GO Classes 2024 | Weekly Quiz 7 | Linear Algebra | Question: 6

Answer 1

Matrix A has 3 vectors and each of them belongs to $R^{4}$ space.

It is given that if solution exists for $Ax=b$ then it will be infinite solution, which means columns of A are linearly dependent and hence $rank (A) \leq 2$

it is also given that for some b no solution exists for $Ax=b$ and this means b is not a zero vector and A is zero matrix

So the possible values of $rank (A) $ are  $0, 1, 2$

Answer 2

Matrix A has 4 rows and 3 columns. A=[x x x

                                                                       x x x

                                                                       x x x

                                                                       x x x]

Looking at the two possibilities given:

     1) For some given b, Ax =b is not solvable: which means there is no solution → [000 | non-zero] row exists in the last row of the echelon form of the augmented matrix [A|b]. Thus the last row of A in its echelon form is [000]

     2) Solutions are not unique when they exist: which means there is an infinite solution → Echelon form of A has at least one free column. Therefore, number of pivot columns A can have is <=2. And the number of pivot columns in echelon A is the same as the rank of A

 

Therefore, Rank(A) < = 2 : List of all possibilities of Rank(A)=0,1,2

Comments

Echelon form of A has at least one free column.

 Therefore, number of pivot columns A can have is <=2.

  How can you conclude the 2nd statement from the 1st statement ? Even if Echelon has at least 1 free column, how does it make number of pivot columns less than equal to 2 ?

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@Swarnava Bose use rank-nullity theorem.It states that no of free variables+no. of pivot variables =no of columns.

We know that rank=no. of linearly independent columns=no of pivot variables in echelon form of matrix.

Here in this question it is 4*3 matrix,hence it has 3 columns.

Since echelon form of matrix has alteast one free columns,using the above theorem,we get

1+no of pivot variables=3

Hence,pivot variables is <=2.

Answer 3

Three possibilities 0,1,2

Rank=0 means A is zero matrix so exist a solution for b if its 0 vector

Rank=1,2 infinite solutions because of free variable column (not unique)

Rank=3 not possible because it will give unique solution.  Also 3 linearly independent vectors in space R4 which won’t give solution for all b . ( [0 0 0] = [non zero] )

Answer 4

A belongs to R4 space ,AX=b is not solvable for some b means  neither A has 4 LI column nor b is linear combination of column of A. and When AX=b solution exist it is not unique means it has infinite many solution i.e. atleast one vector must be Linear dependent.

so rank(A) is not full rank.  
rank of A could be 0,1,2