Question

Let $x$ be a non-zero real number. $A$ determinant is chosen from the set of all determinants of order $2$ with entries $x$ or $-x$ only. The probability that the value of the determinant is non-zero is

• A. $3/16$
• B. $1/4$
• C. $1/2$
• D. $1/8$

Answer 1

Answer: (A), (C)

$3/16$
$1/2$

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=ad-bc$
Each of the element can take two values, therefore total number of ways $=16$
But $ad-bc\ne 0$ if and only if $ad=x^2$ and $bc=-x^2$ or $ad=-x^2$ and $bc=x^2$
But $ad=x^2$ and $bc=-x^2$ $\iff (a=d=x$ or $a=-d=x)$ and $(b=x,c=-x$ or $b=-x,c=x)$
That is there are four ways
Similarly $ad=-x^2$ and $bc=x^2$ in four ways
Thus, required probability $=\frac{4+4}{16}=\frac{1}{2}$