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For the system of equations AX = B, which of the following is correct?

Solution

✅ Correct Option: 3

For the system of equations AX=BAX = B:

A non-singular matrix has determinant ≠0\neq 0 and an inverse A−1A^{-1} exists.

A singular matrix has determinant =0= 0 and no inverse exists.


When AA is non-singular, the system has a unique solution.

Since A−1A^{-1} exists, multiply both sides by A−1A^{-1}:

A−1(AX)=A−1BA^{-1}(AX) = A^{-1}B

(A−1A)X=A−1B(A^{-1}A)X = A^{-1}B

X=A−1BX = A^{-1}B

This gives exactly one solution.


When AA is singular, the system cannot have a unique solution.

Since A−1A^{-1} does not exist, the system may have either infinitely many solutions or no solution at all, depending on BB.


Checking Statement 1: "If AA is singular, then AX=BAX = B has a unique solution."

This is incorrect. A singular matrix cannot produce a unique solution.


Checking Statement 2: "If AA is non-singular, then AX=BAX = B has infinitely many solutions."

This is incorrect. A non-singular matrix produces exactly one solution X=A−1BX = A^{-1}B, not infinitely many.


Checking Statement 3: "If AA is non-singular, then AX=BAX = B has a unique solution."

This is correct. When AA is non-singular, X=A−1BX = A^{-1}B gives a unique solution.


Checking Statement 4: "If AA is singular, then AX=BAX = B is always consistent."

This is incorrect. A singular matrix does not guarantee consistency; the system may have no solution depending on BB.


Therefore, the correct answer is Option 3.

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