Solution 1: Given matrix is.

Solution 1: Given matrix is .
Let’s first find the row reduced echelon form of above matrix. Perform elementary row
operations.
Apply
Apply
a): From above rref(A) matrix, it is observe that the 1st and 3rd columns contains pivot
elements. So, these two columns are linearly independent. And hence, total number of
linearly independent column of matrix A is 2.
b): Solve that is we need to solve
. This gives,
So,
Hence, basis for the null space of A is .
c): The vector b is the sum of the four columns of A that is
Then the general solution to is
, where are arbitrary.
Solution 2: Given vector
And the matrix
Let’s solve
The augmented matrix is
Let’s perform elementary row operations.