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 the matrix of the problem.
Using row operations we obtain a echelon form of the matrix A, that is
Since columns 1,2 and 3 of R have pivots, then a basis for the column space of A is
.
And the rank of A is 3 because are three pivots in R.