I have row reduced the matrix and got
$$\left(\begin{array}{cccc|c} 1 & 1 & -6 & 4 &6\\ 0 & 1 & -3 & 4 & 4\\ 0 & 0& 24& -10 & -10 \end{array}\right)$$

but the general solution is $(2+15t,-33t,5t,1+12t)$ how can it be?


Let $x_3=u$. Then from $24u - 10x_4 = -10$ it follows that $x_4 = 1 + \frac{12u}{5}$. Similarly, you can used your reduced matrix to determine values for $x_1$ and $x_2$. Writing $u=5t$ gets you rid of the denominators.

Why did we start with $x_3$? There is no special reason, but it makes sense to start with an unknown that occurs in the row of the reduced matrix having the most zeroes. We could have started with $x_4$ just as easy.


We start from the last equation and work our way upwards, simply because the last equation involves the least number of unknowns.

Let $x_4=a$. Substituting in the last equation gives us $x_3=\frac{5(a-1)}{12}$

Substituting these two values in the second last equation gives us $x_2=\frac{-11(a-1)}{4}$

Similarly, from the first equation, $x_1=\frac{3+5a}{4}$

Now, observe that 3 of these values are fractions and two of them involve $(a-1)$. So we intuitively see that we can replace $(a-1)$ by something to get an integral value.

Now, what should we replace it by? $a$ should be of the form $x+1$ to make $(a-1)$ a single number we can reduce further. What should this $x$ be? Observe intuitively that the denominators of the expressions having $(a-1)$ have the lcm $12$. So we pick this $x$ as $12t$.

So, $x_4=a=1+12t$ :)


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