Let $AX = B$ be a system of three linear equations in three variables. Then the system has

(A) a unique solutions if $|A| = 0$

(B) a unique solutions if $|A| \neq 0$

(C) no solutions if $|A| = 0$ and (adj A) $B \neq 0$

(D) infinitely many solutions if $|A| = 0$ and (adj A)$B = 0$

Choose the correct answer from the options given below:

For a system of linear equations $AX = B$ with three variables, the determinant $|A|$ determines the nature of solutions.

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Statement (A): Unique solution if $|A| = 0$

When $|A| = 0$, the matrix is singular (non-invertible). A unique solution cannot exist when the determinant is zero.

This statement is incorrect.

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Statement (B): Unique solution if $|A| \neq 0$

When $|A| \neq 0$, the matrix $A$ is invertible.

The solution can be written as:

$X = A^{-1}B$

This gives exactly one unique solution.

This statement is correct.

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Statement (C): No solution if $|A| = 0$ and (adj $A$)$B \neq 0$

Using the property $A \times$ (adj $A$) $= |A| \times I = 0$ when $|A| = 0$

Multiply both sides of $AX = B$ by (adj $A$):

(adj $A$) $\times A \times X =$ (adj $A$) $\times B$

$0 \times X =$ (adj $A$) $\times B$

$0 =$ (adj $A$) $\times B$

If (adj $A$)$B \neq 0$, then $0 =$ (some non-zero value), which is impossible.

Therefore, no solution exists.

This statement is correct.

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Statement (D): Infinitely many solutions if $|A| = 0$ and (adj $A$)$B = 0$

From the same reasoning:

$0 =$ (adj $A$) $\times B$

If (adj $A$)$B = 0$, then $0 = 0$, which is consistent.

When $|A| = 0$ and the system is consistent, the equations are dependent and there exist free variables.

This gives infinitely many solutions.

This statement is correct.

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Statements (B), (C), and (D) are correct.

Statement (A) is incorrect.