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The rank of the matrix $\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$ is

1. $4$
2. $2$
3. $1$
4. $0$

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Rank of this matrix is 1 as the determint of 2nd order matrix is 0 and 1st order matrix is non zero so rank is $1$.

Correct Answer: $C$

If u ever doubt in exam that rank can also be zero or not, then Only Null matrix has rank zero.
if we check no. of non zero rows , then the ranks becomes 1.   which way we should decide rank and what is the correct answer.
Rank of a matrix is also defined as number of non-zero rows in echelon form.

And the given matrix is already in Echelon form.

$\begin{pmatrix} 1 &1 \\ 0 & 0 \end{pmatrix}$
$\mathbf{\mid A_{n\times n} \mid} \neq 0\:\:\textbf{iff}$ rank of the marix $\mathbf{A}$ is $n.$

$\textbf{(OR)}$

$\mathbf{\mid A_{n\times n} \mid} \neq 0\:\:\Longleftrightarrow$ rank of the marix $\mathbf{A}$ is $n.$
Rank Of A Matrix Can be easily calculated by Calculating Order of Submatrix Having Determinant Not Equal to zero in the above submatrix of size 2*2 determinant turns out to be zero hence rank cannot be One but if we see for 1*1 submatrix we have [1] as submatrix whose determinant is not equal to zero hence Rank is 1 .
Eigen Value of a upper triangular matrix = Principal diagonal elements.

The above matrix is a upper triangular matrix, thus the Eigen Value of the above matrix is 1 and 0 (diagnal elemnts).

Number of non zero Eigen values of a matrix = Rank of that matrix.

Therefore the rank of above matrix=1.

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