Find the basis and dimension for the row space of the matrix $A=\left[\begin{array}{ccccccccccccc}1& 2& 3& 42& 4& 6& 83& 6& 9& 124& 8& 12& 16\end{array}\right]$

Step 1 ：First, we perform row operations to put the matrix in row echelon form. We begin by subtracting 2 times the first row from the second row, 3 times the first row from the third row, and 4 times the first row from the fourth row to get $\left[\begin{array}{ccccccccccccc}1& 2& 3& 40& 0& 0& 00& 0& 0& 00& 0& 0& 0\end{array}\right]$

Step 2 ：Next, we can see that the second, third, and fourth rows are all zero vectors, so we eliminate them to get $\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]$

Step 3 ：The basis for the row space of A is the set of all non-zero rows in the row echelon form of A, which in this case is $\{\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]\}$

Step 4 ：The dimension of the row space of A is the number of vectors in its basis, which in this case is 1