test series

Question

@sachinmittal1  ​​​​​sir, please verify this table it is for maybe condition.

Comments

@sachinmittal1

Answer 1

We are given a matric A with m rows and n columns.

Ax$\large ^{\overrightarrow{}}$=B$\large ^{\overrightarrow{}}$ and B is a non zero vector.

Case 1.When m<n :

Lets take m =2 and n=3.We have 2 rows which means the three vectors are in R$\large ^{2}$.Since we have more than n linearly independent vectors in R$\large ^{n}$ space so these vectors are always linearly dependent.
But if we have exactly 2 linearly independent vectors among the three we can fill the entire space of  R$\large ^{2}$ ,then we can genearate any B$\large ^{\overrightarrow{}}$ but at the same time we will have a free variable so we will get infinitely many solutions.But if B$\large ^{\overrightarrow{}}$ is not in the linear combination of the columns of A then no solution exists foor the system.

Case 2.When m=n :

Lets take m=3 and n=3.So we have 3 rows in  R$\large ^{3}$.If these three vectors are linearly independent so they will fill the entire space of  R$\large ^{3}$ and we have a solution for any B$\large ^{{\overrightarrow{}}}$ and it will be a unique solution.If the three vectors aren’t linearly independent then we need to check if B$\large ^{{{}\overrightarrow{}}}$ is inn the linear combination of the columns of A.If it happens then we have infinite solutions else no solution is possible.

Case 3. When m>n:

Lets take m=3 and n=2 So we have 2 columns in R$\large ^{3}$ .So we can never fill the entire space of  R$\large ^{3}$ .But if these two column vectors are linearly independent we can get unique solutions for the system iff B$\large ^{\overrightarrow{}}$ is in the linear combination of the columns of A,But if these vectors are linearly dependent then we will have infinitely many solutions given B$\large ^{\overrightarrow{}}$ is in the linear combination of the columns of A;else we will not have a solution for this system of equations.

 

Comments

Nice explanation.