If the least number of zeroes in a lower triangular matrix is 10, then what is the order of the matrix ?

Options

A. `3 xx 3`
B. `4 xx 4`
C. `5 xx 5`
D. `10 xx 10`

Explanation

To solve the problem, we need to determine the order of a lower triangular matrix given that the least number of zeros in it is 10.

### Step-by-Step Solution:

1. **Understanding Lower Triangular Matrix**:
A lower triangular matrix is a square matrix where all the entries above the main diagonal are zero. For example, in a 3x3 lower triangular matrix, the structure looks like this:
\[ \begin{bmatrix} a_{11} & 0 & 0 \\ a_{21} & a_{22} & 0 \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]
Here, the entries above the diagonal (i.e., \(a_{12}, a_{13}, a_{23}\)) are all zero.

2. **Counting the Zeros**:
The total number of entries in an \(n \times n\) lower triangular matrix is given by the formula:
\[ \text{Total entries} = n^2 \]
The number of non-zero entries in a lower triangular matrix is equal to the number of entries on and below the diagonal, which is:
\[ \text{Non-zero entries} = \frac{n(n + 1)}{2} \]
Thus, the number of zeros in the matrix can be calculated as:
\[ \text{Zeros} = n^2 - \frac{n(n + 1)}{2} \]

3. **Setting Up the Equation**:
We know from the problem that the least number of zeros is 10. Therefore, we can set up the equation:
\[ n^2 - \frac{n(n + 1)}{2} = 10 \]

4. **Simplifying the Equation**:
To eliminate the fraction, multiply the entire equation by 2:
\[ 2n^2 - n(n + 1) = 20 \]
Simplifying this gives:
\[ 2n^2 - n^2 - n = 20 \implies n^2 - n - 20 = 0 \]

5. **Factoring the Quadratic**:
Now we can factor the quadratic equation:
\[ (n + 5)(n - 4) = 0 \]
This gives us two potential solutions:
\[ n + 5 = 0 \quad \text{or} \quad n - 4 = 0 \]
Thus, \(n = -5\) or \(n = 4\).

6. **Determining the Order of the Matrix**:
Since the order of a matrix cannot be negative, we discard \(n = -5\) and accept:
\[ n = 4 \]

### Conclusion:
The order of the matrix is \(4\).