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Determinants of Scaled, Transposed, and Inverted Matrices

A public linear algebra question that chains three determinant properties — scalar multiples in dimension n, transposes, and inverses — into a single evaluation.

Question

Let AA be a 3×33 \times 3 real matrix with det⁡A=5\det A = 5. Evaluate det⁡ ⁣(2ATA−1)\det\!\left(2A^{\mathsf{T}}A^{-1}\right).

  1. a.22
  2. b.88
  3. c.4040
  4. d.200200

Correct answer

B. 88

Full reasoning

  1. 1The determinant is multiplicative, so det⁡ ⁣(2ATA−1)=det⁡(2I3) det⁡ ⁣(AT)det⁡ ⁣(A−1)\det\!\left(2A^{\mathsf{T}}A^{-1}\right) = \det(2I_3)\,\det\!\left(A^{\mathsf{T}}\right)\det\!\left(A^{-1}\right), where the scalar 22 multiplies a 3×33 \times 3 matrix.
  2. 2Scaling: multiplying an n×nn \times n matrix by cc scales each of its nn rows, so the determinant is multiplied by cnc^{n}. Here c=2c = 2 and n=3n = 3, giving 23=82^{3} = 8.
  3. 3Transpose: det⁡ ⁣(AT)=det⁡A=5\det\!\left(A^{\mathsf{T}}\right) = \det A = 5.
  4. 4Inverse: det⁡A⋅det⁡ ⁣(A−1)=det⁡(I)=1\det A \cdot \det\!\left(A^{-1}\right) = \det(I) = 1, so det⁡ ⁣(A−1)=15\det\!\left(A^{-1}\right) = \tfrac{1}{5} (which is defined because det⁡A=5≠0\det A = 5 \neq 0).
  5. 5Multiply the three factors: 8⋅5⋅15=88 \cdot 5 \cdot \tfrac{1}{5} = 8.

Why each choice is right or wrong

Choice A

This scales the determinant by 22 instead of 232^{3}. Multiplying a 3×33 \times 3 matrix by 22 multiplies every one of its three rows by 22, so the determinant picks up a factor of 23=82^{3} = 8, not 22.

Choice B

Correct. det⁡ ⁣(2ATA−1)=23det⁡ ⁣(AT)det⁡ ⁣(A−1)=8⋅5⋅15=8\det\!\left(2A^{\mathsf{T}}A^{-1}\right) = 2^{3}\det\!\left(A^{\mathsf{T}}\right)\det\!\left(A^{-1}\right) = 8 \cdot 5 \cdot \tfrac{1}{5} = 8.

Choice C

This applies the scalar and the transpose but drops det⁡ ⁣(A−1)\det\!\left(A^{-1}\right) entirely, computing 23⋅52^{3} \cdot 5 as if the inverse factor were the identity.

Choice D

This treats A−1A^{-1} as another copy of AA, giving 23⋅5⋅52^{3} \cdot 5 \cdot 5. The determinant of an inverse is the reciprocal, 15\tfrac{1}{5}, not 55.

Related formula

Determinant of a scalar multiple

det⁡(cA)=cndet⁡(A)\det(cA) = c^{n}\det(A)

Assumptions: A is square of size n; the exponent is the matrix size, not the scalar. det(A^T) = det(A) and det(A^-1) = 1/det(A), the latter requiring det(A) nonzero.

Related topic and practice

Eigenvalues and DeterminantsLinear Algebra Domain GuideEigenvalues of a triangular matrixCreate a free account to practice

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Sources

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  1. MIT OCW 18.06SC Linear Algebra, session: Properties of Determinants — MIT OpenCourseWare. Accessed 2026-08-03. MIT OpenCourseWare materials are CC BY-NC-SA; attribute MIT OCW, link the source page, and avoid verbatim reuse beyond short cited references.
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