in Linear Algebra edited by
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16 votes
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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. $0$
in Linear Algebra edited by
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3 Answers

20 votes
20 votes
Best answer
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: $C$
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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.

$\begin{pmatrix} 1 &1 \\ 0 & 0 \end{pmatrix}$
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6
$\mathbf{\mid  A_{n\times n} \mid} \neq 0\:\:\textbf{iff}$ 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 $n.$
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0
20 years ago, life was so simple...:D
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0
2 votes
2 votes
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.

 

1 vote
1 vote
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 .
Answer:

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