  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 matrixIf $$l + 5 = p + 2m$$, 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 triangularLets 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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