A) \[3\times 3\]
B) \[4\times 4\]
C) \[5\times 5\]
D) \[10\times 10\]
Correct Answer: B
Solution :
[b] Number of zeroes in a lower triangular matrix of order \[n\times n\] is \[1+2+3+....+n=\frac{n(n+1)}{2}\] Number of zeroes = 10 \[\Rightarrow \frac{n(n+1)}{2}=10\] \[\Rightarrow {{n}^{2}}+n=20=0\] \[\Rightarrow (n+5)(n-4)=0\] \[\Rightarrow n=4\] or \[-5\] (-5 is meaningless) \[\Rightarrow n=4.\Rightarrow \] order of the matrix is \[4\times 4\]
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