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(B) Given $AB = 0$,where $A$ and $B$ are $3 	imes 3$ non-zero matrices.Taking the determinant on both sides,$|AB| = |0| = 0$.Since $|AB| = |A||B| = 0

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The system of linear equations $AX = 0$ has infinitely many solutions.

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(B) Given $AB = 0$,where $A$ and $B$ are $3 	imes 3$ non-zero matrices.Taking the determinant on both sides,$|AB| = |0| = 0$.Since $|AB| = |A||B| = 0$,it implies that at least one of $|A|$ or $|B|$ must be $0$.If $|A| 
eq 0$,then $A$ is invertible,so $A^{-1}(AB) = A^{-1}(0) \Rightarrow B = 0$,which contradicts the given condition that $B$ is a non-zero matrix.If $|B| 
eq 0$,then $B$ is invertible,so $(AB)B^{-1} = 0(B^{-1}) \Rightarrow A = 0$,which contradicts the given condition that $A$ is a non-zero matrix.Therefore,$|A| = 0$ and $|B| = 0$.Since $|A| = 0$,the matrix $A$ is singular,which means the system of linear equations $AX = 0$ has infinitely many solutions.

Let $A$ and $B$ be two $3 \times 3$ non-zero real matrices such that $AB$ is a zero matrix. Then:

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