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(B) determinant of order $2$ is of the form $\Delta = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$.Each element $a, b, c, d$ can be chosen

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$\frac{3}{8}$

Vedclass · Answer

(B) determinant of order $2$ is of the form $\Delta = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$.Each element $a, b, c, d$ can be chosen in $2$ ways ($0$ or $1$).Total number of possible determinants = $2^4 = 16$.The determinant $\Delta$ is non-zero if $ad - bc 
eq 0$,which means $ad 
eq bc$.Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$.$ad - bc = 0$ occurs when $ad = bc$. This happens in the following cases:$1$. $ad = 0$ and $bc = 0$: $ad=0$ occurs in $3$ ways $((0,0), (0,1), (1,0))$,and $bc=0$ occurs in $3$ ways. Total ways = $3 	imes 3 = 9$.$2$. $ad = 1$ and $bc = 1$: $ad=1$ occurs in $1$ way $((1,1))$,and $bc=1$ occurs in $1$ way. Total ways = $1 	imes 1 = 1$.Total cases where $\Delta = 0$ is $9 + 1 = 10$.Total cases where $\Delta 
eq 0$ is $16 - 10 = 6$.Therefore,the required probability = $\frac{6}{16} = \frac{3}{8}$.

$A$ 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 is non-zero is

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