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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
in Calculus
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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.
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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 $3/16+3/16=3/8$

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.

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I think the meaning of non-zero determinant is either positive or negative but not equal  to zero..... So why determinants not consider???

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