Hello and Good day! I am having a bit of difficulty understanding how to go about this problem. Do I attempt to first get the reduced row echelon form of the matrix? If so, how do I go about finding the values for the real numbers as the questions asks to find 5 real numbers(x1,x2,x3,x4, and x5). Any help is much appreciated. Thank you.
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If you subtract the second row from the third, you're left with the equation $$x_3-x_4+2x_5=0$$ or $$x_3=x_4-2x_5$$
Now you can solve for $x_2$ and $x_1$ in terms of $x_4$ and $x_5$. There are infinitely many solutions to this system of linear equations. You can express $3$ of the variables in terms of the other $2$, but that's the best you can hope for.
You are after all solutions of the system$$\left\{\begin{array}{l}x_1+x_2+x_3-x_4+4x_5=0\\x_2+x_3-x_4+3x_5=0\\x_2+2x_3-2x_4+5x_5=0.\end{array}\right.$$If from the third equation you subtract the second one, you get$$\left\{\begin{array}{l}x_1+x_2+x_3-x_4+4x_5=0\\x_2+x_3-x_4+3x_5=0\\x_3-x_4+2x_5=0.\end{array}\right.$$And now you get from the third equation that $x_3=x_4-2x_5$. Then, from the second equation you get that$$x_2=-x_3+x_4-3x_5=-x_5.$$And now, from the first equation you get that$$x_1=-x_2-x_3+x_4-4x_5=-x_5.$$