How to find an original matrix after it has been changed to reduced echelon form
I'm not sure how to start this problem. If you are given a matrix A that is in reduced echelon form, and given the first, second and forth columns of the original matrix, how do you find the original matrix? I tried using unknowns with the known 1st 2nd and 4th columns, but ended up with a wrong answer. Does anyone have suggestions? For example, lets say
A $= \begin{pmatrix} &1 &0 &0 &-1 &2 \\ &0 &1 &0 &4 &2 \\ &0 &0 &1 &4 &2 \end{pmatrix}$
and we know the first, second and fourth columns of the original are $\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} and \begin{pmatrix} 4 \\ 6 \\ 8 \end{pmatrix}$ respectively. I tried solving it by plugging unknowns into the other columns and row reducing from there and solving for the unknowns, but this didn't work for some reason. Does anyone have any ideas of how to do this?
$\endgroup$3 Answers
$\begingroup$1) Pivot on the third row, fourth column, so that now columns 1,2,4 form the identity matrix. Get matrix B.
2) Form a matrix $C$ from the three given vectors.
3) Multiply CB.
Hint: $\exists E$ so that $E\left(\matrix{2&2&4\\1&1&6\\0&1&8}\right)=\left(\matrix{1&0&-1\\0&1&4\\0&0&4}\right)$. This represents the combination of elementary row operations that were applied. However, we need to apply $E^{-1}$ to the other columns in the result to obtain what they were originally. Where's $E^{-1}$?
$\endgroup$ 3 $\begingroup$I got $$ \left( \begin{matrix} -1/2 \\ 3/4 \\ 1 \end{matrix} \right) \quad\quad \left( \begin{matrix} 7 \\ 11/2 \\ 4 \end{matrix} \right) $$ for the third and fifth columns from using using variables for the unknown columns and bringing this into echelon form, and then comparing the component terms with the result of your echelon form and solving for the unknowns.
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