ISRO CSE 2014 | Question: 72

Question

The rank of the matrix $A=\begin{pmatrix} 1 & 2 & 1 & -1\\ 9& 5& 2& 2\\ 7& 1& 0 & 4 \end{pmatrix}$  is ______ .

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Comments

tell if any mistake there in matrix. i will edit and post answer.

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$\begin{pmatrix} 1 & 2 & 1 & -1\\ 9& 5& 2& 2\\ 7& 1& 0 & 4 \end{pmatrix}$ check it now.

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ccorrect

Answer 1

Rank of Matrix is 2.

That's 2.

Comments

@ cse7 .Check this lecture.

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are you sure that once you once you find  ROW ECHELON FORM no need to reduce further??

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https://gateoverflow.in/exam/161/linear-algebra-gate2020-previous-gate-1

In this exam, the answer is given as option B. Kindly look into it.

@Lakshman Bhaiya

Answer 2

R3 + 2R1 and then R3-R2. We get last row as zero. Hence, rank is 2.

Answer 3

Here is the answer :

Answer 4

Simply...
Do like...
Converting R3 --> (R2 - 2*R1) - R3...!

Which leads us to...
[1  2  1  -1
 9   5  2  2
 0  0  0  0]
Now, no further reduction is possible...! {Because, 2nd row isn't the scalar multiple of 1st one}
Hence,  as The rank of the matrix is the number of non-zero rows of the Echelon Form, the rank of the matrix given is 2

Answer 5

Very simple Question let's see the Explanation 

Answer 6

step 1 : convert in in Echelon form using row reduce method and then you can clearly check that hwo much rank is there