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

Let A,B,C be $n×n$ matrices. If $B=CA{C}^{-1}$ show that $det\left(A\right)=det\left(B\right)$
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Step 1
Consider that A, B, C are $n×n$ matrices.
Given: $B=CA{C}^{-1}$
To Show: $det\left(A\right)=det\left(B\right)$
Apply property of determinants which says if ${A}_{1},{A}_{2},{A}_{3}$, …An are n×n matrices. Then $det\left({A}_{1}{A}_{2}{A}_{3}\dots {A}_{n}\right)=det\left({A}_{1}\right)det\left({A}_{2}\right)det\left({A}_{3}\right)\dots det\left({A}_{n}\right)$. Therefore
$det\left(B\right)=det\left(CA{C}^{-1}\right)$
$=det\left(C\right)det\left(A\right)det\left({C}^{-1}\right)$
Step 2
Apply commutative laws of determinants that is, $det\left({A}_{1}\right)det\left({A}_{2}\right)=det\left({A}_{2}\right)det\left({A}_{1}\right).$
$det\left(B\right)=det\left(CA{C}^{-1}\right)$
$=det\left(C\right)det\left(A\right)det\left({C}^{-1}\right)$
$=det\left(C\right)det\left({C}^{-1}\right)det\left(A\right)$
Apply property of determinants that $det\left({A}_{1}^{-1}\right)=\frac{1}{det\left({A}_{1}\right)}$
$det\left(B\right)=det\left(CA{C}^{-1}\right)$
$=det\left(C\right)det\left(A\right)det\left({C}^{-1}\right)$
$=det\left(C\right)det\left({C}^{-1}\right)det\left(A\right)$
$=det\left(C\right)\frac{1}{det\left(C\right)}det\left(A\right)$
$=det\left(A\right)$
Jeffrey Jordon