please send handwritten solution for Q 6.3 Read the following carefully, and then answer the questions that follow.We say that two n xn matrices are cousins, if there is some invertible matrix P so thatA = PBP-1"Notice that if A and B are cousins, then B = PAP = (PA(P)-. Thus, the definition is symmetric" (ie the order we say itdoesn't matterk if A and B are csasins then B and A are coesins.One of the first things you might notice about the notion of cousins, it that it is clearly connected todiagonalizability:Theorem 1A matrix A is diagonalizable if and only if there is a diagonal matrix D so that A and D arecousins.One might think from the definition that the main use of the definition of cousin matrices is toreframe diagonalizability. However, it actually allows us to generalize it in useful ways - in thefollowing example and the first question below, we'll see matrices A and B which are cousins,but where neither of which is diagonal, and for which techniques from our study of diagonalizablematrices can be applied.-7 -5]and let B = Ry/2 =7Let A =be the matrix of rotation by 3n/2.10Then with P=, if you compute PBP1, you will see that the end result is A.Thus, A and B are cousins.The next theorem shows us that cousins share a lot of important properties in common:Theorem 2If A and B are cousins, then:1. det(A) = det(B), and2. CA(x) = c8(x) (and hence A and B have the same eigenvalues.)Proof: Let A = PBP for some invertible matrix P.1. det(A) = det(PBP-1) = det(P) - det(B) - det(P') = det(B) (using that det (P) = 1/det(P~1).)2. CA(x) = det(xl - A) = det (xl – PBP-1). But we can rewrite I as PIP-1 and substitutethis in; so, continuing we have: .= det(xPIPp-1- PBP-1).But then you can check that XPIP – PBP = P(xl – B)P, by multiplying out theright-hand side. This allows us to continue with:. = det (P(xl – B)P) = det(P) - det (xl – B) - det(P-1) = det(xl – B) = cs(x)Continues on the next page.MAT223 - Winter 2022Midterm 26.1Use the result of the Example above to compute A100 (for A the matrix from the Example).Hint: consider (PBP-)100, What is Bl00,6.2 () True or False: If A and B are nx n matrices so that det(A) = det(B), then A and B arecousins. Hint: part of Theorem 2 should be useful.6.3) True or False: If A and B are cousins and v is an eigenvector of A, then v is aneigenvector of B.

Question

please send handwritten solution for Q 6.3 Read the following carefully, and then answer the questions that follow.We say that two n xn matrices are cousins, if there is some invertible matrix P so thatA = PBP-1&#34;Notice that if A and B are cousins, then B = PAP = (PA(P)-. Thus, the definition is symmetric&#34; (ie the order we say itdoesn't matterk if A and B are csasins then B and A are coesins.One of the first things you might notice about the notion of cousins, it that it is clearly connected todiagonalizability:Theorem 1A matrix A is diagonalizable if and only if there is a diagonal matrix D so that A and D arecousins.One might think from the definition that the main use of the definition of cousin matrices is toreframe diagonalizability. However, it actually allows us to generalize it in useful ways - in thefollowing example and the first question below, we'll see matrices A and B which are cousins,but where neither of which is diagonal, and for which techniques from our study of diagonalizablematrices can be applied.-7 -5]and let B = Ry/2 =7Let A =be the matrix of rotation by 3n/2.10Then with P=, if you compute PBP1, you will see that the end result is A.Thus, A and B are cousins.The next theorem shows us that cousins share a lot of important properties in common:Theorem 2If A and B are cousins, then:1. det(A) = det(B), and2. CA(x) = c8(x) (and hence A and B have the same eigenvalues.)Proof: Let A = PBP for some invertible matrix P.1. det(A) = det(PBP-1) = det(P) - det(B) - det(P') = det(B) (using that det (P) = 1/det(P~1).)2. CA(x) = det(xl - A) = det (xl – PBP-1). But we can rewrite I as PIP-1 and substitutethis in; so, continuing we have: .= det(xPIPp-1- PBP-1).But then you can check that XPIP – PBP = P(xl – B)P, by multiplying out theright-hand side. This allows us to continue with:. = det (P(xl – B)P) = det(P) - det (xl – B) - det(P-1) = det(xl – B) = cs(x)Continues on the next page.MAT223 - Winter 2022Midterm 26.1Use the result of the Example above to compute A100 (for A the matrix from the Example).Hint: consider (PBP-)100, What is Bl00,6.2 () True or False: If A and B are nx n matrices so that det(A) = det(B), then A and B arecousins. Hint: part of Theorem 2 should be useful.6.3) True or False: If A and B are cousins and v is an eigenvector of A, then v is aneigenvector of B.

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Answered: Image /qna-images/answer/013d5bbf-75f4-490e-aeae-4ab62afe1c96.jpg

) True or False: If A and B are cousins and v is an eigenvector of A, then v is an eigenvector of B.