If A is n x n, what is the determinant of scalar k times A?

Question

If A: n x n, prove det(kA) =k^n det(A).

Answer 1

Recall that for a square matrix A, if B is a matrix obtained by multiplying a single row of A by a scalar k, then det(B)=k*det(A). kA is obtained by multiplying all n rows of A by k, so det(kA)=k*k*...*k*det(A)=k^n*det(A).

Alternatively, recall that for any two nxn matrices A and B, det(AB)=det(BA)=det(A)*det(B). Then observe that kA=(kI)A (where I is the nxn identity matrix). det(kI)=k^n so det((kI)A)=det(kA)=(k^n)(det(A)).

Answer 2

Determinants have several properties.

If any row or column is multiplied by a constant k, then the determinant will also be multiplied by that constant.

The proof for a 2 * 2 matrix is straightforward and can be extended to n * n matrices.

A = [ a₁₁ a₁₂..... a_1n

a₂₁ a₂₂ .....a_2n

.................

a_n1 a_n2 .....a_nn]

k*A = [ k*a₁₁ k*a₁₂..... k*a_1n

k*a₂₁ k*a₂₂ .....k*a_2n

.................

k*a_n1 k*a_n2 ....k*a_nn ]

We will gradually extract the k's from each columns, and since there are n columns:

det(k*A) = k* det (Ak1) = k² * det(Ak2) =.... kⁿ * det(A)

with the intermediate matrices

Ak1 = [ a₁₁ k*a₁₂..... k*a_1n

a₂₁ k*a₂₂ .....k*a_2n

.................

a_n1 k*a_n2 ....k*a_nn ]

Ak2 = [ a₁₁ a₁₂..... k*a_1n

a₂₁ a₂₂ .....k*a_2n

a_n1 a_n2 ....k*a_nn ]