Gaussian ellimination2 on the matrix A a1 a2a6 1 0 2 2 0 1

Gaussian ellimination^2 on the matrix A = (a_1, a_2,...,a_6) = (1 0 2 2 0 1 0 0 3 1 6 6 -1 1 -1 -1 1 2 1 2 -2 1 -1 -1) produces the matrix B = (b_1,b_2,...,b_6) = (1 0 0 0 0 1 0 0 3 1 0 0 0 0 1 0 0 0 0 1 0 -2 3 0) In this question, you do not need to explain your answers. rank(A) =^3dim(N(A)) = dim(N(A^T)) dim(Row(A)) = dim(Row(B)) = Is Row(A) = Row(B)? Is Col(A) = Col(B)? List a basics B for the column space of A. (Use the name of the conditions rather than writing the entries.) Find the coordinate vector [a_6]D Find a basis for N(A)

Solution

To answer the questions relatedd to matrix A we need to change the matrix A into Row reduced Echelon Form doing the following Elelmentary Row operations as follows :

The last matrix Obtained is in Row reduced Echelon form i.e.

The rank of Matrix A = number of leading ones in Row reduced echelon form.

Here number of leading ones are 4.

So rank(A) = 4

b) To find dimension of Null Space :

We first augment the matrix with a column containing all zeros.

The reduced row echelon form of the augmented matrix is (doing by above method) :

which corresponds to the system

A leading entry on the (i,j) position indicates that the j-th unknown will be determined using the i-th equation.

Those columns in the coefficient part of the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary.

The system has infinitely many solutions:

The solution can be written in the vector form:

c₃ +

c₆ +

c₇

The nullity of the matrix A is 3. This is the dimension of the null space. It equals the number of vectors in null space of A.

Row Operation 1:

1 0 3 -1 1 -2 0 1 1 1 2 1 2 0 6 -1 1 -1 2 0 6 -1 2 -1

add -2 times the 1st row to the 3rd row

1 0 3 -1 1 -2 0 1 1 1 2 1 0 0 0 1 -1 3 2 0 6 -1 2 -1