Question

The minimum number of zeros in an upper triangular matrix of order $n\times n$ is
(a) $\dfrac{n\left( n-1 \right)}{2}$
(b) $\dfrac{n\left( n+1 \right)}{2}$
(c) $\dfrac{n\left( n-1 \right)\left( n+1 \right)}{2}$
(d) None of these

Answer 1

Hint: Triangular matrix is a special kind of square matrix where all the elements above or below the principal diagonal are zero. A matrix that has all the elements below the principal diagonal as zero is an upper triangular matrix. An upper triangular matrix as $\left[ {{A}_{ij}} \right]=0$ for all $i>j$ , where I is the row and j is the column. We have to find the number of zeros in each row and add them to get the required answer.

Complete step by step answer:
We know that a triangular matrix is a special kind of square matrix where all the elements above or below the principal diagonal are zero. A matrix that has all the elements below the principal diagonal as zero is an upper triangular matrix.
We can represent an upper triangular matrix as $\left[ {{A}_{ij}} \right]=0$ for all $i>j$ , where i is the row and j is the column.
An upper triangular matrix is represented as
 \[A=\left[ \begin{matrix}
   {{a}_{1,1}} & {{a}_{1,2}} & {{a}_{1,3}} & {{a}_{1,4}} & ... & {{a}_{1,n}} \\
   0 & {{a}_{2,2}} & {{a}_{2,3}} & {{a}_{2,4}} & ... & {{a}_{2,n}} \\
   0 & 0 & {{a}_{3,3}} & {{a}_{3,4}} & ... & {{a}_{3,n}} \\
   .. & .. & .. & .. & .. & .. \\
   .. & .. & .. & .. & .. & .. \\
   0 & 0 & 0 & 0 & 0 & {{a}_{n,n}} \\
\end{matrix} \right]\]
Therefore, number of zeroes in the first row =0
Number of zeroes in the second row $={{A}_{2,1}}=1$
Number of zeroes in the second row $={{A}_{3,1}},{{A}_{3,2}}=2$
Likewise, the number of zeroes in the ${{n}^{th}}$ row $=\left( n-1 \right)$
Therefore, total number of zeroes $=1+2+3+...+\left( n-1 \right)$
We know that the sum of n natural numbers is given by $\dfrac{n\left( n+1 \right)}{2}$ .
Therefore, we can find the sum of $\left( n-1 \right)$ natural numbers by substituting n as $\left( n-1 \right)$ in the above formula.
$\begin{align}
  & \Rightarrow S=\dfrac{\left( n-1 \right)\left( n-1+1 \right)}{2} \\
 & \Rightarrow S=\dfrac{n\left( n-1 \right)}{2} \\
\end{align}$
Therefore, the minimum number of zeros in an upper triangular matrix of order $n\times n$ is $\dfrac{n\left( n-1 \right)}{2}$ .

So, the correct answer is “Option a”.

Note: Students have a chance of getting confused with upper and lower triangular matrices. A lower triangular matrix is a matrix where all the elements above the principal diagonal are 0. We can verify whether the minimum number of zeros in an upper triangular matrix of order $n\times n$ is $\dfrac{n\left( n-1 \right)}{2}$ by considering a square matrix of order, say $3\times 3$ .
$A=\left[ \begin{matrix}
   2 & 8 & 4 \\
   0 & 3 & 7 \\
   0 & 0 & 9 \\
\end{matrix} \right]$
Let us substitute $n=3$ in the formula $\dfrac{n\left( n-1 \right)}{2}$ .
$\begin{align}
  & \Rightarrow \text{Number of zeroes}=\dfrac{3\left( 3-1 \right)}{2} \\
 & \Rightarrow \text{Number of zeroes}=\dfrac{3\times 2}{2} \\
 & \Rightarrow \text{Number of zeroes}=3 \\
\end{align}$
Hence, verified.