GATE CSE 1994 | Question: 1.9

Question

The rank of matrix $\begin{bmatrix} 0 & 0 & -3 \\ 9 & 3 & 5 \\ 3 & 1 & 1 \end{bmatrix}$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: $C$

Determinant comes out to be $0$. So, rank cannot be $3$. The minor $\begin{bmatrix} 3 & 5 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}\neq0.$ So, rank is $2.$

(OR)

If we do elementary row operations on the given matrix then we get

$\begin{bmatrix}0&0&-3\\9&3&5\\3&1&1\end{bmatrix} \overset{R_2\leftarrow R_2 - 3R_3}{\to} \begin{bmatrix}0&0&-3\\0&0&2\\3&1&1\end{bmatrix}$

$\overset{R_1 \leftarrow R_1 + \frac{3}{2}R_2}{\to} \begin{bmatrix}0&0&0\\0&0&2\\3&1&1\end{bmatrix}\overset{R_1 \leftrightarrow R_3}{\to} \begin{bmatrix}3&1&1\\0&0&2\\0&0&0\end{bmatrix}$

As the number of non zero rows is $2$, the rank of the matrix is also $2.$

Comments

R1-->3/2R2+ R1

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Chk column wise, will give rank 2 in single step

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Rank is also number of LI column vectors.

col1 = 3*col2 and we can’t get col3 using any linear combination of col1 & col2.
(x*col1 + y*col2 = [0 9x 3x]^T + [0 3y y]^T = [0 9x+3y 3x+y]^T = 3*[0 3x+y 3x+y]^T, where x & y are scalar quantity)

So, rank is 2, because there are 2 LI column vectors.

Answer 2

$C_1 \rightarrow C_1- 3C_2$

$\begin{bmatrix} 0&0 &-3 \\ 0&3 &5 \\ 0 &1 &1 \end{bmatrix}$

$Rank = 2$

Answer 3

Answer: 2

Rank can be defined as:

$\equiv $ Linearly Independent Rows

$\equiv $ Linearly Independent Columns

$\equiv $ Pivot elements (count) in Echelon form of the Matrix

$\equiv $ Non-zero rows of Echelon form of the Matrix

Given matrix:

$\begin{bmatrix}
0 & 0 & -3\\ 
9 & 3 & 5\\ 
3 & 1& 1 
\end{bmatrix}$

Let’s convert it to Echelon Form.

$\equiv \begin{bmatrix}
0 & 0 & -3\\ 
9 & 3 & 5\\ 
3 & 1& 1 
\end{bmatrix} \xrightarrow{R_1 \leftrightarrow R_3} \begin{bmatrix}
3 & 1 & 1\\ 
9 & 3 & 5\\ 
0 & 0 & -3 
\end{bmatrix}$

$\equiv \begin{bmatrix}
3 & 1 & 1\\ 
9 & 3 & 5\\ 
0 & 0 & -3 
\end{bmatrix} \xrightarrow[]{R_2 \rightarrow R_2 - 3R_1} \begin{bmatrix}
3 & 1 & 1\\ 
0 & 0 & 2\\ 
0 & 0 & -3 
\end{bmatrix}$

$\equiv \begin{bmatrix}
3 & 1 & 1\\ 
0 & 0 & 2\\ 
0 & 0 & -3 
\end{bmatrix} \xrightarrow[]{R_3 → 3R_2 + 2R_3} \begin{bmatrix}
3 & 1 & 1\\ 
0 & 0 & 2\\ 
0 & 0 & 0 
\end{bmatrix}$

In the resultant matrix, we have two Pivot elements as marked with $\color{Red} red$:

$\begin{bmatrix}
\color{Red} 3& 1 & 1\\ 
0 & 0 & \color{Red} 2\\ 
0 & 0 & 0 
\end{bmatrix}$

Therefore, rank is 2.

Comments

@Sachin Mittal 1 sir we are getting rank as 2 by converting the matrix into echelon form using gauss method .

but if we look at the original matrix then column 1 is 3 times of column 2 . and column 3 is not linearly dependent at all .

Rank = Number of LI column =1 

plz sir clear my doubt ??

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With that approach, we will still have 2 LI Columns in the matrix, isn’t it? Hence the rank=2.

Edit:

As column 1 will be dependent on column2, we can say one LI Column would be column 2   $...(1)$

And column 3 is LI itself  $...(2)$

Answer 4

Do row operation:

R₂→R₂-3R₃

R₁→R₁+(3/2)R₂

Then you a get a matrix whose first row is [0 0 0] and other two row are LI

. Hence rank 2...

 

 

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