# 2×2matrix having elements 0 and 1 is selected at random.probability that it's a non singular matrix?

May 3, 2018

$\frac{3}{8}$

#### Explanation:

First of all, we must observe that there are 16 possible matrices: we have to fill ${a}_{1 , 1} , {a}_{1 , 2} , {a}_{2 , 1} , {a}_{2 , 2}$ with either $0$ or $1$, so we have four spots and two choices for each spot, leading to ${4}^{2} = 16$ matrices in total.

Let's think about the number of zero elements:

1. if all elements are zero, then the determinant is surely zero
2. if three elements are zero, then we have a zero row and a zero column. Both of these conditions are sufficient for the determinant to be zero
3. if two elements are zero, for the same reason, they can't be aligned vertically or horizontally.
4. If one element is zero, the matrix is invertible
5. If all elements are ones, then the matrix has two identical rows/columns, and the determinant is again zero

So, by the point (4), you have to chose the spot to be zero, and you have four choices:

$01$
$11$

$10$
$11$

$11$
$01$

$11$
$10$

Plus, the two diagonal matrix in point (3):

$10$
$01$

$01$
$10$

So, $6$ matrices out of the $16$ possible combinations means $\frac{6}{16}$ probability, which simplifies to $\frac{3}{8}$.