Let `A` be a square matrix of order `n`.
`l`= maximum number of different entries if `A` is a upper triangular matrix.
`m`= minimum number of zeros if `A` is a triangular matrix.
`p` = minimum number of zeros if `A` is a diagonal matrix.
If `l+2m=2p+1`, then `n` is :

Options

A. (a) `1`
B. (b) `2`
C. (c) `3`
D. (d) `4`

Answer

To solve the problem, we need to find the values of \( l \), \( m \), and \( p \) based on the definitions provided and then use the equation \( l + 2m = 2p + 1 \) to find \( n \).

### Step 1: Calculate \( l \)
\( l \) is the maximum number of different entries in an upper triangular matrix of order \( n \).

In an upper triangular matrix, all entries below the diagonal are zero. Therefore, the non-zero entries can occupy the following positions:
- First row: \( n \) entries
- Second row: \( n-1 \) entries
- Third row: \( n-2 \) entries
- ...
- Last row: \( 1 \) entry

The total number of non-zero entries is:
\[ l = n + (n-1) + (n-2) + \ldots + 1 = \frac{n(n + 1)}{2} \]
Adding one for the zero entry, we have:
\[ l = \frac{n(n + 1)}{2} + 1 \]

### Step 2: Calculate \( m \)
\( m \) is the minimum number of zeros in a triangular matrix.

In a triangular matrix, all entries above the diagonal are zero. Thus, the number of zeros is:
- First row: \( 0 \) zeros
- Second row: \( 1 \) zero
- Third row: \( 2 \) zeros
- ...
- Last row: \( n-1 \) zeros

The total number of zeros is:
\[ m = 0 + 1 + 2 + \ldots + (n-1) = \frac{(n-1)n}{2} \]

### Step 3: Calculate \( p \)
\( p \) is the minimum number of zeros in a diagonal matrix.

In a diagonal matrix, all entries except the diagonal can be zero. If all diagonal entries are non-zero, the number of zeros is:
\[ p = n^2 - n \]
This is because there are \( n^2 \) total entries and \( n \) of them are non-zero (the diagonal entries).

### Step 4: Set up the equation
Now we substitute \( l \), \( m \), and \( p \) into the equation \( l + 2m = 2p + 1 \):
\[ \left( \frac{n(n + 1)}{2} + 1 \right) + 2\left( \frac{(n-1)n}{2} \right) = 2(n^2 - n) + 1 \]
Simplifying the left side:
\[ \frac{n(n + 1)}{2} + 1 + (n-1)n = \frac{n(n + 1)}{2} + 1 + \frac{2(n^2 - n)}{2} \]
Combining terms:
\[ \frac{n(n + 1) + 2(n^2 - n) + 2}{2} = 2(n^2 - n) + 1 \]
This simplifies to:
\[ \frac{n^2 + n + 2n^2 - 2n + 2}{2} = 2n^2 - 2n + 1 \]
Combining like terms:
\[ \frac{3n^2 - n + 2}{2} = 2n^2 - 2n + 1 \]

### Step 5: Solve for \( n \)
Cross-multiplying gives:
\[ 3n^2 - n + 2 = 4n^2 - 4n + 2 \]
Rearranging:
\[ 0 = n^2 - 3n \]
Factoring:
\[ n(n - 3) = 0 \]
Thus, \( n = 0 \) or \( n = 3 \).

Since \( n \) must be a positive integer, we have:
\[ n = 3 \]

### Final Answer
The value of \( n \) is \( 3 \).