Find the basis and dimension for the row space of the matrix \[A=\begin{bmatrix} 2 & 4 & -2 \cr 1 & 2 & -1 \cr 3 & 6 & -3 \end{bmatrix}\]

Step 1 ：Step 1: We begin by row reducing the given matrix. We can subtract half of the first row from the second row and subtract the first row from the third row to get: \[\begin{bmatrix} 2 & 4 & -2 \cr 0 & 0 & 0 \cr 1 & 2 & -1 \end{bmatrix}\]

Step 2 ：Step 2: To simplify further, we can divide the first row by 2 and subtract it from the third row to get: \[\begin{bmatrix} 1 & 2 & -1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \end{bmatrix}\]

Step 3 ：Step 3: We can see that the row reduced matrix has only one non-zero row. So, the basis for the row space of the matrix is the set of vectors formed by the non-zero rows. Hence, the basis of the row space of matrix A is: \[\left\{\begin{bmatrix} 1 \cr 2 \cr -1 \end{bmatrix}\right\}\]

Step 4 ：Step 4: The dimension of the row space is the number of vectors in its basis. So, the dimension of the row space of matrix A is 1.