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Determine the value of $b$ for which the system

$$\begin{align} x_1 + 4x_2 − 3x_3 + 2x_4 &= 2\\ 2x_1 + 7x_2 − 4x_3 + 4x_4 &= 3\\ −x_1 − 5x_2 + 5x_3 − 2x_4 &= b\\ 3x_1 +10x_2 − 5x_3 + 6x_4 &= 4 \end{align}$$

is soluble, and determine the solution set.

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    $\begingroup$ Second question on very related subject, and you show no effort or self work, ideas, etc. Not good. $\endgroup$ – DonAntonio Nov 24 '13 at 14:30
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We have this system:

$$\begin{align} x_1 + 4x_2 − 3x_3 + 2x_4 &= 2\\ 2x_1 + 7x_2 − 4x_3 + 4x_4 &= 3\\ −x_1 − 5x_2 + 5x_3 − 2x_4 &= b\\ 3x_1 +10x_2 − 5x_3 + 6x_4 &= 4 \end{align}$$

If you notice that by multiplying the first equation by $(-1)$ and the second one by $2$ and adding these two things together you get precisely the fourth equation $3x_1 +10x_2 − 5x_3 + 6x_4 = 4$, you see, that you can ignore the fourth equation and the original system is equivalent to:

$$\begin{align} x_1 + 4x_2 − 3x_3 + 2x_4 &= 2\\ 2x_1 + 7x_2 − 4x_3 + 4x_4 &= 3\\ −x_1 − 5x_2 + 5x_3 − 2x_4 &= b \end{align}$$

Now if you notice that by multiplying the first row by $(-3)$ and adding the second row you get the equation $−x_1 − 5x_2 + 5x_3 − 2x_4 = -3$ you can see that the system cannot be solvable unless $b=-3$.


Anyway, (one possibility of) the standard solution without guesswork would be using elementary row operation to get $$\begin{pmatrix} 1&4&-3&2&2\\ 2&7&-4&4&3\\ -1&-5&5&-2&b\\ 3&10&-5&6&4 \end{pmatrix} \sim \begin{pmatrix} 1& 0& 5& 2&-2\\ 0& 1&-2& 0& 1\\ 0& 0& 0& 0&b+3\\ 0& 0& 0& 0& 0 \end{pmatrix}$$ (I suppose you are able to make the computations yourself.)

What can you say from the last matrix whether the system is solvable or not (depending on $b$)?

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