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(B) $2 	imes 2$ determinant has $4$ positions,each can be filled by $0$ or $1$. Thus,the total number of possible determinants is $2^4 = 16$.Let the

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$\frac{3}{8}$

Vedclass · Answer

(B) $2 	imes 2$ determinant has $4$ positions,each can be filled by $0$ or $1$. Thus,the total number of possible determinants is $2^4 = 16$.Let the determinant be $\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$.The determinant 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 $1$.The condition $ad 
eq bc$ holds in the following cases:$1$. $ad = 1$ and $bc = 0$: This implies $a=1, d=1$ and $(b=0$ or $c=0)$. The pairs $(b, c)$ can be $(0, 0), (0, 1), (1, 0)$. There are $3$ such matrices.$2$. $ad = 0$ and $bc = 1$: This implies $b=1, c=1$ and $(a=0$ or $d=0)$. The pairs $(a, d)$ can be $(0, 0), (0, 1), (1, 0)$. There are $3$ such matrices.Total number of matrices with non-zero determinant $= 3 + 3 = 6$.Therefore,the required probability $= \frac{6}{16} = \frac{3}{8}$.

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

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