Question

A second order determinant is written randomly using the numbers 1 and - 1 as its elements, the probability that the value of the determinant is non zero, is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{8}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$n\left(S\right)=2^4=16$
($\because$ There are four places to fill with 1 and -1 i.e., every place can be filled by 2 ways).
$n\left(A\right)=$ Number of determinants whose value is zero
$\therefore n\left(A\right)=$ It can be happened if all entry are +ve = 1 case
only two entries are negative $={}^4{C}_2=6$ cases
all four entries are negative $={}^4{C}_4=1$ case
$n\left(A\right)=8$ like cases

$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{8}{16}=\frac{1}{2}$