GATE CSE 1998 | Question: 2.1

Question

The rank of the matrix given below is:

$$\begin{bmatrix} 1 &4 &8 &7\\ 0 &0& 3 &0\\ 4 &2& 3 &1\\ 3 &12 &24 &21 \end{bmatrix}$$

• A. $3$
• B. $1$
• C. $2$
• D. $4$

Answer 1

$\begin{bmatrix} 1 &4 &8 &7\\ 0 &0& 3 &0\\ 4 &2& 3 &1\\ 3 &12 &24 &21 \end{bmatrix} = 3\begin{bmatrix} 1 &4 &8 &7\\ 0 &0& 3 &0\\ 4 &2& 3 &1\\ 1 &4 &8 &7 \end{bmatrix} $

$R_1$ and $R_4$ are the same and hence we can remove $R_4$ making the rank surely less than $4$.

$\text{Taking 3 out from $R_2$} \implies  9\begin{bmatrix} 1 &4 &8 &7\\ 0 &0& 1 &0\\ 4 &2& 3 &1 \end{bmatrix} $

${R_{1} \leftarrow R_{1}-8R_{2} \atop{R_{3} \leftarrow R_{3}-3R_{2}} } \implies  9\begin{bmatrix} 1 &4 &0 &7\\ 0 &0& 1 &0\\ 4 &2& 0 &1 \end{bmatrix} $

None of the rows are linearly dependent (we cannot make any of them all $0's$. 

So, Rank will be $\textbf{3}$.

(A) is correct option!

Comments

see this..

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If you have a look at your solution again, you will find that you've jotted down the question wrong.The element at (2,4) is 21( not 2) which would make the first and second rows linearly dependent. Hence rank can't be 4.

It is 3.

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If we take 2 then it would be rank =4 and if we consider 21 then rank would  be 3

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Perfect :)

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Everything is fine except “ None of the rows are linearly independent “. It should be
”none of the rows are linearly dependent “ so the rank is three because if rank is equal to :- number of linearly independent rows = number of linearly independent columns.

@Lakshman Patel RJIT Please edit this answer.

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Fixed now 👍

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As per my understanding, above solution not correct.

common can only be taken from all elements of a matrix.

In determinant , common can be taken out from a row or column.

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@Sachin Mittal 1 Sir in the solution, is this way of taking common from a row of matrix is correct ??

Answer 2

Correct Question -->

The rank of the matrix given below is:

    1    4    8     7
    0    0    3     0
    4    2    3     1
    3   12  24    21

 

 

 

Since R4=3R1 Then Rank != 4

now try for rank of 3

                1  4  8          
                0  0  3   = -3 *   1   4    = -3 * -14 =52
                4  2  3               4   2
  
      here 52 != 0
    So, Rank of the given matrix is = 3

Answer 3

Answer D. 4

To calculate matrix rank transform matrix to upper triangular form using elementary row operations.

	A1	A2	A3	A4
1	1	4	8	7
2	0	0	3	0
3	4	2	3	1
4	3	12	24	2

Multiply the 1st row by 4.  R1->R1×4

	A1	A2	A3	A4
1	4	16	32	28
2	0	0	3	0
3	4	2	3	1
4	3	12	24	2

Subtract the 1st row from the 3rd row and restore it       R3->R3-R1

	A1	A2	A3	A4
1	1	4	8	7
2	0	0	3	0
3	0	-14	-29	-27
4	3	12	24	2

Multiply the 1st row by 3.  R1-->R1×3

	A1	A2	A3	A4
1	3	12	24	21
2	0	0	3	0
3	0	-14	-29	-27
4	3	12	24	2

Subtract the 1st row from the 4th row and restore it.    R4-->R4-R1

	A1	A2	A3	A4
1	1	4	8	7
2	0	0	3	0
3	0	-14	-29	-27
4	0	0	0	-19

Swap the 2nd and the 3rd rows R2<->R3

	A1	A2	A3	A4
1	1	4	8	7
2	0	-14	-29	-27
3	0	0	3	0
4	0	0	0	-19

Calculate the number of linearly independent rows

	A1	A2	A3	A4
1	1	4	8	7
2	0	-14	-29	-27
3	0	0	3	0
4	0	0	0	-19

 

Answer 4

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