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I seem to be having trouble with solving the following system of equations so that an infinite number of solutions would arise. The question is "Find the values of $p$ and $q$ for which the following system of equations has an infinite number of solutions and clearly explain my reasoning:" $$2x+y+z=5, x-y+z=3, -2x+py+2z=q$$ What I've managed to do is convert it as an augmented matrix and tried to solve from there: $$\left[\begin{array}{rrr|r} 2 & 1 & 1 & 5\\ 1 & -1 & 1 & 3 \\ -2 & p & 2 & q\end{array}\right]$$ However, when I tried to solve the augmented matrix, I ended up with a solution that has $p$ on both sides in the third row. The answer in my textbook says that: "$t(p+10)=q+2$ has infinitely many solutions for $t$ when $p+10=0$ and $q+2=0, {\therefore}{p=-10, q=-2}$."

I can't seem to find that solution, no matter what I tried, so any help with the matrix to reach that solution would be greatly appreciated!!

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  • $\begingroup$ Can you show us your work? $\endgroup$ – Rumplestillskin Jun 23 '17 at 3:04
  • $\begingroup$ find p such that the third row (to the left of the | ) find a combination of rows 1 and 2, that equals the fixed values in row 3. i.e. $-4(2,1,1|5) + 6(1,-1,1|3) = (-2,-10,2|-2) = (-2,p,2|q)$ $\endgroup$ – Doug M Jun 23 '17 at 3:14
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Begin with the augmented matrix $$ \left( \begin{array}{ccc|c} 2 & 1 & 1 & 5\\ 1 & -1 & 1 & 3 \\ -2 & p & 2 & q\end{array} \right) $$ and row reduce this by

  1. multiplying the first row by $-1/2$ and adding it to the second row, and
  2. adding the first row to the last row

to get:
$$ \left( \begin{array}{ccc|c} 2 & 1 & 1 & 5\\ 0 & -\frac{3}{2} & \frac{1}{2} & \frac{1}{2} \\ 0 & p+1 & 3 & 5+q\end{array} \right). $$ Then multiply the second row by $-2/3$ to get: $$ \left( \begin{array}{ccc|c} 2 & 1 & 1 & 5\\ 0 & 1 & -\frac{1}{3} & -\frac{1}{3} \\ 0 & p+1 & 3 & 5+q\end{array} \right). $$ Next,

  1. multiply the second row by $-1$ and add it to get first row, and
  2. multiply the second row by $-(p+1)$ and add it to the third row:

$$ \left( \begin{array}{ccc|c} 2 & 0 & \frac{4}{3} & \frac{16}{3}\\ 0 & 1 & -\frac{1}{3} & -\frac{1}{3} \\ 0 & 0 & 3+\frac{p+1}{3} & 5+q+\frac{p+1}{3} \end{array} \right). $$ The equation corresponding to the last row in the augmented matrix is $$\left(3+\frac{p+1}{3}\right)z= 5+q+\frac{p+1}{3}.$$ Multiply both sides by three to obtain: $$ (p+10)z = 15+3q+p+1, $$ which is $$ (p+10)z = p+3q+16. $$ So what we want to do is find $p$ and $q$ so that $p+10=0$ and $p+3q+16=0$. This would mean that the system of equations will have an infinite number of solutions.

So $p=-10$ and $q=-2$.

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  • $\begingroup$ Could you show me how you managed to put it into the reduced row echelon form? Thanks! $\endgroup$ – Maths Matador Jun 23 '17 at 4:33
  • $\begingroup$ @W.Nguyen I have updated my solution. $\endgroup$ – Mee Seong Im Jun 23 '17 at 20:43

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