Linear algebra: please solve q4, 6 and 8 correctly and handwritten example 2 are attached In Exercises 1-12, determine whether the given matrixA is diagonalizable. If A is diagonalizable, calculate A³using the method of Example 2.3. A =5. A =7. A =8. A =[-3 2-2 12]3 -2 -48-7-16-3 3 7-1 -1 -4-8-3-16121 010 274. A =4.7 Si13[8³]016. A = [-17]01 HEEXAMPLE 1SolutionTherefore, S-¹AS has the formS-¹AS =0 0 0and we have shown that if A has n linearly independent eigenvectors, then A is similarto a diagonal matrix.λ₁ 002200Now suppose that C-¹AC = D, where C is nonsingular and D is a diagonal matrix.Let us write C and D in column form asC = [C1, C₂, ..., C₁] and D= [d₁e₁, d₂e2, ..., dnen].From C-¹AC = D, we obtain AC = CD, and we write both of these in column form asAC = [AC₁, AC2, ..., ACn]CD= [d₁ Ce₁, d₂Ce2, ..., dn Cen].But since Cek = Ck for k = 1, 2, ..., n, we see that AC = CD impliesACk = dk Ck, k = 1, 2, ...,n.Since C is nonsingular, the vectors C₁, C2, ...,Cn are linearly independent (and inparticular, no Ck is the zero vector). Thus the diagonal entries of D are the eigenvaluesof A, and the column vectors of C are a set of n linearly independent eigenvectors.EXAMPLE 2Note that the proof of Theorem 19 gives a procedure for diagonalizing an (n × n)matrix A. That is, if A has n linearly independent eigenvectors u₁, U2, ..., un, then thematrix S = [u₁, u₂, ..., u₂] will diagonalize A.SolutionShow that A is diagonalizable by finding a matrix S such that S-¹AS = D:5 -63 -4Forming S = [u₁, u₂], we obtain=It is easy to verify that A has eigenvalues λ₁ = 2 and ₂ = -1 with correspondingeigenvectorsA =0023--8---8-and u₂ =000:an5-6 -44]-[21]S =2Next, forming S-¹(AS), we obtainS-¹ (AS) =As a check on the calculations, we form S-¹AS. The matrix AS is given by5-62 1AS = [3] [} } ] [3]=-41=D;4.7 Similarity Transformations and Diagonalization−1 2D¹0S-14[][][]Use the result of Example 1 to calculate A¹0, where5 -63 -4- [ 20⁰Hence A¹0 = SD¹0 S-¹ is given by4 10.A =A¹0(-1) ¹⁰=As was noted in Eq. (2), D¹0 = S-¹A¹0 S. Therefore, A ¹⁰ SD10 S-1. Now byExample 1,]-[1024]2047 -20461023-10222 00 -1= D.329Sometimes complex arithmetic is necessary to diagonalize a real matrix.

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Linear algebra: please solve q4, 6 and 8 correctly and handwritten example 2 are attached In Exercises 1-12, determine whether the given matrixA is diagonalizable. If A is diagonalizable, calculate A³using the method of Example 2.3. A =5. A =7. A =8. A =[-3 2-2 12]3 -2 -48-7-16-3 3 7-1 -1 -4-8-3-16121 010 274. A =4.7 Si13[8³]016. A = [-17]01 HEEXAMPLE 1SolutionTherefore, S-¹AS has the formS-¹AS =0 0 0and we have shown that if A has n linearly independent eigenvectors, then A is similarto a diagonal matrix.λ₁ 002200Now suppose that C-¹AC = D, where C is nonsingular and D is a diagonal matrix.Let us write C and D in column form asC = [C1, C₂, ..., C₁] and D= [d₁e₁, d₂e2, ..., dnen].From C-¹AC = D, we obtain AC = CD, and we write both of these in column form asAC = [AC₁, AC2, ..., ACn]CD= [d₁ Ce₁, d₂Ce2, ..., dn Cen].But since Cek = Ck for k = 1, 2, ..., n, we see that AC = CD impliesACk = dk Ck, k = 1, 2, ...,n.Since C is nonsingular, the vectors C₁, C2, ...,Cn are linearly independent (and inparticular, no Ck is the zero vector). Thus the diagonal entries of D are the eigenvaluesof A, and the column vectors of C are a set of n linearly independent eigenvectors.EXAMPLE 2Note that the proof of Theorem 19 gives a procedure for diagonalizing an (n × n)matrix A. That is, if A has n linearly independent eigenvectors u₁, U2, ..., un, then thematrix S = [u₁, u₂, ..., u₂] will diagonalize A.SolutionShow that A is diagonalizable by finding a matrix S such that S-¹AS = D:5 -63 -4Forming S = [u₁, u₂], we obtain=It is easy to verify that A has eigenvalues λ₁ = 2 and ₂ = -1 with correspondingeigenvectorsA =0023--8---8-and u₂ =000:an5-6 -44]-[21]S =2Next, forming S-¹(AS), we obtainS-¹ (AS) =As a check on the calculations, we form S-¹AS. The matrix AS is given by5-62 1AS = [3] [} } ] [3]=-41=D;4.7 Similarity Transformations and Diagonalization−1 2D¹0S-14[][][]Use the result of Example 1 to calculate A¹0, where5 -63 -4- [ 20⁰Hence A¹0 = SD¹0 S-¹ is given by4 10.A =A¹0(-1) ¹⁰=As was noted in Eq. (2), D¹0 = S-¹A¹0 S. Therefore, A ¹⁰ SD10 S-1. Now byExample 1,>]-[1024]2047 -20461023-10222 00 -1= D.329Sometimes complex arithmetic is necessary to diagonalize a real matrix.

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In Exercises 1-12, determine whether the given matrix A is diagonalizable. If A is diagonalizable, calculate A5 using the method of Example 2. 3. A = 5. A = 7. A = 8. A = [ -3 2 -2 1 1 0 10 2 ] 3 -2 -4 8-7-16 -3 3 7 -1 -1 -4 -8-3-16 12 7 4. A = 4.7 Si 13 [6³] 01 6. A = [17] A= 01