Step-by-step explanation:
To prove it we just use the definition of similar matrices and properties of determinants:
If
are similar matrices, then there is an invertible matrix
, such that
(that's the definition of matrices being similar). And so we compute the determinant of such matrix to get:
![det(A)=det(C^{-1}BC)=det(C^{-1})det(B)det(C)](https://tex.z-dn.net/?f=det%28A%29%3Ddet%28C%5E%7B-1%7DBC%29%3Ddet%28C%5E%7B-1%7D%29det%28B%29det%28C%29)
![=\frac{1}{det(C)}det(B)det(C)=det(B)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7Bdet%28C%29%7Ddet%28B%29det%28C%29%3Ddet%28B%29)
(Determinant of a product of matrices is the product of their determinants, and the determinant of
is just
)
Answer:
25.1 cm
Step-by-step explanation:
Answer:
No, and this isn't the place to put an ad
Step-by-step explanation:
Answer:
m = 1/5
Step-by-step explanation:
Answer:
free points btw
Step-by-step explanation: