Final answer:
The basis for the row space of matrix A can be found by reducing it to RREF. The non-zero rows of this form are independent and span the row space, thus forming the basis for that space. For matrix A given, the basis is the non-zero rows after row-reduction.
Step-by-step explanation:
To find a basis for the row space of matrix A, we need to reduce the matrix to its row echelon form (REF) or reduced row echelon form (RREF). The non-zero rows of the RREF will form a basis for the row space since they are independent and span the row space of A.
The given matrix A is:
A = [ 1 -3 0 0 -2]
[ 5 -15 0 0 -10]
[ 0 0 -1 2 -7]
[ 0 0 2 -2 6]
We can start by performing row operations: R2 → R2 - 5 × R1, leaving R1 as is because it contains a leading 1.
The resulting matrix will be:
[ 1 -3 0 0 -2]
[ 0 0 0 0 0]
[ 0 0 -1 2 -7]
[ 0 0 2 -2 6]
We continue row operations until we achieve an RREF. Since the second row has become all zeros, we can ignore it in our basis. Moreover, R3 and R4 already have leading 1s at different columns, this indicates they are independent from each other and R1.
Therefore, the basis for the row space of A is given by the non-zero rows of the RREF:
Basis = [ 1 -3 0 0 -2]
[ 0 0 -1 2 -7]
[ 0 0 2 -2 6]
The vectors represented by these rows are independent and span the row space, which means they form the basis. Each vector is a combination of the original row vectors that preserves the row space of matrix A.