If A is a 3 × 3 matrix, then what will be the value of k if Det(A^-1) = (Det A)^k?

We are given that,

Order of matrix = 3 × 3

Det(A^-1) = (Det A)^k

An n-by-n square matrix A is called invertible if there exists an n-by-n square matrix B such that where I_n denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication.

We know that,

If A and B are square matrices of same order, then

Det (AB) = Det (A).Det (B)

Since, A is an invertible matrix, this means that, A has an inverse called A^-1.

Then, if A and A^-1 are inverse matrices, then

Det (AA^-1) = Det (A).Det (A^-1)

By property of inverse matrices,

AA^-1 = I

∴, Det (I) = Det (A).Det (A^-1)

Since, Det (I) = 1

⇒ 1 = Det (A).Det (A^-1)

⇒ Det (A^-1) = Det (A)^-1

Since, according to question,

Det(A^-1) = (Det A)^k

⇒ k = -1

Thus, the value of k is -1.