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(B) For a homogeneous system $AX=O$,if the system has non-trivial solutions (infinite solutions),then the determinant of the matrix $A$ must be zero,i

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$2$

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(B) For a homogeneous system $AX=O$,if the system has non-trivial solutions (infinite solutions),then the determinant of the matrix $A$ must be zero,i.e.,$|A| = 0$. This implies that the rank of $A$ must be less than the number of variables,which is $3$. Given that $X = [l, m, 0]^T$ is a solution with $l, m 
eq 0$,the solution space contains at least one non-zero vector. Since the solution is of the form $k[l, m, 0]^T$,the dimension of the null space (nullity) is at least $1$. By the Rank-Nullity Theorem,$	ext{rank}(A) + 	ext{nullity}(A) = n$,where $n=3$. If the nullity is $1$,then $	ext{rank}(A) = 3 - 1 = 2$. If the nullity were $2$,the solution space would be a plane,but the given form suggests a line of solutions. Thus,the rank of $A$ is $2$.

Consider a homogeneous system of three linear equations in three unknowns represented by $AX=O$. If $X=\left[\begin{array}{c}l \\ m \\ 0\end{array}\right]$,where $l \neq 0, m \neq 0, l, m \in \mathbb{R}$,represents an infinite number of solutions of this system,then the rank of $A$ is:

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