(30 pts) If we let A be the matrix 1 2 3 11 1 4 0 1 2 0 2 3 0 1 I 0 0 0 0 find bases for the image of A, the kernel of A and the row space of A. Verify the rank nullity theorem in this case. . Given matrix A : 12 3' A= 1 4 O 1 2 O O 0 finding the PREF . AReduced Row Echelon form) of A. performing now operations to transform A into its reduced row echelon form CRREF ) osubtracting the first row from the second row: . . 2 0 2 30 0 O Osubtracting the first jow from the fourth dow . 2 -3 0 2 - 3 O O - 2 -3 -1 -! subtracting the Second row from the third down 1423 1 1 in .0 0 . ol 0 0 10 - 2, 1-3 -1 -11- finding bases for the image of A. the Pivot columns correspond , to the tint, second and fourth columns of A: So, a bases for the image of A is the set of first , second , and fourth columns of A: Basis for Image of A: = finding bases for the kernel of A: The free variables correspond to the third and fifth columns of A. So, a bases for the bemel of A is the set of the third and fifth columns of ? . ' .. Bases for Kemel of A: 1 Bases for the Row space of A : The non-zero rows in the RRef of A correspond To a bases for the row space of 1 . so , the basis for the yow space of A! is the Set of the first and second row of" A: Base for Row space of A: s( 23 1 1), (0 . 2. 73 0 1) 3 - :, verifying the Rank - Nullity theorem The rank of A is the number of pivot columns in the RREF , which is 3 . The nullity of A is the number of free variables , which is 2 . . The sum of the rank and nullity of A is 3+2 =5. which is the total number of columns in 4 . + Therefore; the rank- nullity theorem holds true for the matrix A