Question

# Let A,B,C be n times n matrices. If B=CAC^{-1} show that det(A)=det(B)

Matrices
Let A,B,C be $$n \times n$$ matrices. If $$B=CAC^{-1}$$ show that $$det(A)=det(B)$$

2020-11-10
Step 1
Consider that A, B, C are $$n \times n$$ matrices.
Given: $$B=CAC^{-1}$$
To Show: $$det(A)=det(B)$$
Apply property of determinants which says if $$A_1, A_2, A_3$$, …An are n×n matrices. Then $$det(A_1A_2A_3 \dotsc A_n)= det(A_1)det(A_2)det(A_3) \dotsc det(A_n)$$. Therefore
$$det(B)=det(CAC^{-1})$$
$$=det(C)det(A)det(C^{-1})$$
Step 2
Apply commutative laws of determinants that is, $$det(A_1)det(A_2)= det(A_2)det(A_1).$$
$$det(B)=det(CAC^{-1})$$
$$= det(C)det(A)det(C^{-1})$$
$$= det(C)det(C^{-1})det(A)$$
Apply property of determinants that $$det(A_1^{-1})=\frac{1}{det(A_1)}$$
$$det(B)=det(CAC^{-1})$$
$$=det(C)det(A)det(C^{-1})$$
$$=det(C)det(C^{-1})det(A)$$
$$=det(C)\frac{1}{det(C)}det(A)$$
$$=det(A)$$