edited by
3,944 views

3 Answers

Best answer
20 votes
20 votes
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: $C$
edited by
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 votes
1 votes
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:

Related questions

28 votes
28 votes
5 answers
1
Kathleen asked Sep 15, 2014
4,550 views
Obtain the eigen values of the matrix$$A=\begin {bmatrix} 1 & 2 & 34 & 49 \\ 0 & 2 & 43 & 94 \\ 0 & 0 & -2 & 104 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$
39 votes
39 votes
8 answers
2
Kathleen asked Sep 15, 2014
17,664 views
The maximum number of edges in a $n$-node undirected graph without self loops is$n^2$$\frac{n(n-1)}{2}$$n-1$$\frac{(n+1)(n)}{2}$
32 votes
32 votes
5 answers
3
Kathleen asked Sep 15, 2014
13,673 views
In the absolute addressing mode:the operand is inside the instructionthe address of the operand in inside the instructionthe register containing the address of the operan...
21 votes
21 votes
1 answer
4
Kathleen asked Sep 15, 2014
8,014 views
The optimal page replacement algorithm will select the page thatHas not been used for the longest time in the pastWill not be used for the longest time in the futureHas b...