Question

Find a basis for the column space and rank of the matrix ((-2,-2,-4,1),(7,-3,14,-6),(2,-2,4,-2)(2,-6,4,-3))

Answer 1

Answer:  

• B=\{\left[\begin{array}{c}-2\\-2\\-4\\1\end{array}\right], \left[\begin{array}{c}7\\-3\\14&-6\end{array}\right], \left[\begin{array}{c}2\\-2\\4\\-2\end{array}\right]  \}[/tex]  is a basis for the column space of A.  
• The rank of A is 3.  

Step-by-step explanation:

Remember, the column space of A is the generating subspace by the columns of A and if R is a echelon form of the matrix A then the column vectors of A, corresponding to the columns of R with pivots, form a basis for the space column. The rank of the matrix is the number of pivots in one of its echelon forms.

 Let  $A=\left[\begin{array}{cccc}-2&7&2&2\\-2&-3&-2&-6\\-4&14&4&4\\1&-6&-2&-3\end{array}\right]$  the matrix of the problem.  

Using row operations we obtain a echelon form of the matrix A, that is

$R=\left[\begin{array}{cccc}1&-6&-2&-3\\0&9&2&0\\0&0&-8&-21\\0&0&0&0\end{array}\right]$

Since columns 1,2 and 3 of R have pivots, then a basis for the column space of A is

$B=\{\left[\begin{array}{c}-2\\-2\\-4\\1\end{array}\right], \left[\begin{array}{c}7\\-3\\14&-6\end{array}\right], \left[\begin{array}{c}2\\-2\\4\\-2\end{array}\right] \}$.  

And the rank of A is 3 because are three pivots in R.