Question

A determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only. The probability that the determinant chosen in non-zero is

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

Answer 1

The correct option is B $3/8$

A determinant of order 2 is of the form $\Delta =\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$

It is equal to $ad-bc.$

The total number of ways of choosing a, b, c and d is $2\times 2\times 2\times 2=16$.

Now $\Delta \ne 0$ if and only if either $ad=1,bc=0$ or $ad=0,bc=1$.

But $ad=1,bc=0$ if and only if $a=d=1$ and at least one of b, c is zero.

Thus, $ad=1,bc=0$ in three cases$\left[\right(a,b,c,d)=(1,0,0,1\left)\right(1,0,1,1),(1,1,0,1\left)\right]$.

Similarly, $ad=0,bc=1$ in three cases$\left[\right(a,b,c,d)=(0,1,1,0\left)\right(1,1,1,0),(0,1,1,1\left)\right]$.

Thus, the probability of the required event is $\frac{6}{16}=\frac{3}{8}$.