Question

Let $AX=B$ be a system of n-linear equations in n unknowns then

• A. System is consistent and infinitely many solution if $|A|=0.$
• B. System is inconsistent and finitely many solution if $|A|=0$.
• C. System is consistent and unique solution if $|A|\ne0.$
• D. System is consistent if $|A|=0$ and (adj A) $B\ne0$.

Hint:

Cramer's Rule and Matrix Inverse method only work when $|A| \neq 0$. If you see $|A| \neq 0$ in a question about consistency, "unique solution" is almost always the intended answer.

Answer 1

The Correct Option is C

We are analyzing the nature of solutions for a square system of linear equations $AX=B$ based on the determinant of the coefficient matrix $A$.

Step 1: \color{redAnalyze the case where $|A| \ne 0$
When the determinant of the coefficient matrix is non-zero, the matrix $A$ is non-singular and invertible.
We can pre-multiply the equation $AX=B$ by $A^{-1}$ to get $X = A^{-1}B$.
Since the inverse is unique, the system will always have exactly one unique solution.
A system that has at least one solution is called consistent.
Therefore, if $|A| \ne 0$, the system is consistent and has a unique solution.

Step 2: \color{redAnalyze the case where $|A| = 0$ and $(adj~A)B \ne 0$
When $|A|=0$, the matrix is singular and does not have an inverse.
In this scenario, we check the product of the adjoint of A and the constant matrix B.
If $(adj~A)B \ne 0$, the equations are contradictory, and the system has no solution.
A system with no solution is called inconsistent.
Statement (4) is therefore incorrect because it claims the system is consistent in this case.

Step 3: \color{redAnalyze the case where $|A| = 0$ and $(adj~A)B = 0$
If both the determinant is zero and the product $(adj~A)B$ is the zero matrix, the system is usually consistent and has infinitely many solutions.
However, statement (1) is incomplete because it only mentions $|A|=0$ without the adjoint condition.
Statement (2) is incorrect because an inconsistent system cannot have "finitely many" solutions; it must have zero solutions.

Step 4: \color{redConclusion
Comparing our findings with the options, statement (3) is a fundamental theorem of linear algebra.
If $|A| \neq 0$, the system is guaranteed to be consistent and have a unique solution.
This matches Option (3).