recategorized by
619 views
1 votes
1 votes

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

  1. $\frac{3}{16}$
  2. $\frac{3}{8}$
  3. $\frac{1}{4}$
  4. none of these
recategorized by

3 Answers

3 votes
3 votes
Answer will be there are total 16 possible determinants but there are only 6 possible determinants which are not equal to 0 so answer is 6/16 so 3/8.
2 votes
2 votes

let the matrix be $\begin{bmatrix} a & b\\ c&d \end{bmatrix}$

determinant of matrix$=ad-bc$

since each of a,b,c,d can be 0,1.So determinant can be one of -1,0,1.

we want the determinant to be non zero.

so Required probability=probability that determinant is -1 $+$ probability that determinant is 1.

Also each element has 2 choices.

$\therefore$ There are 16 possible matrices.

probability that determinant is -1:

for the determinant to be -1,$ad=0,bc=1$.

for bc=1,b=1 and c=1 is the only possibility.

for ad=0,there are 3 possibilities:$a=0,d=1$ and $a=1,d=0$and $a=0,d=0$(at least one of a or d has to be 0).

Therefore, there are 3 matrices with determinant=-1.

Hence, probability that determinant$=-1$ is $3/16$

Similarly, there are 3 matrices with determinant=1.(here ad=1,bc=0)

Hence, probability that determinant$=1$ is $3/16$.

Total probability is $3/16+3/16=3/8$

 

0 votes
0 votes

Answer: $A$

With $0$ and $1$, the number of determinants possible $= 2^{4} = 16$ as for every location of $0$ and $1$ there are $4$ choices in total.

Now, There are only $3$ determinants with positive values:

$\begin{bmatrix}1 &0\\0 &1 \end{bmatrix}$, $\begin{bmatrix}1 &1\\0 &1 \end{bmatrix}$, $\begin{bmatrix}1 &0\\1 &1 \end{bmatrix}$

So, the probability of choosing a non-zero determinant $= \frac{\text{Total non-zero determinants possible}}{\text{Total determinants possible}}= \frac{3}{16}$

$\therefore A$ is the correct answer.

Related questions

2 votes
2 votes
2 answers
1
gatecse asked Sep 18, 2019
514 views
If $\begin{vmatrix} 10! & 11! & 12! \\ 11! & 12! & 13! \\ 12! & 13! & 14! \end{vmatrix} = k(10!)(11!)(12!)$, then the value of $k$ is$1$$2$$3$$4$
1 votes
1 votes
1 answer
2
gatecse asked Sep 18, 2019
554 views
If $f(x) = \begin{vmatrix} 2 \cos ^2 x & \sin 2x & – \sin x \\ \sin 2x & 2 \sin ^2 x & \cos x \\ \sin x & – \cos x & 0 \end{vmatrix},$ then $\int_0^{\frac{\pi}{2}} [...
1 votes
1 votes
1 answer
3
gatecse asked Sep 18, 2019
459 views
Let $A_1,A_2,A_3, \dots , A_n$ be $n$ independent events such that $P(A_i) = \frac{1}{i+1}$ for $i=1,2,3, \dots , n$. The probability that none of $A_1, A_2, A_3, \dots ,...