let A= set of 3x3 determinat having entries +1 or -1, if a determinat A₁ is chosen randomly from the set of A,then the probability that the product of the elements of any row or any column of A₁ is -1 is _

I am not sure, but here is my best guess.

Sample Space = 2^9

(There are 9 positions, each position has two possiblities (-1 or +1) )

Favourable Outcomes :

Number of -1 in each row or coloumn should be odd.

So we get two cases here.

(i) No. of -1 in each row and coloumn is 3

(ii) No. of -1 in each row and coloumn is 1

Case 1:

This case can occur only when all the elements in the matrix is -1.

No. of favourable cases = 1

Case 2:

Consider the matrix row wise.

In the first row, you can put the -1 in any of the 3 places.

In the second row, you can put the -1 in two places (no below the previous one)

In the third row, you can put the -1 in only one place (remaining one, not below the other two).

So you get 3*2*1 = 6 possibilities in this case.

Total no. of favourable outcome = 6+1 = 7

So required probability is 7/2⁹ ..

Okay I missed a case.

All digits in a single row are -1. Rest of the elements are 1.

So we get 3 more cases.

So Probability = 10/2^9