First 10 prime numbers are

={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Let A is a 2 $$\times$$ 2 matrix,

$$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$$

Given that matrix A is singular.

$$\therefore$$ | A | = 0

$$ \Rightarrow \left| {\matrix{ a & b \cr c & d \cr } } \right| = 0$$

$$ \Rightarrow ad = bc$$

Case I :

ad = bc condition satisfy when a = b = c = d.

For ex when a = 2, b = 2, c = 2, d = 2, then ad = bc satisfy.

Now there are 10 prime numbers.

We can choose any one of the 10 prime number in $${}^{10}C_{1}$$ = 10 ways and put them in the four positions of the matrix and matrix will be singular.

$$\therefore$$ In this case, total favorable case = 10

Case 2 :

ad = bc condition satisfies when

(1)

a = 2, d - 3 then

(a) b = 2, c = 3

(b) b = 3, c = 2

or

a = 3, d = 2 then

(a) b = 2, c = 3

(b) b = 3, c = 2

So you can see for two different prime number for a and d there are 4 possible value of b and c which satisfy ad = bc condition.

Two different values of a and d can be chosen from 10 prime numbers = $${}^{10}C_{2}$$ ways

And for each combination of a and d there are 4 possible values of b and c.

$$\therefore$$ Total possible values = $${}^{10}C_{2}$$ $$\times$$ 4

From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition

= 10 + $${}^{10}C_{2}$$ $$\times$$ 4

For sample space,

Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number = $${}^{10}C_{1}$$ ways

Similarly,

For element b of matrix A = $${}^{10}C_{1}$$ ways

For element c of matrix A = $${}^{10}C_{1}$$ ways

For element d of matrix A = $${}^{10}C_{1}$$ ways

$$\therefore$$ Sample space = $${}^{10}C_{1}$$ $$\times$$ $${}^{10}C_{1}$$ $$\times$$ $${}^{10}C_{1}$$ $$\times$$ $${}^{10}C_{1}$$ = 10⁴

$$\therefore$$ Probability $$ = {{10 + {{}^{10}C_{2}} \times 4} \over {{{10}^4}}}$$

$$ = {{10 + 180} \over {{{10}^4}}}$$

$$ = {{190} \over {{{10}^4}}}$$

$$ = {{19} \over {{{10}^3}}}$$