ISI2018-MMA-12

Question

The rank of the matrix

$\begin{bmatrix} 1 &2 &3 &4 \\ 5& 6 & 7 & 8 \\ 6 & 8 & 10 & 12 \\ 151 & 262 & 373 & 484 \end{bmatrix}$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

$\begin{bmatrix} 1 & 2 &3 &4 \\ 5&6 & 7 &8 \\ 6& 8 & 10 & 12\\ 151& 262 &373 & 484 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 0& 0 &0 &0 \\ 5&6 & 7 &8 \\ 6& 8 & 10 & 12\\ 151& 262 &373 & 484 \end{bmatrix} \left [ R_{1} \leftarrow R_{1}-\left ( R_{3}-R_{2} \right )\right ]$

$\Rightarrow \begin{bmatrix} 0& 0 &0 &0 \\ 5&1 & 1 &1 \\ 6& 2 & 2 & 2\\ 151& 111 &111 & 111 \end{bmatrix} \left [ C_{1} \leftarrow C_{1}-C_{2}, C_{2} \leftarrow C_{2}-C_{3}, C_{3} \leftarrow C_{3}-C_{4}\right ]$

$\Rightarrow \begin{bmatrix} 0& 0 &0 &0 \\ 5&1 & 0 &0 \\ 6& 2 & 0 & 0\\ 151& 111 &0 & 0 \end{bmatrix} \left [ C_{3} \leftarrow C_{3}-C_{2},C_{4} \leftarrow C_{4}-C_{3}\right ]$

Ans will be $B)2$

Answer 2

There are two dependencies in the matrix

1) R4=101*R1+10*R2

    so R4 could be made zero by the following operation

    $R4\leftarrow R4-(101*R1+10*R2)$

    $\begin{vmatrix} 1 &2 & 3 &4 \\ 5 & 6 & 7 &8 \\ 6& 8 & 10 &12 \\ 0& 0 &0 &0 \end{vmatrix}$

2)  R3=R1+R2

     $R3\leftarrow R3-(R1+R2)$

     $\begin{vmatrix} 1 &2 & 3 &4 \\ 5 & 6 & 7 &8 \\ 0& 0 & 0 &0 \\ 0& 0 &0 &0 \end{vmatrix}$

     Now no more rows can be made zero

    So there are two independent rows

    Rank=2

Answer 3

Elementary row and column operations do not change the rank of the matrix.

Rank of a matrix is the number of linearly independent rows( or columns) in a matrix.

We have,

$\begin{bmatrix} 1& 2& 3& 4\\ 5& 6& 7& 8\\ 6& 8& 10& 12\\ 151& 262& 373& 484\end{bmatrix}$

Applying C4 $\leftarrow$ C4 - C3

$\begin{bmatrix} 1& 2& 3& 1\\ 5& 6& 7& 1\\ 6& 8& 10& 2\\ 151& 262& 373& 111\end{bmatrix}$

Applying C3 $\leftarrow$ C3 - C2

$\begin{bmatrix} 1& 2& 1& 1\\ 5& 6& 1& 1\\ 6& 8& 2& 2\\ 151& 262& 111& 111\end{bmatrix}$

Applying C2 $\leftarrow$ C2 - C1

$\begin{bmatrix} 1& 1& 1& 1\\ 5& 1& 1& 1\\ 6& 2& 2& 2\\ 151& 111& 111& 111\end{bmatrix}$

Applying C4 $\leftarrow$ C4 - C2 and C3 $\leftarrow$ C3 - C2

$\begin{bmatrix} 1& 1& 0& 0\\ 5& 1& 0& 0\\ 6& 2& 0& 0\\ 151& 111& 0& 0\end{bmatrix}$

$\because$ there are only $2$ Linearly Independent columns(C1 and C2) so rank =$2$

$\therefore$ Option $B$ is the right answer.

Comments

Ans will be $2).$ Plz chk my ans :)

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$C_4\leftarrow C_4-C_3$

$C_3\leftarrow C_3-C_2$

$C_2\leftarrow C_2-C_1$

Apply these transformations one by one then u will get
$\begin{bmatrix}1&1&1&1\\5&1&1&1\\6&2&2&2\\151&111&111&111\end{bmatrix}$

$\implies rank=2$

@Satbir

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@Verma Ashish

@srestha

Can you guys please check and tell where I have gone wrong. I am not able to find my mistake.

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I think this line 

Applying R4 ← R4 - (R1+ 50*R2 )

Answer 4

The Rank of the matrix is equal to the number of linear independent columns in the matrix.

We know that , the pivot column (leading entry of non zero row) is linearly independent column.

So we will convert this matrix into Echelon form using Gauss elimination method.

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              $\begin{bmatrix} 1 & 2& 3& 4\\ 5 & 6& 7& 8\\ 6& 8& 10&12 \\ 151& 262& 373&484 \end{bmatrix}$

Steps:-

           1.We will make all the entries in column below a leading entry(here it is 1) ,zero

                  R2 $\leftarrow$ R2 – 5R1

                  R3 $\leftarrow$ R3 – 6R1

                  R4 $\leftarrow$ R4 – 151R1

           

            $\begin{bmatrix} 1 &2 &3 &4 \\ 0&-4 &-8 &-12 \\ 0 & -4 & -8 &-12 \\ 0 & 0& 0& 0 \end{bmatrix}$

 

2. Now make all the entries in column below a leading entry (here it is -4) zero

              R3$\leftarrow$ R3 – R2

 

              $\begin{bmatrix} 1 &2 &3 &4 \\ 0 &-4 &-8 &-12 \\ 0& 0 & 0 &0 \\ 0& 0 &0 &0 \end{bmatrix}$

 

         Now, since there are two Pivot columns (i.e C1 and C2) 

         therefore there are two linear independent coloumn which means the Rank is 2

        $\therefore$ B) 2