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

The correct option is D $\frac{1}{2}$

$n\left(S\right)=$ sample space $=2^4=16$
(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
only two entries are negative ${}^4{\text{C}}_2=6$ case
all four entries are negative =${}^4{\text{C}}_2=1$case
$n\left(A\right)=8$ like cases
$\stackrel{\underset{}{\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)}}{onecase},\stackrel{\underset{}{\left(\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right)}}{onecase},\stackrel{\underset{}{\left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)}}{6case}$ so on.
$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{8}{16}=\frac{1}{2}$