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(A) determinant of order $2$ is given by $\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$. Each element can be either $0$ or $1$. Since there

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$3/16$

Vedclass · Answer

(A) determinant of order $2$ is given by $\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$. Each element can be either $0$ or $1$. Since there are $4$ positions,the total number of possible determinants is $2^4 = 16$. The value of the determinant is $ad - bc$. For this to be positive,$ad - bc > 0$,which means $ad > bc$. Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$. $ad > bc$ is possible only if $ad = 1$ and $bc = 0$. This implies $a=1, d=1$ and $bc=0$. The condition $bc=0$ is satisfied if $(b, c)$ is $(0, 0), (0, 1),$ or $(1, 0)$. Thus,the favorable cases are: $1. \begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} = 1(1) - 0(0) = 1$ $2. \begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix} = 1(1) - 0(1) = 1$ $3. \begin{vmatrix} 1 & 1 \ 0 & 1 \end{vmatrix} = 1(1) - 1(0) = 1$ There are $3$ favorable outcomes. Therefore,the probability is $\frac{3}{16}$.

$A$ determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only. The probability that the value of the determinant chosen is positive is:

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