Find the basis and dimension for the row space of the following matrix: \[\begin{pmatrix} 1 & 2 & 3 \ 2 & 4 & 6 \ 3 & 6 & 9 \end{pmatrix}\]
Answer
Step 3: The remaining row is a basis for the row space. The dimension is the number of vectors in the basis, which is 1 in this case.
Step 1 ：Step 1: Reduce the matrix to row-echelon form. This can be achieved by subtracting the first row from the second and third rows, and then dividing the second and third rows by 1 and 2 respectively: \[\begin{pmatrix} 1 & 2 & 3 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}\]
Step 2 ：Step 2: Remove the rows of zeros, since they do not contribute to the basis of the row space: \[\begin{pmatrix} 1 & 2 & 3 \end{pmatrix}\]
Step 3 ：Step 3: The remaining row is a basis for the row space. The dimension is the number of vectors in the basis, which is 1 in this case.
