Determinants & Their Properties

Every square matrix hides one number that tells you almost everything you’d want to know about it. Can it be undone? By how much does it stretch space? Are its rows secretly saying the same thing? One number — the determinant — answers all three. (In practice it is the tripwire for a singular matrix — the situation where a linear model has no unique fit because two predictors carry the same information.)

Intuition — the area scale factor

Think of a 2x2 matrix as a transformation of the plane. It takes the unit square (the box spanned by (1,0) and (0,1)) and maps it to a parallelogram. The determinant is the area of that parallelogram — the factor by which the transformation scales area. A determinant of 3 triples areas; a determinant of 1 preserves them; a determinant of 0 collapses the square flat onto a line, so its area vanishes. In 3D the same story holds with volume instead of area.

Drag the entries of the matrix below and watch the parallelogram breathe. Its signed area is the determinant — make the columns collinear and the area collapses to zero, exactly the singular case. A negative reading means the transform flipped orientation.

Mechanism — how to compute it

For a 2x2 matrix the formula is one line — main diagonal product minus the anti-diagonal product:

For a 3x3 matrix, use cofactor expansion along the first row. Each entry is multiplied by the 2x2 determinant of the rows and columns that remain, with the sign pattern + − +:

det([[a, b, c],      | e  f |        | d  f |        | d  e |
     [d, e, f],  = a·| h  i |  −  b· | g  i |  +  c· | g  h |
     [g, h, i]])

  = a(ei − fh) − b(di − fg) + c(dh − eg)

You can expand along any row or column (matching the sign pattern) — pick the one with the most zeros to save work.

The properties that actually get tested

These few facts answer most exam questions without any heavy computation:

• Product rule: det(AB) = det(A)·det(B). Determinants turn matrix products into ordinary multiplication.
• Transpose: det(Aᵀ) = det(A) — transposing never changes the determinant.
• Powers: det(Aᵏ) = det(A)ᵏ, a direct consequence of the product rule.
• Scalar multiple: for an n×n matrix, det(cA) = cⁿ · det(A) — the scalar comes out once per row, so it is raised to the nth power.
• The big equivalence: det(A) = 0 if and only if the matrix is singular (non-invertible), if and only if its rows/columns are linearly dependent, if and only if 0 is an eigenvalue. All four statements say the same thing.

That last bullet is the heart of the topic. A zero determinant means the transformation flattened space — it lost a dimension — so no inverse can undo it.

How GATE asks this

Usually a NAT: either compute a 2x2 or 3x3 determinant directly, or reason about a det(AB)-style product without grinding out a full expansion. The 2024 paper gave a 3x3 matrix M with linearly dependent rows and asked for det(M² + 12M) — a question that is trivial once you spot that dependent rows force det(M) = 0. The trick is always to factor a determinant out, never to multiply the matrices.

Worked example

Part A — a clean 2x2. Compute det([[2, 1], [3, 4]]).

det = (2)(4) − (1)(3) = 8 − 3 = 5

The parallelogram spanned by the columns has area 5, and since it is nonzero the matrix is invertible.

Part B — the 2024-style trap. A 3x3 matrix M has linearly dependent rows (say, row 3 is row 1 plus row 2). Find det(M² + 12M).

Do not multiply anything. Factor M out of the expression first:

M² + 12M = M·M + 12M = M·(M + 12I)

det(M² + 12M) = det( M·(M + 12I) )
              = det(M) · det(M + 12I)      ← product rule

Because M has linearly dependent rows, it is singular, so det(M) = 0. Therefore the whole product is:

det(M² + 12M) = 0 · det(M + 12I) = 0

The answer is 0, and we never needed the entries of M. Spotting the dependent rows — hence det(M) = 0 — and factoring with the product rule is the entire solution.

Quick check