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I have a system of linear equations that has no solutions.

It looks like this in augmented matrix form.

\begin{array}{ccc|c}1&-1&2&5\\2&1&0&1\\1&8&-1&3\\-1&-5&-12&41\end{array}

I need to replace right hand value in equation 4 (41) with a value that will give me a unique solution when in reduced row echelon form. How do I approach this problem in a logical way to find the possible values?

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    $\begingroup$ Replace $41$ by a parameter $a$, and then compute the reduced row-echelon form. Then we obtain $a+37=0$ for a unique solution. $\endgroup$ – Dietrich Burde Jul 6 '17 at 14:49
  • $\begingroup$ @DietrichBurde thank you! $\endgroup$ – Max Jul 6 '17 at 15:17
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solving the first three equation we obtain $$x_1=\frac{4}{27},x_2=\frac{19}{27},x_3=\frac{25}{9}$$ plugging this in the left-hand side of our System in the last equation we get $$-\frac{4}{27}-5\cdot \frac{19}{27}-12\cdot \frac{25}{9}=...$$ Can you finish?

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  • $\begingroup$ Wow, the solution is a lot simpler than I expected! Thank you! $\endgroup$ – Max Jul 6 '17 at 15:18
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Replacing the value 41 with an arbitrary value $x$ the augmented matrix would be: $$\begin{array}{ccc|c}1&-1&2&5\\2&1&0&1\\1&8&-1&3\\-1&-5&-12&x\end{array}$$ Now reduce the matrix to its echelon form: $$\begin{array}{ccc|c}1&-1&2&5\\0&3&-4&-9\\0&0&9&25\\0&0&0&x+37\end{array}$$ For this system to be consistent, the rank of the coefficient matrix should equal to rank of augmented matrix (the rank is the maximum number of non zero rows in echelon form, obviously), and furthermore, to have unique solution, the number of variables involved should also be equal to the rank of coefficient/augmented matrix.

For this, $x+37 = 0$ which implies $x = -37$.

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  • $\begingroup$ So in cases like this one, it is enough to just reduce it until the row in question has all zeroes? $\endgroup$ – Max Jul 6 '17 at 15:33
  • $\begingroup$ that would not be always zero. For instance, if your question contained only first three rows and the entry you had to replace was $a_{3,4}$, then you would say the entry $a_{3,4}$ in the echelon form would not be zero, as that row in the coefficient matrix is not all zero. It would be better to say there would be equal number of non-zero rows in the coefficient matrix as in augmented matrix which in turn is equal to the number of variables involved. $\endgroup$ – bikalpa Jul 6 '17 at 15:44

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