Question

The number of elements that a square matrix of order $n$ has below its leading diagonal is

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n(n-1)}{2}$
• C. $\frac{(n-1)(n-2)}{2}$
• D. $\frac{(n+1)(n+2)}{2}$

Answer 1

Answer: (B)

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

Given that matrix is a square matrix of order $n$

No.of elements in the first column that are below the leading diagonal$=n-1$

No.of elements in the second column that are below the leading diagonal$=n-2$

No.of elements in the third column that are below the leading diagonal$=n-3$

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No.of elements in the $n^{th}$ column that are below the leading diagonal$=n-n$

$\Rightarrow$Therefore total no.of elements$=(n-1)+(n-2)+(n-3)+....(n-n)$

$\Rightarrow (n+n+n+...ntimes)-(1+2+\mathrm{3....}n)$

$\Rightarrow (n\times n)-\frac{n(n+1)}{2}$

$\Rightarrow n^2-\frac{n^2+n}{2}$

$\Rightarrow \frac{2n^2-n^2-n}{2}$

$\Rightarrow \frac{n^2-n}{2}$

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

Therefore, the total no.of elements in the square matrix of order $n$ that are below the leading diagonal$=\frac{n(n-1)}{2}$