$$a_1x+b_1y+c_1z=0\\ a_2x+b_2y+c_2z=0\\ a_3x+b_3y+c_3z=0$$
I'm new to Linear Algebra. But from what i understand isn't there only one possible RREF:
$$\left(\begin{array}{ccc|c} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{array}\right)$$
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$\begingroup$ No. There may be other possible RREF's, for example, the whole matrix could be the $0$ matrix, only first row could be non zero etc. $\endgroup$– voldemortCommented Jan 20, 2015 at 15:44
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$\begingroup$ @voldemort then wouldn't there be a lot of possibilities? $\endgroup$– DanxeCommented Jan 20, 2015 at 15:46
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$\begingroup$ Yes- and that's what the problem asks you. It wouldn't be "lot" as you are dealing with only 3 by 3 matrix. Read the definition and see what matrices are possible. $\endgroup$– voldemortCommented Jan 20, 2015 at 15:48
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2 Answers
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Note that since the RHS is homogenous, it will always be $0$. Also note that we can make all non-diagonal elements $0$ if there is a non-zero pivot, so any RREF will be of the form $$\left(\begin{array}{ccc|c} \ast&.&.&0\\ 0&\ast&.&0\\ 0&0&\ast&0 \end{array}\right)$$ Now if a diagonal element is non-zero, we can scale it to $1$. If it is zero however, we can only swap rows to move it. This means the diagonal is defined by the number of $1$s in it (This is called the rank of the matrix). Therefor all possible RREFs are $$\left(\begin{array}{ccc|c} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{array}\right)\\ \left(\begin{array}{ccc|c} 1&0&.&0\\ 0&1&.&0\\ 0&0&0&0 \end{array}\right)\\ \left(\begin{array}{ccc|c} 1&.&.&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right)\\ \left(\begin{array}{ccc|c} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right)$$
Note that the last is only possible if the original matrix is already the zero-matrix. The $.$ can be any real number.
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Not really, on some conditions you may get RREF with only two or even one pivot, which depends on the entries of coefficient matrix.
