MXB106 Linear Algebra Workbook 2 - Semester 1, 2019 - Solution

This document presents the solutions to the Linear Algebra Workbook 2 for the MXB106 course, Semester 1, 2019. The assignment focuses on two main problems. The first problem involves finding the row reduced echelon form of a given matrix, determining the number of linearly independent co...Read More

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.

Apply
Apply
From last matrix we get
So,
Hence, basis for null space is .
b): The null space of A is . Let be the vector orthogonal to null
space of A, that is
Solve we get the set of orthogonal vector as
Now consider a vector fro null space of A as and a vector from orthogonal set
of vector as . Given
Suppose
So,
Since, above system is inconsistent. So we can not the sum of a vector in the null space of
A and a vector orthogonal to the null space of A.