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

I've put it into augmented matrix form and used Gauss-Jordan elimination to reduce it into rref and got $$ \left [\begin{array}{rrr|r} 1&0&1&\frac{1}{2}b_2\\ 0&1&-1&\frac{1}{2}b_1-\frac{1}{2}b_2\\ 0&0&0&\frac{1}{3}b_3-\frac{1}{6}b_1+\frac{1}{4}b_2 \end{array}\right]\;,$$ It shows that it's inconsistent for pretty much all $b$ in $\mathbb{R}^3$ except when all $(b_1,b_2,b_3)=(0,0,0)$

Someone please explain why it's not consistent for all $b$. My row reducing might be a bit off but nevertheless you will get an equation in terms of $b$ in the last column and row of the augmented matrix.

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Your last row is a red flag for consistency.

We have $\displaystyle 0x_1+0x_2+0x_3=\frac{b_3}3-\frac{b_1}6+\frac{b_2}4.$

We know that the left side is ALWAYS equal to $0$, so the right side must be equal to $0$.

Therefore, if $\displaystyle \frac{b_3}3-\frac{b_1}6+\frac{b_2}4 \neq 0$, then this equation is inconsistent and has no solutions.

Notice that means that $b$ cannot lie on to plane $\displaystyle \frac{x_3}3-\frac{x_1}6+\frac{x_2}4=0$, in $\mathbb R^3$.


If $b = (0,0,0)$, then we have $x_1+x_3=0$, and $x_2-x_3=0$. Expressing in terms of $x_3$, which is free, we have $x_1=-x_3$, and $x_2=x_3$, describing the solutions WHEN $b=(0,0,0)$.

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  • $\begingroup$ Thanks I think I get it now $\endgroup$ – TheMaui999 Jun 17 '17 at 4:40
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Your last equation in the final system is $$0x+0y+0z=\frac{b_3}3-\frac{b_1}6+\frac{b_2}4.$$ This can only have a solution when $b_2/3-b_1/6+b_2/4=0$. So the system is inconsistent otherwise, for instance, if $(b_1,b_2,b_3)=(1,0,0)$.

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  • $\begingroup$ but what about when b=(0,0,0) the system is consistent? $\endgroup$ – TheMaui999 Jun 17 '17 at 4:31
  • $\begingroup$ @TheMaui999 A homogeneous system of linear equations is always consistent. $\endgroup$ – Angina Seng Jun 17 '17 at 4:36

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