A 3×3 matrix 𝐴 has det(𝐴)=2020. Find det(Adj(𝐴)).

Answer 1

The relationship between a matrix A, its adjugate matrix Adj(A) and its determinant det(A) is given by

A * Adj(A) = det(A) * I, <1>

where I is the identity matrix (in our case, it is 3 x 3).

Taking the determinant of both sides of <1> yields

det( A * Adj(A) ) = det( det(A) * I ). <2>

The determinant function satisfies det(B*C) = det(B) * det(C) for all square matrices B and C of the same dimensions.

The determinant also satisfies det(k*A) = k^dim(A) * det(A) for any square matrix A and any scalar k, where dim is the dimension (in our case, the number of rows or the number of columns).

Therefore, we can simplify <2> to become

det(A) * det(Adj(A)) = det(A)^dim(A) * det(I). <3>

Substituting det(A) = 2020, dim(A) = 3 and det(I) = 1 yields

2020 * det(Adj(A)) = 2020³ * 1

So det(Adj(A)) = 2020² = 4080400.