Question

In a upper triangular matrix $n\times n$, minimum number of zeroes is

• A. $\frac{n(n-1)}{2}$
• B. $\frac{n(n+1)}{2}$
• C. $\frac{2n(n-1)}{2}$
• D. None of these

Answer 1

Answer: (A)

$\frac{n(n-1)}{2}$

As we know a square matrix $A=\left[a_{ij}\right]$ is called an upper triangular matrix, if
$a_{ij}=0$ for all $i>j$
Such as $A={\left[\begin{array}{cccc}1& 2& 4& 3\\ 0& 5& 1& 3\\ 0& 0& 2& 9\\ 0& 0& 0& 5\end{array}\right]}_{4\times 4}$
Therefore, number of zeroes
$=\frac{4(4-1)}{2}=6$

i. e. If $A$ is an upper triangle,

$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots .& a_{1n}\\ 0& a_{22}& ....& a_{2n}\\ ....& ....& ....& ....\\ 0& 0& ....& a_{nn}\end{array}\right]$

then, number of minimum zero is given by $\frac{n(n-1)}{2}$