Question

A second order determinant is written down at random using the numbers 1, - 1 as elements. The probability that the value of the determinant is non zero is

• A. $\frac{1}{2}$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{1}{8}$

Answer 1

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

Let S be the set of second order determinants formed using 1 and -1 as elements. $n\left(S\right)=2^4=16$ (because each of the four places can be filled in 2 ways). The zero determinants are
$\left|\begin{array}{cc}1& 1\\ 1& 1\end{array}\right|,\left|\begin{array}{cc}1& 1\\ -1& -1\end{array}\right|,\left|\begin{array}{cc}-1& -1\\ 1& 1\end{array}\right|,\left|\begin{array}{cc}-1& 1\\ -1& 1\end{array}\right|,\left|\begin{array}{cc}1& -1\\ 1& -1\end{array}\right|,\left|\begin{array}{cc}-1& 1\\ -1& 1\end{array}\right|,\left|\begin{array}{cc}1& -1\\ -1& 1\end{array}\right|,\left|\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right|$
n(E) = 8
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{8}{16}=\frac{1}{2}$