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12 Determinant Properties That Appear on Every Exam: det(kA), det(A⁻¹), det(adj A) and the Traps

Most determinant exam questions do not ask you to compute from numbers. They give a single value of det⁡A\det A and ask for the determinant of a matrix built from AA. Every such question is a one-liner once you know these 12 properties. Throughout, AA and BB are n×nn\times n and kk is a constant.

By Ajarn ModUpdated
Three formulas to know by heart
  • det⁡(kA)=kndet⁡A\det(kA) = k^n\det A — the exponent is the size nn
  • det⁡(AB)=det⁡Adet⁡B\det(AB) = \det A\det B, hence det⁡A−1=1/det⁡A\det A^{-1} = 1/\det A and det⁡Am=(det⁡A)m\det A^m = (\det A)^m
  • det⁡(adj⁡A)=(det⁡A)n−1\det(\operatorname{adj}A) = (\det A)^{n-1}

Why the properties hold: the determinant is an area factor

A 2×2 matrix turns the unit square into a parallelogram, and ∣det⁡A∣|\det A| is its area (for 3×3, a volume). Seen this way most properties are obvious: stretch every side by 2 and the area grows by 222^2, the volume by 232^3.

012340123area 5A = [3 1; 1 2] · det A = 5012340123×4det(2A) = 2²·det A3×3 (volume): det(2A) = 2³ det A
Left: det⁡A=5\det A = 5 is the area the unit square becomes · right: scaling every row by 2 multiplies the area by 222^2.

Properties 1–6: row operations and special matrices

  1. Transpose: det⁡AT=det⁡A\det A^T = \det A — so every row rule also works for columns
  2. Swap two rows: the sign changes
  3. Multiply one row by kk: the determinant is multiplied by kk (factors come out one row at a time)
  4. Add a multiple of another row Ri←Ri+kRjR_i \leftarrow R_i + kR_j: no change — the main tool of row reduction
  5. A zero row, two equal rows or proportional rows: det⁡=0\det = 0
  6. Triangular or diagonal matrix: det⁡\det = product of the diagonal entries
Example

If ∣abcdefghi∣=5\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = 5, find ∣2a2b2cghid+3ae+3bf+3c∣\begin{vmatrix} 2a & 2b & 2c \\ g & h & i \\ d + 3a & e + 3b & f + 3c \end{vmatrix}

  1. Take 2 out of row 1 (property 3) · subtract a multiple of row 1 from row 3 to get (d,e,f)(d, e, f) without changing the value (property 4)

  2. The rows are now (a,b,c),(g,h,i),(d,e,f)(a, b, c), (g, h, i), (d, e, f); swapping rows 2 and 3 flips the sign (property 2)

2×(−5)=−102\times(-5) = -10

Properties 7–10: constants, products, inverses and powers

  1. det⁡(kA)=kndet⁡A\det(kA) = k^n\det A — because kk sits in all nn rows
  2. det⁡(AB)=det⁡Adet⁡B=det⁡(BA)\det(AB) = \det A\det B = \det(BA), even though AB≠BAAB \ne BA
  3. det⁡(A−1)=1det⁡A\det(A^{-1}) = \dfrac{1}{\det A} — and AA is invertible exactly when det⁡A≠0\det A \ne 0
  4. det⁡(Am)=(det⁡A)m\det(A^m) = (\det A)^m
Example

AA is 3×33\times3 with det⁡A=4\det A = 4. Find det⁡(2A−1AT)\det(2A^{-1}A^T)

  1. det⁡(2A−1AT)=23⋅14⋅4=8\det(2A^{-1}A^T) = 2^3\cdot\frac14\cdot4 = 8

8

Property 11: the adjugate, where students slip most

Taking determinants of Aadj⁡A=(det⁡A)IA\operatorname{adj}A = (\det A)I gives det⁡A⋅det⁡(adj⁡A)=(det⁡A)n\det A\cdot\det(\operatorname{adj}A) = (\det A)^n:

det⁡(adj⁡A)=(det⁡A)n−1,adj⁡(kA)=kn−1adj⁡A,adj⁡(adj⁡A)=(det⁡A)n−2A\det(\operatorname{adj}A) = (\det A)^{n-1},\qquad \operatorname{adj}(kA) = k^{n-1}\operatorname{adj}A,\qquad \operatorname{adj}(\operatorname{adj}A) = (\det A)^{n-2}A
Example

AA is 3×33\times3 with det⁡A=3\det A = 3. Find det⁡(2adj⁡A)\det(2\operatorname{adj}A) and det⁡(adj⁡(adj⁡A))\det(\operatorname{adj}(\operatorname{adj}A))

  1. det⁡(2adj⁡A)=23⋅32=72\det(2\operatorname{adj}A) = 2^3\cdot3^{2} = 72
  2. det⁡(adj⁡(adj⁡A))=det⁡(3A)=33⋅3=81\det(\operatorname{adj}(\operatorname{adj}A)) = \det(3A) = 3^3\cdot3 = 81

72 and 81

Property 12: what has no formula

Traps that appear every year
  • det⁡(A+B)≠det⁡A+det⁡B\det(A + B) \ne \det A + \det B — add the matrices first (only row-by-row linearity exists)
  • det⁡(kA)≠kdet⁡A\det(kA) \ne k\det A and det⁡(adj⁡A)≠det⁡A\det(\operatorname{adj}A) \ne \det A — the exponent depends on nn
  • det⁡(−A)=(−1)ndet⁡A\det(-A) = (-1)^n\det A — unchanged for 2×2 but sign-flipped for 3×3
  • ∣A∣|A| is not an absolute value; determinants can be negative

A bonus trick: if every row has the same sum, add all columns into the first column and take that sum out as a factor. For example, the matrix with aa on the diagonal and bb elsewhere has det⁡=(a−b)n−1(a+(n−1)b)\det = (a - b)^{n-1}(a + (n - 1)b).

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Frequently asked questions

Is det(2A) equal to 2 det A?

No. Multiplying a matrix by 2 multiplies every row by 2, and a factor 2 can be taken out of each row once. For an n×n matrix det(2A) = 2ⁿ det A, so a 3×3 matrix gives 8 det A.

How do you find det(adj A)?

Use det(adj A) = (det A)ⁿ⁻¹, which follows from A adj A = (det A)I: taking determinants of both sides gives det A · det(adj A) = (det A)ⁿ. For a 3×3 matrix with det A = −2, det(adj A) = 4.

Is det(A + B) equal to det A + det B?

No. Determinants are not additive over whole matrices; add the matrices first and then take the determinant. Linearity only holds one row at a time.

Which row operation leaves the determinant unchanged?

Adding a multiple of one row to another (R_i ← R_i + kR_j) leaves it unchanged. Swapping two rows changes the sign, and multiplying one row by k multiplies the determinant by k.

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