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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.


Solution

$$A$$ 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 $$L=1+2+3+.........+n-{ \dfrac { n(n+1) }{ 2 }  }$$
$$m=$$ 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 $$m=\underbrace { 1+1+1+........+1 }_{ n\ numbers } $$
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 $$0's={ n }^{ 2 }-\dfrac { n(n+1) }{ 2 } \\ P=\dfrac { 2{ n }^{ 2 }-{ n }^{ 2 }-n }{ 2 } \\ P=\dfrac { { n }^{ 2 }-n }{ 2 } \\ L+5=P+2m\\ \dfrac { n(n+1) }{ 2 } +5=\dfrac { { n }^{ 2 }-n }{ 2 } +2n\\ \dfrac { { n }^{ 2 }+n+10 }{ 2 } =\dfrac { { n }^{ 2 }-n+4n }{ 2 } \\ { n }^{ 2 }+n+10={ n }^{ 2 }+3n\\ 2n=10\\ n=5$$ 
So order of the matrix $$=5$$

Mathematics

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