Consider the system of equations ax+by=0, cx+ | vedclass

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(B) The system of equations is homogeneous: $ax+by=0$ and $cx+dy=0$.$A$ homogeneous system always has the trivial solution $(x=0, y=0)$.Therefore,the

Vedclass · Accepted Answer

$Statement-1$ is True,$Statement-2$ is True; $Statement-2$ is $NOT$ a correct explanation for $Statement-1$.

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(B) The system of equations is homogeneous: $ax+by=0$ and $cx+dy=0$.$A$ homogeneous system always has the trivial solution $(x=0, y=0)$.Therefore,the probability that the system has a solution is $1$,which makes $Statement-2$ True.For a unique solution,the determinant of the coefficient matrix must be non-zero: $\left|\begin{array}{ll}a & b \ c & d\end{array}ight| 
eq 0$,which implies $ad - bc 
eq 0$.Since $a, b, c, d \in \{0, 1\}$,the total number of possible outcomes is $2^4 = 16$.The condition $ad - bc 
eq 0$ implies $ad 
eq bc$.Possible cases for $(ad, bc)$ are $(1, 0)$ or $(0, 1)$.If $ad=1$,then $a=1$ and $d=1$. $bc$ must be $0$. $bc=0$ occurs if $(b,c) \in \{(0,0), (0,1), (1,0)\}$. This gives $3$ cases.If $ad=0$,then $bc$ must be $1$,so $b=1$ and $c=1$. $ad=0$ occurs if $(a,d) \in \{(0,0), (0,1), (1,0)\}$. This gives $3$ cases.Total favorable cases $= 3 + 3 = 6$.Probability $= 6/16 = 3/8$. Thus,$Statement-1$ is True.Since $Statement-2$ states that the system always has a solution (which is true for any homogeneous system),it does not explain why the probability of a unique solution is $3/8$. Therefore,$Statement-2$ is not the correct explanation for $Statement-1$.

Consider the system of equations $ax+by=0, cx+dy=0$,where $a, b, c, d \in \{0, 1\}$.
$STATEMENT-1$: The probability that the system of equations has a unique solution is $3/8$.
$STATEMENT-2$: The probability that the system of equations has a solution is $1$.

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Consider the system of equations $ax+by=0, cx+dy=0$,where $a, b, c, d \in \{0, 1\}$.$STATEMENT-1$: The probability that the system of equations has a unique solution is $3/8$.$STATEMENT-2$: The probability that the system of equations has a solution is $1$.