Does a zero determinant mean the system has no solution?

Here is a question very close to JEE Advanced 2024 Paper 1 Q6, which tests this misconception head-on:

The misconception answer reasons "there's a determinant condition lurking, so maybe it's singular and unsolvable." The correct move is to compute the determinant of the coefficient matrix $\begin{pmatrix} a & b \\ b & c \end{pmatrix}$:

A non-zero determinant means the matrix is non-singular, so the system has exactly one solution. The determinant condition here guarantees a unique solution rather than denying one.

The rule "$\det A \ne 0$ gives a unique solution" is genuinely true, so students over-extend it into a tidy-looking opposite: "$\det A = 0$ gives no solution." It is half-right, which is exactly why it survives. The missing third of the picture is the infinite-solution family: a singular system whose equations are consistent has infinitely many solutions, not none.

Kazunga and Bansilal (2020), analysing matrix-algebra misconceptions through an APOS lens, found students routinely conflate invertibility with solvability and never construct the infinite-solution family for themselves. Because the two-branch rule ("unique or none") feels symmetric and complete, simply being told the three-branch rule rarely dislodges it. What dislodges it is building two systems with the same zero determinant — one with infinitely many solutions, one with none — and seeing the determinant cannot tell them apart.

If $\det A \ne 0$, the matrix is non-singular and there is exactly one solution. If $\det A = 0$, the matrix is singular and the solution is not unique — now check consistency. The consistency check is simple: scale one equation to match another's coefficients and see whether the constants agree. If they agree, the second equation is redundant and you have infinitely many solutions; if they contradict, you have none.

Formally, for $A\mathbf{x} = \mathbf{b}$ the system is consistent precisely when $\operatorname{rank}(A) = \operatorname{rank}([A\,|\,\mathbf{b}])$. With a singular $A$, consistency gives infinitely many solutions; inconsistency gives none. A homogeneous system $A\mathbf{x} = \mathbf{0}$ is always consistent (the zero vector works), so a singular homogeneous system always has infinitely many solutions — never none.

Build two systems that share the same singular coefficient matrix $\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$, whose determinant is $(1)(4) - (2)(2) = 0$.

1. System 1 (consistent). $x + 2y = 3$, $2x + 4y = 6$. Doubling the first equation gives $2x + 4y = 6$, exactly the second. The equations agree, so there are infinitely many solutions: $(3,0)$, $(1,1)$, $(-1,2)$, and so on.
2. System 2 (inconsistent). Keep the same matrix but change the constant: $x + 2y = 3$, $2x + 4y = 7$. Doubling the first gives $2x + 4y = 6$, but the second demands $7$. Since $2x + 4y$ cannot equal both 6 and 7, there is no solution.
3. Compare. Both coefficient matrices have determinant zero, yet System 1 has infinitely many solutions and System 2 has none. The determinant is identical across the two, so it cannot be what decides between them. Only the constants — the consistency check — separate the branches.
4. Conclude. A zero determinant means "not unique", not "no solution". Reach for
   $$\det A \ne 0 \Rightarrow \text{unique}; \qquad \det A = 0 \Rightarrow \text{not unique, then check consistency.}$$

The same framework settles JEE Advanced 2024 Paper 1 Q14: a $3\times3$ skew-symmetric matrix has $\det = 0$ (odd order), so it is singular; the homogeneous-type system is consistent, giving an infinite solution set — never "no solution".