Determinants & Their Properties

The last lesson left us wanting one number, computed once, that tells whether a square matrix’s columns are independent — whether Ax = b has its single unique answer. There is such a number. It tells you more besides:

• whether this matrix can be undone;
• by how much it stretches space; and
• whether its rows are secretly saying the same thing.

One number — the determinant — answers all of it.

What it measures — a scaling of area

Think of a 2×2 matrix as a transformation of the plane. It takes the unit square (the box spanned by (1,0) and (0,1)) and sends 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;
• 1 leaves them alone; and
• 0 collapses the square flat onto a line, so the area vanishes.

In 3-D the same story holds with volume.

One wrinkle trips up nearly everyone, and it is easier to meet now than later: the determinant can come out negative, and area cannot. The determinant is signed area. Its size is the area scale factor; its sign is a separate bit of information. It tells you whether the transformation kept the plane the right way round or mirrored it, the way your left hand becomes a right hand in a mirror. So |det A| is the area, and det A = −3 means “areas tripled, orientation flipped”. Nothing has gone wrong.

Drag the matrix entries below and watch the parallelogram breathe. Its signed area is the determinant. Make the two columns line up, and the area collapses to zero, exactly the singular case from last lesson (rank below full). A negative reading just means the transform flipped orientation, like a mirror.

How to compute it

2×2

For a 2×2 matrix the rule is one line — the main-diagonal product minus the off-diagonal product:

3×3

For a 3×3 matrix, use cofactor expansion along the first row: multiply each entry by the 2×2 determinant of the rows and columns that remain, alternating signs + − +:

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)

Worth running once on real numbers. For M = [[1, 2, 3], [0, 1, 4], [5, 6, 0]], expanding along the first row:

det(M) = 1·(1·0 − 4·6) − 2·(0·0 − 4·5) + 3·(0·6 − 1·5)
       = 1·(−24)      − 2·(−20)      + 3·(−5)
       = −24 + 40 − 15 = 1

Non-zero, so M is invertible. Notice how the minus sign in the middle term flipped −20 into +40 — that alternating pattern is where most arithmetic slips happen.

You may expand along any row or column (keeping the sign pattern), so pick the one with the most zeros to save work.

The properties that get tested

A handful of facts answer most exam questions with almost no computation:

• Product rule: det(AB) = det(A)·det(B) — determinants turn a matrix product into ordinary multiplication.

• Transpose: det(Aᵀ) = det(A) — transposing never changes it.

• Powers: det(Aᵏ) = det(A)ᵏ, straight from the product rule.

• Scalar: 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
  • A is singular (no inverse)
  • its rows/columns are linearly dependent
  • 0 is an eigenvalue

  These are four ways to say the same thing. One of them is a sum of multiples of the others, so it carries no new information. The transformation flattened space and lost a dimension, so nothing can undo it.

A worked example — the 2024 trap

Part A — a clean 2×2. 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 non-zero the matrix is invertible.

Part B — the 2024 style. A 3×3 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 first:

M² + 12M = M·(M + 12I)

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

Dependent rows make M singular, so det(M) = 0, and the whole product is 0 — the entries of M never mattered:

det(M² + 12M) = 0 · det(M + 12I) = 0

Spotting the dependent rows (hence det(M) = 0) and factoring with the product rule is the entire solution.

A question to carry forward

When det(A) ≠ 0, the matrix scales space without flattening it, so its action can be undone — there is a matrix that reverses it.

Here is the thread onward:

• What is that reversing matrix called?
• How do we actually build it?
• Why does the determinant sit right in the denominator of the formula?

In one breath

• The determinant is one number per square matrix: the signed area/volume scale factor of its transformation.
• Compute: 2×2 is ad − bc; 3×3 is cofactor expansion a(ei−fh) − b(di−fg) + c(dh−eg) (expand along the row/column with the most zeros).
• Properties: det(AB)=det(A)det(B), det(Aᵀ)=det(A), det(Aᵏ)=det(A)ᵏ, det(cA)=cⁿdet(A). No sum rule — det(A+B) ≠ det(A)+det(B).
• The equivalence: det(A)=0 ⇔ singular ⇔ rows/cols dependent ⇔ 0 is an eigenvalue (space flattened, no inverse).
• The exam move: factor the determinant out with the product rule; never multiply the matrices.

Practice