A2 = [1 2 3; 4 5 6; 7 8 9; 3 2 4; 6 5 4; 9 8 7]b2 = [1 1 1 1 1 1]′ (a) Find a basis for the row space of A2. (b) How many solutions does A2 ?

Answer:Remember, a basis for the row space of a matrix A is the set of rows different of zero of the echelon form of A. We need to find the echelon form of the matrix augmented matrix of the system A2x=b2[tex]B=\left[\begin{array}{cccc}1&2&3&1\\4&5&6&1\\7&8&9&1\\3&2&4&1\\6&5&4&1\\9&8&7&1\end{array}\right][/tex]We apply row operations:1. To row 2 we subtract row 1, 4 times.To row 3 we subtract row 1, 7 times.To row 4 we subtract row 1, 3 times.To row 5 we subtract row 1, 6 times.To row 6 we subtract row 1, 9 times.We obtain the matrix[tex]\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&-6&-12&-6\\0&-4&-5&-2\\0&-7&-14&-5\\0&-10&-20&-8\end{array}\right][/tex]2.We subtract row two twice to row three of the previous matrix.we subtract 4/3 from row two to row 4.we subtract 7/3 from row two to row 5.we subtract 10/3 from row two to row 6.We obtain the matrix[tex]\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&0&0\\0&0&3&2\\0&0&0&2\\0&0&0&2\end{array}\right][/tex]3.we exchange rows three and four of the previous matrix and obtain the echelon form of the augmented matrix.[tex]\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&3&2\\0&0&0&0\\0&0&0&2\\0&0&0&2\end{array}\right][/tex]Since the only nonzero rows of the augmented matrix of the coefficient matrix are the first three, then the set [tex]\{\left[\begin{array}{c}1\\2\\3\end{array}\right],\left[\begin{array}{c}0\\-3\\-6\end{array}\right],\left[\begin{array}{c}0\\0\\3\end{array}\right] \}[/tex]is a basis for Row (A2)Now, observe that the last two rows of the echelon form of the augmented matrix have the last coordinate different of zero. Then, the system is inconsistent. This means that the system has no solutions.