5

I am given the following :

Let $A$ be a $4 \times 4$ matrix with RREF given by: $$ U = \begin{bmatrix} 1 & 0 & 2 & 1 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}. $$ If $$ a_1 = \begin{bmatrix} -3 \\ 5 \\ 2 \\ 1 \end{bmatrix} \text{ and } a_2 = \begin{bmatrix} 4 \\ -3 \\ 7 \\ -1 \end{bmatrix}, $$ find $a_3$ and $a_4$. [Note: $a_1,a_2,a_3,a_4$ are the columns of $A$.]

This is in the section with column, row, and null space. I don't know how those apply no this problem though... I have absolutely no idea how to do this. Can anyone help point me in the right direction?

Bill Cook
  • 29,244
rphello101
  • 345

1 Answers1

5

The (homogeneous) linear relations which hold between the columns of the RREF also hold among the columns of the original matrix.

So your final two columns are precisely $2a_1+a_2$ and $a_1+4a_2$.

$$A = \begin{bmatrix} a_1 & a_2 & a_3 & a_4 \end{bmatrix} = \begin{bmatrix} a_1 & a_2 & 2a_1+a_2 & a_1+4a_2 \end{bmatrix} = \begin{bmatrix} -3 & 4 & -2 & 13 \\ 5 & -3 & 7 & 7 \\ 2 & 7 & 11 & 30 \\ 1 & -1 & 3 & -3 \end{bmatrix}$$

As you preform row operation, the columns are never mixed. So if column 3 is 2 times column 1 plus column 2 in the RREF, it must have been that way in the original matrix.

This means that given the RREF and a list of pivot columns (the columns corresponding to leading non-zero row entries) one can easily reconstruct the original matrix.

This also explains why the pivot columns form a basis for the column space.

Bill Cook
  • 29,244
  • Many texts do a very poor job of explaining this "linear correspondence between columns". To fix this I made the following handout for my intro to linear algebra classes: http://mathsci2.appstate.edu/~cookwj/courses/math3510-spring2013/math2240-fall2012-RREF.pdf – Bill Cook Oct 17 '13 at 19:58
  • Thanks for the response. What is the homogeneous linear relation? – rphello101 Oct 17 '13 at 19:58
  • Thanks for the update. That actually makes a lot of sense... – rphello101 Oct 17 '13 at 20:00
  • Homogenous means each term has the same degree. So for example: $2x+3x=-z+5w$ is homogeneous (& linear) while $x+2y=5$ is not (because of the constant).. My handout addresses this in more detail (starting at page 6). – Bill Cook Oct 17 '13 at 20:01