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3 and 4 .Determinants and Matrices
normal
In a upper triangular matrix $n \times n$, minimum number of zeros is
A
$n(n - 1)/2$
B
$n(n + 1)/2$
C
$2n(n - 1)/2$
D
None of these
Solution
(a) As we know, a square matrix $A=[a_ij]$ is called an upper triangular matrix if ${a_{ij}} = 0$ for all $i > j$.
Such as, $A = {\left[ {\begin{array}{*{20}{c}}1&2&4&3\\0&5&1&3\\0&0&2&9\\0&0&0&5\end{array}} \right]_{4 \times 4}}$
Number of zeros = $\frac{{4(4 – 1)}}{2} = 6 = \frac{{n(n – 1)}}{2}$.
Standard 12
Mathematics
