Question

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

Matrices
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asked 2020-11-09
Let A,B,C be \(n \times n\) matrices. If \(B=CAC^{-1}\) show that \(det(A)=det(B)\)

Answers (1)

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 \(det(A_1^{-1})=\frac{1}{det(A_1)}\)
\(det(B)=det(CAC^{-1})\)
\(=det(C)det(A)det(C^{-1})\)
\(=det(C)det(C^{-1})det(A)\)
\(=det(C)\frac{1}{det(C)}det(A)\)
\(=det(A)\)
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