plz provide answer for 6.2 6.1Use the result of the Example above to compute Al00 (for A the matrix from the Example).Hint: consider (PBP-1)100, What is B100?6.2 (cousins. Hint: part of Theorem 2 should be useful.) True or False: If A and B are n xn matrices so that det(A) = det(B), then A and B are63True or False: If A and B are cousins and v is an eigenvector of A, then v is aneigenvector of B. 6.Read the following carefully, and then answer the questions that follow.We say that two n x n matrices are cousins, if there is some invertible matrix P so thatA = PBP-1"Notice that if A and B are cousins, then B = P-AP = (P|A(P)-1. Thus, the definition is "symmetric" (ie. the order we say itdoesn't matterk if A and B are cosusins then B and A are cousins.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]Let A =and let B = R3/2 =be the matrix of rotation by 3n/2.10-1 01Then with P=-12,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) = cB(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(P1) = det(B) (using that det(P) = 1/det(P1).)2. CA(x) = det(xi – A) = det(xI – PBP-1). But we can rewrite I as PIP-1 and substitutethis in; so, continuing we have: .. = det(XPIP-1- PBP-1).But then you can check that XPIP-1 - PBP-1 = P(xI – B)P1, by multiplying out theright-hand side. This allows us to continue with:... = det (P(xl – B)P1) = det(P) det (xl – B) det(P1) = det(xl - B) = cB(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 Bl0026.2 () True or False: If A and B are nxn matrices so that det(A) = det(B), then A and B arecousins. Hint: part of Theorem 2 should be useful.63 ****) 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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plz provide answer for 6.2 6.1Use the result of the Example above to compute Al00 (for A the matrix from the Example).Hint: consider (PBP-1)100, What is B100?6.2 (cousins. Hint: part of Theorem 2 should be useful.) True or False: If A and B are n xn matrices so that det(A) = det(B), then A and B are63True or False: If A and B are cousins and v is an eigenvector of A, then v is aneigenvector of B. 6.Read the following carefully, and then answer the questions that follow.We say that two n x n 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 = P-AP = (P|A(P)-1. Thus, the definition is &#34;symmetric&#34; (ie. the order we say itdoesn't matterk if A and B are cosusins then B and A are cousins.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]Let A =and let B = R3/2 =be the matrix of rotation by 3n/2.10-1 01Then with P=-12,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) = cB(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(P1) = det(B) (using that det(P) = 1/det(P1).)2. CA(x) = det(xi – A) = det(xI – PBP-1). But we can rewrite I as PIP-1 and substitutethis in; so, continuing we have: .. = det(XPIP-1- PBP-1).But then you can check that XPIP-1 - PBP-1 = P(xI – B)P1, by multiplying out theright-hand side. This allows us to continue with:... = det (P(xl – B)P1) = det(P) det (xl – B) det(P1) = det(xl - B) = cB(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 Bl0026.2 () True or False: If A and B are nxn matrices so that det(A) = det(B), then A and B arecousins. Hint: part of Theorem 2 should be useful.63 ****) 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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62 ( ) True or False: If A and B are nxn matrices so that det(A) = det(B), then A and B are %3D cousins. Hint: part of Theorem 2 should be useful.