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18 votes

The rank of the matrix given below is:

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

  1. $3$
  2. $1$
  3. $2$
  4. $4$
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3 Answers

27 votes
27 votes
Best answer

$\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!

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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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4 votes
4 votes
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
3 votes
3 votes
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:

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