Please do parts 1-4 Problem I Let A denote the matrix given below.(i) Find the rank and the nullity of the matrix A.(ii) Find a basis for the row space of A.(iii) Find a basis for the column space of A, then extend it to a basis for Rª.4.7.10.13.16.19.112211231121 -2 01−1211 -2 02020-11 -2 1 02 -1 010-11 -22-1 -2 1-121 −11 -2 2 1-122 -11−1 1 2 -111 1 -2 12 1 2-2 13 -1 3 2 −110 1 -220 -1 2 -120 1 -2 110 -1 212-1-2 1-1 2 -2 11 1 212 -1 22 -2 1 412 −1 222 -1 412.5.8.11.14.17.20.1 1 -2 -1 11-12 1-121 -2 -12211 -1211211231 -221 -2212112 -121-11 -2-1−1 2 -11 -2 -22-2111 -11 -2-1 -3 -2-1 -2-11 -2 -12 -12 -1-12 -2 -1-2 -2 12 -1-12-2-1 -2 -2-2 -2 -4-1 2-21 -1 -1 2 12-2 1 4 -11 -1 -2 2 2-22 −1 -4 13.6.9.12.15.18.1121-2 10-1 2 -1 01-2 102 -11-111−1122 -22-201-21 22-1 2-2 1 422 0 -11-21 1 -1 222 1 -2 -13 3 −12 -1-2 020-1 21 4 -1-1 4-1-2-1 -1-2 2222422

"Since you have posted multiple questions with multiple sub parts, we will provide the solution only to the three sub parts of the first question as per our Q&A guidelines. Please repost the remaining question by specifying the sub parts need to be answered." The matrix is A=11-2011-120-121-2012-120-1 Changing this to the row reduced echelon form by elementary row transformations: A=11-2011-120-121-2012-120-1 ~11-2010-240-20-120-10-360-3 ~11-2010-120-10-120-10-120-1 ~11-20101-2010000000000Thus, we got two pivot elements as indicated by bold digits. ~11-20101-2010000000000 Hence, Rank of matrix A=Number of pivot elements in row reduced echelon form of matrix A =2 Nullity of matrix A=Number of columns in matrix A -Rank of matrix A =5-2=3The row reduced echelon form of matrix A is 11-20101-2010000000000 First two rows of this matrix are linearly independent. Hence, these two rows will form the basis for the row space of A So, the basis for the row space of A={11-201,0-120-1} As the row reduced echelon form of matrix A has two pivot columns i.e. column first and column second, therefore first two columns of original matrix will form the basis for the column space of A. So, the basis for the column space of A={1122,1-11-1} Extending this basis to a basis for R4: As 11221-11-100100001≠0, hence these rows are linearly independent. So, Extended basis for R4={1122,1-11-1,0010,0001}