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I have been trying to bring a matrice to RREF but I have run into trouble, and I can't really see where I mess up:

Let a = -3/2. Bring the matrix to RREF and use it to solve the system of equation

So I have my Matrix:

$\left[\begin{array}{ccc|c} 2 & 0 & -\frac{3}{2}a+4 & \frac{1}{2}a+2 \\ 0 & 1 & -4 + \frac{1}{2}a & -\frac{1}{2}a + 2 \\ 0 & 0 & 3-2a & 0 \end{array}\right]$

Then I have tried to do the following:

Insert value for a:

$\left[\begin{array}{ccc|c} 2 & 0 & -\frac{3}{2}(-\frac{3}{2})+4 & \frac{1}{2}(-\frac{3}{2})+2 \\ 0 & 1 & -4 + \frac{1}{2}(-\frac{3}{2}) & -\frac{1}{2}(-\frac{3}{2}) + 2 \\ 0 & 0 & 3-2(-\frac{3}{2}) & 0 \end{array}\right]$ = $\left[\begin{array}{ccc|c} 2 & 0 & \frac{25}{4} & \frac{5}{4} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 6 & 0 \end{array}\right]$

Multiply row 1 with $\frac{1}{2}$

$\left[\begin{array}{ccc|c} 2*(\frac{1}{2}) & 0 &\frac{25}{4}*(\frac{1}{2}) & \frac{5}{4}*(\frac{1}{2}) \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4}\\ 0 & 0 & 6 & 0 \end{array}\right]$ = $\left[\begin{array}{ccc|c} 1 & 0 & \frac{25}{8} & \frac{5}{8} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 6 & 0 \end{array}\right]$

Multiply row 3 with $\frac{1}{6}$

$\left[\begin{array}{ccc|c} 1 & 0 & \frac{25}{8} & \frac{5}{8} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 6(\frac{1}{6}) & 0 \end{array}\right]$ = $\left[\begin{array}{ccc|c} 1 & 0 & \frac{25}{8} & \frac{5}{8} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 1 & 0 \end{array}\right]$

This is where I am stuck. I am sure I must have messed up somewhere, but I cant seem to figure it out.

If I just replace row 1 and 2 with row 3 times the given fraction, then I get all zeroes and I should be good.

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  • $\begingroup$ When you plug the value $a=-3/2$ you have two mistakes: $-3/2(-3/2)+4=25/4$ and $-1/2(-3/2)+2=11/4$. $\endgroup$
    – Edu
    Jul 31, 2017 at 15:16
  • $\begingroup$ Oh of course. Not negative! $\endgroup$
    – DonutSteve
    Jul 31, 2017 at 15:18

1 Answer 1

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Putting $a=-3/2$ gives $$ A=\left[\begin{array}{rrr|r} 2 & 0 & \frac{25}{4} & \frac{5}{4} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 6 & 0 \end{array}\right] $$ Now, consider the row-reductions \begin{align*} \left[\begin{array}{rrr|r} 2 & 0 & \frac{25}{4} & \frac{5}{4} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 6 & 0 \end{array}\right] &\xrightarrow{R_1\to (1/2)\cdot R_1}\left[\begin{array}{rrr|r} 1 & 0 & \frac{25}{8} & \frac{5}{8} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 6 & 0 \end{array}\right] \\ &\xrightarrow{R_3\to(1/6)\cdot R_3}\left[\begin{array}{rrr|r} 1 & 0 & \frac{25}{8} & \frac{5}{8} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 1 & 0 \end{array}\right] \\ &\xrightarrow{R_1\to R_1+(-25/8)\cdot R_3}\left[\begin{array}{rrr|r} 1 & 0 & 0 & \frac{5}{8} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 1 & 0 \end{array}\right] \\ &\xrightarrow{R_2\to R_2+(19/4)\cdot R_3}\left[\begin{array}{rrr|r} 1 & 0 & 0 & \frac{5}{8} \\ 0 & 1 & 0 & \frac{11}{4} \\ 0 & 0 & 1 & 0 \end{array}\right] \end{align*} Hence $$ \DeclareMathOperator{rref}{rref}\rref\left[\begin{array}{rrr|r} 2 & 0 & \frac{25}{4} & \frac{5}{4} \\ 0 & 1 & -\frac{19}{4} & \frac{11}{4} \\ 0 & 0 & 6 & 0 \end{array}\right]=\left[\begin{array}{rrr|r} 1 & 0 & 0 & \frac{5}{8} \\ 0 & 1 & 0 & \frac{11}{4} \\ 0 & 0 & 1 & 0 \end{array}\right] $$

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  • $\begingroup$ Yes I figured that out in the end! Thank you though! $\endgroup$
    – DonutSteve
    Jul 31, 2017 at 20:26

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