GATE CSE 2021 Set 2 | Question: 24

Question

Suppose that $P$ is a $4 \times 5$ matrix such that every solution of the equation $\text{Px=0}$ is a scalar multiple of $\begin{bmatrix} 2 & 5 & 4 &3 & 1 \end{bmatrix}^T$. The rank of $P$ is __________

Comments

Detailed Video Solution – Here

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What I understand is that rank(A)= # pivots in the echelon form of a matrix.

As, the given matrix is of 5 unknown variable but with 4 equations, so,  $\exists$ 1 free variable.

So, the # pivot is (5 – 1)= 4. The rank is 4.

Q. How this # LI or LD eigenvector is relevant to this question?

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see simple apply rank nullity theorem :

rank + nullity = n(no of Dim)

okay, now as to question there's matrix 4*5 mean m*n as to that n is 5 also remember no.of column also represent dimension.
then in question there's has been mentioned that there only of scalar multiple which you can call only one vector space .
remember :

no.of vector space = no.of Null space(Nullity)

okay so we get 1 nullity.
then now apply theorem:-
we get,
r + 1 = 5
r = 5 - 1
r=4

THIS IS ANOTHER WAY OF SOLVING THIS QUESTION HOPE IT HELPED YOU!!

Answer 1

Every solution to $Px = 0$ is scalar multiple of $\begin{bmatrix} 2 & 5 & 4 &3 & 1 \end{bmatrix}^T$, It means out of 5 column vectors of matrix $P$,  $4$  are linearly independent as we have only one line in NULL Space (along the given vector).

Rank is nothing but the number of linearly independent column vectors in a matrix which is $4$ here.

Comments

When we say that Ax=0 has some non trivial solution then that means c1v1+c2v2+...+cnVn=0 and here at least one of the ci is not zero as solution is non trivial. So A will be a linearly dependent matrix.

@Anuj2000

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Concept of the question-

“Every solution is a scalar multiple of $[2 \ 5\  4\  3\  1]^T$”

It means, $[2 \ 5\  4\  3\  1]^T$” is the only linearly independent solution.

There is a difference between Linearly Independent Solution and Linearly Independent Column.

# of LI Solution = # free variables = Nullity = 1 (here)

While, #LI Columns = # pivot elements = Rank

Hence, Rank = $n – \ nullity = 5-1 = 4$ (Ans).

 

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Nice Solution

Answer 2

Rank $+$ Nullity = Number of Columns

Here, Nullity is $1$. (Nullity is the dimension of the null space)

Rank : $5\ – 1 = 4$

Comments

Nullity is . (Nullity is the dimension of the null space)

How please explain??

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I am think that way please check its correct or not:

Here matrix is 4×5

He ask Rank of matrix:

Minimizing the matrix, doing operations.

Rank = minimum ( row, column)

         = minimum (5,4)

        = 4

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it‘s not always right.

Because Rank(A) <= min(row,column) is the right condition.

So, we can’t presume that rank will always be equals to min(row, column).

here it can be anything less than or equal to 4. so, we need to consider other properties and take up to the right answer. Here, by considering nullity to get the rank of the matrix in the given question.

here, Nullity is 1, and by considering the dimension theorem

rank(P)  + nullity(P) = number of dimension(P)

rank(P) = number of dimension(P) – nullity(P)

rank(P) = 5 – 1 = 4.

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Very well explained. But I have one suggestion - along with “Nullity is the dimension of the null space”, you can also mention the fact that “Nullity of a matrix A is no. of possible column vectors X satisfying the equation AX = 0”. It will be more intuitive for some people.

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how nullity is one

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@Anuj,

Nullity is the dimension of the null space

in other words,

Nullity of a matrix A is no. of possible column vectors X satisfying the equation AX = 0

In the given question, we are interested in finding the nullity of P. So we need to find out the number of possible column vectors X satisfying the equation PX = 0. In the question, it is also given that

every solution of the equation PX=0 is a scalar multiple of [25431]$^T$

this means that X is of the form k [25431]$^T$ where k is a scalar. Clearly, the Null Space of P contains only 1 vector. Hence Nullity is 1.

You can also refer to this video for better understanding.

Answer 3

This problem is based on Linear Homogeneous equations (System of Linear Equations)

Comments

simple and nice explanation

Answer 4

it is in form PX=0.

P is of order 4X5.  4 EQUATIONS AND 5 UNKNOWNS(5 variables)

n=5.

solution of PX=0 is

k  [2 5 4 3 1] transpose

here we have only 1 arbitary value.(that is only 1 scalar i.e.,K)

no. of arbitary values(scalars)= n-r

1=5-r

r=4