Question

$A$ is a square matrix of order $n$.
$l=$ maximum number of distinct entries if $A$ is a triangular matrix
$m=$ maximum number of distinct entries if $A$ is a diagonal matrix
$p=$ minimum number of zeroes if $A$ is a triangular matrix
If $l+5=p+2m$, find the order of the matrix.

Answer 1

$A$ is a square matrix of order $n$

$L=$ maximum number of distinct entries if $A$ is triangular.

Lets assume all the non-zero elements in the triangular matrix are distinct.

So maximum number is $=1+2+3+.........+n$

because in the first column there is one non zero elements, in second column there are two elements and in the $nth$ column $n$ elements.

So here $L=1+2+3+.........+n-\frac{n(n+1)}{2}$

$m=$ maximum number of distinct entries if $A$ is diagonal.

Lets assume all the non zero elements are distinct. So in each column there is one non-zero element so $m=\underset{nnumbers}{\underset{}{1+1+1+........+1}}$

So $m=n$

$P=$ minimum number of zeros in $A$ if $A$ is triangular

Lets assume all the elements in and above or below of the diagonal in the triangular matrix is non-zero.

So minimum number of $0^{\prime }s=n^2-\frac{n(n+1)}{2}P=\frac{2n^2-n^2-n}{2}P=\frac{n^2-n}{2}L+5=P+2m\frac{n(n+1)}{2}+5=\frac{n^2-n}{2}+2n\frac{n^2+n+10}{2}=\frac{n^2-n+4n}{2}{n}^2+n+10=n^2+3n2n=10n=5$

So order of the matrix $=5$