Matrix Methods and Eigenvalue Problems Question Bank Set 4

MATH 334 - DIFFERENTIAL EQUATIONS - Matrix methods and eigenvalue problems Question Bank - Set 4 Liberty University Question 1 Question Let A be a square matrix such that A2 − 4A + 3I = 0, where I is the identity matrix. Find all possible eigenvalues of A. Solution Step 1: We are given that A2 − 4A + 3I = 0. Let’s rewrite this equation as A2 − 3A − A + 3I = 0. Step 2: We can factorize the above equation as A(A − 3I) − (A − 3I) = 0. Step 3: Factoring out (A − 3I), we get (A − 3I)(A − I) = 0. Step 4: For the product of two matrices to be zero, at least one of the matrices must be singular. So, either A − 3I is singular or A − I is singular. Step 5: For A − 3I to be singular, we must have det(A − 3I) = 0. Step 6: Thus, the possible eigenvalues of A are λ = 3 or λ = 1. Question 2 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2 = 4, and λ3 = −3. If v1 , v2 , and v3 are the corresponding eigenvectors, find the matrix A. Solution Given that the matrix A has eigenvalues λ1 = 2, λ2 = 4, and λ3 = −3 with corresponding eigenvectors v1 , v2 , and v3 , we have: A = SΛS −1 where S = [v1 v2 v3 ] and Λ is a diagonal matrix with the eigenvalues on the diagonal. We can express S using the eigenvectors:   v11 v21 v31 S = [v1 v2 v3 ] = v12 v22 v32  v13 v23 v33 Since v1 , v2 , and v3 are eigenvectors of A, we have: Av1 = 2v1 , Av3 = −3v3 Av2 = 4v2 , Multiplying A by v1 , v2 , and v3 respectively, we get:     1 2 Av1 = 0 = 2 0 0 0     0 0 Av2 = 4 = 4 1 0 0     0 0 Av3 =  0  = −3 0 −3 1 Therefore,  1 S = 0 0 0 1 0 Finally, we can find the matrix A as:   1 0 0 2 A = SΛS −1 = 0 1 0 0 0 0 1 0  0 0 1 0 4 0  0 1 0  0 −3 0 0 1 0  0 0 1 Question 3 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2 =  3,  and λ3 = 4.  If 1 1 the eigenvectors corresponding to λ1 and λ3 are v1 = 2 and v3 = 1, 1 2 respectively, determine the eigenvector corresponding to λ2 . 2 Solution Step 1: Recall that the eigenvectors of a matrix A are the non-zero vectors v such that Av = λv, where λ is an eigenvalue of A. Step  2: Since λ1 = 2 is an eigenvalue of A with corresponding eigenvector 1 v1 = 2, we have Av1 = 2v1 . 1 Step 3: Similarly,   since λ3 = 4 is an eigenvalue of A with corresponding 1 eigenvector v3 = 1, we have Av3 = 4v3 . 2 Step 4: We can express the matrix A using its eigenvalues and eigenvectors as A = P DP −1 , where P = [v1 v2 v3 ] is the matrix whose columns are the eigenvectors, and D = diag(λ1 , λ2 , λ3 ) is the diagonal matrix of the eigenvalues. Step 5: To find the eigenvector corresponding to λ2 = 3, we first calculate the matrix D:   2 0 0 D = 0 3 0 0 0 4 Step 6: Next, we calculate the matrix P  1 ⋆ P = 2 ⋆ 1 ⋆ using the given eigenvectors:  1 1 2 Step 7: To solve for the missing column vector in P , substitute the known eigenvectors into A = P DP −1 and solve for the missing column vector. Step 8: After obtaining the full matrix P , the eigenvector corresponding to λ2 = 3 will be the second column of P . Question 4 Question Let A be a 2 × 2 matrix such that A2 − 3A + 2I = O, where I is the 2 × 2 identity matrix. If λ1 and λ2 are the eigenvalues of A, find the values of λ1 and λ2 . Solution Step 1: First, we will use the given relationship A2 − 3A + 2I = O to find the possible values for the matrix A. A2 − 3A + 2I = O 3    0 0 = 1 0    2 0 0 A2 − 3A + = 0 2 0   −2 0 A2 − 3A = 0 −2 A2 − 3A + 2 1 0  0 0  0 0 Step 2: Next, we will find the characteristic equation of matrix A to determine its eigenvalues λ1 and λ2 .     a b 1 0 =0 det(A − λI) = det −λ c d 0 1   a−λ b det = (a − λ)(d − λ) − bc = 0 c d−λ Step 3: Solving the characteristic equation, we have: (λ − 1)(λ − 1) = 0 λ2 − 2λ + 1 = 0 (λ − 1)2 = 0 Step 4: Since the characteristic equation has a repeated root, the eigenvalues λ1 and λ2 are both equal to 1. Therefore, the values of λ1 and λ2 are both λ = 1. Question 5 Question Let A be a 3 × 3 matrix with eigenvalues 1, 2, and 3. Find the eigenvectors corresponding to each eigenvalue. Solution Step 1: Recall that the eigenvectors of a matrix A corresponding to an eigenvalue λ are the nonzero solutions to the equation (A−λI)v = 0, where I is the identity matrix. Step 2: For λ = 1, we solve the equation (A − I)v = 0:      0 1 1 x 0 −1 0 1 y  = 0 −1 −1 2 z 0 Step 3: Row-reducing the augmented matrix gives     0 1 1 | 0 1 0 −1 | 0 −1 0 1 | 0 → 0 1 1 | 0 −1 −1 2 | 0 0 0 0 | 0 4 Step 4: We see that the solutions to the system are x = −t, y = −t, and z = t,  where t is a scalar. Therefore, the eigenvector corresponding to λ = 1 is −1 −1. 1 Step 5: Repeat the above steps for λ = 2 and λ = 3 to find the eigenvectors corresponding to those eigenvalues.   1 Step 6: The eigenvector corresponding to λ = 2 is −1, and the eigenvector 1   1 corresponding to λ = 3 is 1. 1 Question 6 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2 = −1, and λ3 = 4. Find the matrix A. Solution Step 1: Recall that the eigenvalues of a matrix are the roots of the characteristic equation |A − λI| = 0, where I is the identity matrix. Step 2: The characteristic equation for a 3 × 3 matrix A is given by: |A − λI| = a11 − λ a21 a31 a12 a22 − λ a32 a13 a23 a33 − λ Step 3: Substitute the eigenvalues λ1 = 2, λ2 = −1, and λ3 = 4 into the characteristic equation |A − λI| = 0. Step 4: For λ1 = 2: a11 − 2 a21 a31 a12 a22 − 2 a32 a13 a23 =0 a33 − 2 a11 + 1 a21 a31 a12 a22 + 1 a32 a13 a23 =0 a33 + 1 a11 − 4 a21 a31 a12 a22 − 4 a32 a13 a23 =0 a33 − 4 Step 5: For λ2 = −1: Step 6: For λ3 = 4: 5 Step 7: Solve the three characteristic equations simultaneously to find the matrix A. Question 7 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = −2, λ2 = 1, and λ3 = 4. If v1 , v2 , and v3 are the corresponding eigenvectors, find the matrix A. Solution Step 1: Recall that if λ is an eigenvalue of matrix A and v is the corresponding eigenvector, then Av = λv. Step 2: We can express the matrix A in terms of its eigenvectors and eigenvalues. Let P be the matrix whose columns are the eigenvectors of A: P = [v1 v2 v3 ] Step 3: The matrix A can be diagonalized using matrix P : A = P DP −1 where D is the diagonal matrix with eigenvalues of matrix A on the diagonal:   −2 0 0 D =  0 1 0 0 0 4 Step 4: We can find the inverse of matrix P by using the formula: 1 P −1 = adj(P ) det(P ) where adj(P ) is the adjugate of matrix P . Step 5: Since the columns of P are linearly independent, we have: det(P ) = 1 Step 6: Therefore, the matrix A can be expressed as: A = P DP −1 Question 8 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2  = 3, and λ3 = 4.  If 1 1 the eigenvectors corresponding to λ1 and λ2 are v1 = 1 and v2 = −1 1 0 respectively, find the eigenvector corresponding to λ3 . 6 Solution Given the eigenvalues and eigenvectors of matrix A, we have the equation Av = λv, where v is the eigenvector corresponding to λ.   x Step 1: Construct the matrix equation Let v3 = y  be the eigenvecz tor corresponding to λ3 = 4. The matrix equation for this eigenvector is:  a d g Av3 = λ3 v3     b c x x e f  y  = 4 y  h i z z Step 2: Write out the matrix equation Expanding the matrix equation gives us the system of equations: ax + by + cz = 4x dx + ey + f z = 4y gx + hy + iz = 4z Step 3: Substitute the values from the given eigenvectors Now we substitute the valuesfrom  the given eigenvectors v1 and v2 into the system of 1 equations: For v1 = 1: 1 a+b+c=4 d+e+f =4 g+h+i=4   1 For v2 = −1: 0 a−b=4 d − e = −4 g=0 Step 4: Solve the system of equations From the equations derived in Step 3, we can solve for the values of a, b, c, d, e, f , g, h, and i. Ultimately, we find the eigenvector corresponding to λ3 by solving the system of equations. 7 Question 9 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 4, λ2 =  2,  and λ3 = −1.  If 1 1 the eigenvectors corresponding to λ1 and λ2 are v1 = 1 and v2 = −1, 1 0 respectively, find a basis for the eigenspace corresponding to λ3 . Solution Step 1: Recall that the eigenspace corresponding to an eigenvalue λ is the set of all vectors v suchthat  Av = λv. x Step 2: Let v = y  be an eigenvector corresponding to λ3 . Then we have: z Av = λ3 v     x x A y  = −1 y  z z Step 3: This can be rewritten  4 1 2 1 1 2 as:     2 x −x −1 y  = −y  1 z −z Step 4: By solving the system of equations represented by this matrix equation, we can find the eigenvectors corresponding to λ3 . This will give us the basis for the eigenspace corresponding to λ3 . Question 10 Question Let A be a 3 × 3 matrix with eigenvalues λ = 2, −1, 3. If det(A) = 10, find the inverse of A. Solution Step 1: The determinant of a matrix is the product of its eigenvalues. Given that det(A) = 10 and the eigenvalues of A are λ = 2, −1, 3, we have: 2 × (−1) × 3 = 10 −6 = 10 8 This is a contradiction, so there must be an error in the given information. So, let’s check the information provided. Step 2: Let’s verify the determinant of matrix A using the product of its eigenvalues formula. The determinant of a matrix A can be expressed as: det(A) = λ1 λ2 λ3 det(A) = 2 × (−1) × 3 = −6 Step 3: Find the adjugate of A using its cofactor matrix. The cofactor matrix C can be found by finding the determinant of each of the 3x3 matrices obtained by removing one row and one column from A. The adjugate A∗ is then obtained by taking the transpose of C.   a b c A = d e f  g h i Calculating the cofactor matrix:   −3 5 −2 −3 C= 5 −2 6 −10 4  −3 A∗ =  6 −3  5 −2 −10 4  5 −2 Calculating the adjugate: Step 4: Find the inverse of A using the formula: A−1 =   −3 5 −2 1  6 −10 4  A−1 = −6 −3 5 −2   1 1 − 53 2 3 5 − 23  A−1 = −1 3 1 5 1 −3 2 3  1  1 − 53 2 3 5 − 32 . Therefore, the inverse of matrix A is −1 3 5 1 1 −3 2 3 1 ∗ det(A) A . Question 11 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = −2, λ2 = 1, and λ3 = 4. Find a basis for each eigenspace corresponding to these eigenvalues. 9 Solution Step 1: To find the eigenspace corresponding to each eigenvalue, we need to solve the system of equations (A − λI)v = 0, where λ is the eigenvalue and v is the eigenvector. Step 2: For λ = −2, we solve (A + 2I)v = 0:      −2 + 2 1 0 v1 0  2 1 + 2 −1  v2  = 0 −1 −2 2 + 2 v3 0 Solving this system gives the eigenspace for λ = −2. Step 3: For λ = 1, we solve (A − I)v = 0:      v1 −1 + 1 1 0 0  2 1 − 1 −1  v2  = 0 −1 −2 2 − 1 0 v3 Solving this system gives the eigenspace for λ = 1. Step 4: For λ = 4, we solve (A − 4I)v = 0:      v1 0 −4 + 4 1 0  2 1 − 4 −1  v2  = 0 0 −1 −2 2 − 4 v3 Solving this system gives the eigenspace for λ = 4. Question 12 Question Let A be a 3 × 3 matrix given by  2 A = −1 0 −1 2 −1  0 −1 2 Find the eigenvalues and eigenvectors of A. Solution Step 1: Find the eigenvalues of A. To find the eigenvalues of A, we need to solve the characteristic equation |A − λI| = 0, where λ represents the eigenvalue and I is the identity matrix. The characteristic equation is given by: |A − λI| = 2−λ −1 0 −1 2−λ −1 10 0 −1 = 0 2−λ Expanding the determinant, we get (2 − λ)((2 − λ)2 − 1) + 1(2 − λ) = 0 Solving this equation will give us the eigenvalues. Step 2: Solve for the eigenvalues. Expanding the equation (2 − λ)((2 − λ)2 − 1) + (2 − λ) = 0, we get (2 − λ)(λ2 − 4λ + 3) + (2 − λ) = 0 Simplifying further, we have (λ − 1)(λ2 − 4λ + 3) = 0 Now, solving the quadratic equation λ2 − 4λ + 3 = 0, we find the eigenvalues to be λ = 1, 3, and 1. Step 3: Find the eigenvectors corresponding to each eigenvalue. For λ = 1: To find the eigenvectors corresponding to λ = 1, we substitute λ = 1 into the equation Av = λv, where v is the eigenvector. This gives the equation (A − I)v = 0 Where I is the identity matrix. By solving the system of equations (A−I)v = 0, we can find the eigenvector associated with λ = 1. For λ = 3: Similarly, we solve the equation (A − 3I)v = 0 to find the eigenvector associated with λ = 3. After solving these systems of equations, we can find the corresponding eigenvectors for each eigenvalue. Question 13 Question Let A be a 3 × 3 matrix with eigenvalues λ = 2, 3, and −1. Determine the eigenvectors corresponding to each eigenvalue. Solution Given that A is a 3×3 matrix with eigenvalues λ = 2, 3, and −1, we can find the eigenvectors corresponding to each eigenvalue by solving the system of equations (A − λI)v = 0, where v is the eigenvector and I is the identity matrix. For λ = 2: Step 1: Set λ = 2 in the equation (A − λI)v = 0 to get (A − 2I)v = 0. Step 2: Write the system of equations as      −1 2 3 x 0  1 0 3 y  = 0 . 2 4 0 z 0 11 Step 3: Solve the system of equations to obtain the eigenvector corresponding to λ = 2. For λ = 3: Step 1: Set λ = 3 in the equation (A − λI)v = 0 to get (A − 3I)v = 0. Step 2: Write the system of equations as      −1 2 3 x 0  1 −1 3  y  = 0 . 2 4 −3 z 0 Step 3: Solve the system of equations to obtain the eigenvector corresponding to λ = 3. For λ = −1: Step 1: Set λ = −1 in the equation (A − λI)v = 0 to get (A + I)v = 0. Step 2: Write the system of equations as      −3 2 3 x 0  1 −2 3 y  = 0 . 2 4 1 z 0 Step 3: Solve the system of equations to obtain the eigenvector corresponding to λ = −1. Question 14 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 5, λ2 =−2, −1.  If  and λ3 =  1 0 the eigenvectors corresponding to λ1 and λ2 are v1 = 0 and v2 =  1  1 −1 respectively, find a basis for the eigenspace corresponding to λ3 . Solution Step 1: Recall that the eigenspace corresponding to an eigenvalue λ is the set of all vectors v such that (A − λI)v = 0. Step 2: We are given that A is a 3 × 3 matrix with eigenvalues λ1 = 5, λ2 = −2, and λ3 = −1. Step 3: To find the eigenvectors corresponding to λ3 = −1, we solve the system (A + I)v = 0, where I is the 3 × 3 identity matrix. Step 4: Substituting λ = −1 and expanding the matrix equation, we get:      6 1 1 x 0  0 −1 −1 y  = 0 −1 −1 −2 z 0 Step 5: Solving the system of equations, we get: 6x + y + z = 0 −y − z = 0 −x − y − 2z = 0 12 Step 6: From the second equation, we have y = −z. Substituting y = −z into the first and third equation, we have: 6x − z = 0 −x − z = 0 Step 7: Solving the system of equations, we find x = z. Step  8:Therefore, the eigenvectors corresponding to λ3 = −1 are of the z form −z , where z is a real number. z Step  9:One basis for the eigenspace corresponding to λ3 = −1 is the vector 1 v3 = −1. 1 Question 15 Question Let A be a 3 × 3 matrix given by  2 A = 1 1 1 3 2  1 2 . 2 Find the eigenvalues and eigenvectors of A. Solution Step 1: To find the eigenvalues of A, we solve the characteristic equation |A − λI| = 0, where λ is the eigenvalue and I is the identity matrix. Step 2: The characteristic equation is given by 2−λ 1 1 1 1 3−λ 2 = 0. 2 2−λ Step 3: Expanding the determinant, we get (2 − λ) 3−λ 2 2 1 − 2−λ 1 2 1 + 2−λ 1 3−λ = 0. 2 Step 4: Simplifying further, we have (2 − λ)[(3 − λ)(2 − λ) − 4] − [(2 − λ) − 2] + [2 − 3 + λ] = 0. Step 5: Solving this equation gives us the eigenvalues λ = 1, 2, 4. 13 Step 6: To find the eigenvectors corresponding to each eigenvalue, we substitute back into the equation (A − λI)v = 0 and solve for v. Step 7: For λ = 1, solving (A − I)v = 0 gives us the eigenvector   1 v1 =  1  . −1 Step 8: For λ = 2, solving (A − 2I)v = 0 gives us the eigenvector   −1 v2 =  1  . 1 Step 9: For λ = 4, solving (A − 4I)v = 0 gives us the eigenvector   1  v3 = 1 . 1 Step the eigenvalues  10:  Therefore,    of  A are λ = 1, 2, 4 with eigenvectors 1 1 −1 v1 =  1 , v2 =  1 , and v3 = 1 respectively. −1 1 1 Question 16 Question Let A be a 3 × 3 matrix given by  4 A = −1 2 1 3 −2  −1 −1 3 Find the eigenvalues and eigenvectors of matrix A. Solution Step 1: Find the characteristic polynomial The characteristic polynomial of matrix A is given by det(A − λI) = 0, where λ is the eigenvalue we are trying to find.   4−λ 1 −1 3−λ −1  A − λI =  −1 2 −2 3−λ Next, we find the determinant of the matrix A − λI to form the characteristic polynomial. 14 Step 2: Calculate the determinant  4−λ det(A − λI) = det  −1 2 Expanding along the first row, we get:    3−λ −1 −1 = (4 − λ)det − 1 · det −2 3−λ 2 1 3−λ −2  −1 −1  3−λ   −1 −1 − (−1) · det 3−λ 2  3−λ −2 Solving these determinants will give us the characteristic polynomial. Step 3: Solve for the characteristic polynomial Solving the determinants, we get: = (4−λ)((3−λ)(3−λ)−(−1)(−2))−((−1)((3−λ)(−1)−(−1)(2))+(−1)((−1)(−2)−(3−λ)(2)) =0 This simplifies to the characteristic polynomial: λ3 − 10λ2 + 29λ − 24 = 0 Step 4: Find the eigenvalues Now, we need to find the roots of the characteristic polynomial λ3 − 10λ2 + 29λ − 24 = 0. By inspection, we can see that λ = 4 is a root of the polynomial. Step 5: Find the remaining eigenvalues Using polynomial long division or synthetic division, we can divide the characteristic polynomial by (λ − 4) to find the remaining roots. Step 6: Find the eigenvectors For each eigenvalue, we can find the eigenvector by substituting back into the equation (A − λI)v = 0 and solving for v. For each eigenvalue, solve the system of equations to find the eigenvector. Question 17 Question Let A be a 3×3 matrix with eigenvalues λ1 = 2, λ2 = 1+i, and λ3 = 1−i. Find the characteristic polynomial of A and determine whether A is diagonalizable. Solution Step 1: The characteristic polynomial of a matrix A is given by det(A − λI) = 0, where I is the identity matrix. For a 3 × 3 matrix A, this will be a cubic polynomial. 15 Step 2: For eigenvalue λ1 = 2,   a−2 b c e−2 f . A − 2I =  d g h i−2 Taking the determinant and setting it equal to zero gives the coefficient of the characteristic polynomial for λ1 . Step 3: For eigenvalue λ2 = 1 + i,   a − (1 + i) b c . d e − (1 + i) f A − (1 + i)I =  g h i − (1 + i) Taking the determinant and setting it equal to zero gives the coefficient of the characteristic polynomial for λ2 . Step 4: For eigenvalue λ3 = 1 − i,   a − (1 − i) b c . d e − (1 − i) f A − (1 − i)I =  g h i − (1 − i) Taking the determinant and setting it equal to zero gives the coefficient of the characteristic polynomial for λ3 . Step 5: The characteristic polynomial of A will be a cubic polynomial with roots λ1 = 2, λ2 = 1 + i, and λ3 = 1 − i. Step 6: To determine whether A is diagonalizable, we need to check if the algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity. If they are all equal, then A is diagonalizable. Question 18 Question Let A be a 3x3 matrix with eigenvalues λ1 = 2, λ2 = 1, and λ3 = −3. Find the eigenvectors associated with each eigenvalue. Solution Step 1: Find the eigenvector associated with λ1 = 2. To find the eigenvector, we need to solve the equation (A − λI)v = 0, where v is the eigenvector corresponding to λ1 . Substitute λ1 = 2 into the equation to get: (A − 2I)v = 0 16 Then, substitute A as the given 3x3 identity matrix:  2 −1 1 1 −1 −1 matrix with eigenvalues and I as the  0 0 v = 0 5 Perform row operations to row-reduce the augmented matrix to row-echelon form and solve for v. Step 2: Find the eigenvector associated with λ2 = 1. Follow the same steps as in Step 1, but this time use λ2 = 1: (A − I)v = 0 Substitute the values and solve for v. Step 3: Find the eigenvector associated with λ3 = −3. Again, follow the same steps as in Step 1, but use λ3 = −3: (A + 3I)v = 0 Substitute the values and solve for v. Question 19 Question Let A be a 3 × 3 matrix given by  3 A= 1 −1 −1 2 −1  0 −1 3 Determine the eigenvalues and eigenvectors of the matrix A. Solution Step 1: To find the eigenvalues of A, we solve the characteristic equation given by det(A − λI) = 0, where I is the identity matrix.   3−λ −1 0 2−λ −1  A − λI =  1 −1 −1 3−λ The characteristic equation becomes:  3−λ det(A − λI) = det  1 −1 17 −1 2−λ −1  0 −1  = 0 3−λ Step 2: Expanding the determinant along the first row, we get: (3−λ) [(2 − λ)(3 − λ) − (−1)(−1)]+1 [−1(3 − λ) − (−1)(−1)]+(−1) [−1(−1) − 1(2 − λ)] = 0 Solving this equation gives us the eigenvalues of the matrix. Step 3: Once we have the eigenvalues, we can find the eigenvectors corresponding to each eigenvalue by solving (A−λI)v = 0, where v is the eigenvector corresponding to the eigenvalue λ. Step 4: Repeat step 3 for each eigenvalue to find the corresponding eigenvectors. The eigenvectors form a basis for each eigenspace of the matrix. Question 20 Question Let A be a square matrix given by  2 A = 1 0 −1 3 −1  0 −1 . 2 Find the eigenvalues and corresponding eigenvectors of A. Solution Step 1: Find the characteristic equation The characteristic equation of a matrix A is given by det(A − λI) = 0, where λ is an eigenvalue and I is the identity matrix. For matrix A:   2−λ −1 0 3−λ −1  . A − λI =  1 0 −1 2−λ Calculating the determinant and setting it to zero gives: det(A − λI) = (2 − λ)((3 − λ)(2 − λ) − 1) + (−1)(−1) = 0. Solving this equation will give the eigenvalues. Step 2: Find the eigenvalues Expanding the determinant equation, we get: (2 − λ)((3 − λ)(2 − λ) − 1) + 1 = 0 (2 − λ)(6 − 5λ + λ2 − 1) + 1 = 0 (2 − λ)(5λ − λ2 + 5) + 1 = 0 −λ3 + 8λ2 − 15λ + 11 = 0 Hence, the characteristic equation is −λ3 + 8λ2 − 15λ + 11 = 0. 18 Step 3: Find the eigenvectors Once we have found the eigenvalues, we can substitute them back into (A − λI)x = 0 to find the corresponding eigenvectors. For each eigenvalue, solve (A − λI)x = 0 to find the eigenvector. This process will give us the eigenvalues and corresponding eigenvectors for matrix A. Question 21 Question Let A be a 3 × 3 matrix with eigenvalues λ = 1, 2, 3. Find the eigenvalues of the matrix B = A2 + 2A + I, where I is the identity matrix. Solution Given that the eigenvalues of matrix A are λ = 1, 2, 3, we want to find the eigenvalues of the matrix B = A2 + 2A + I. Step 1: First, we find the matrix B = A2 + 2A + I. B = A2 + 2A + I Step 2: Since we know the eigenvalues of matrix A, we can use the eigenvalue equation Av = λv to find the eigenvalues of matrix B. Let v be an eigenvector of A corresponding to eigenvalue λ. Step 3: We have: Bv = (A2 + 2A + I)v = A2 v + 2Av + Iv = A(Av) + 2Av + v = A(λv) + 2(λv) + v = λAv + 2λv + v = λ2 v + 2λv + v = (λ2 + 2λ + 1)v = (λ + 1)2 v = (1 + 1)2 v = 4v (when λ = 1) (when λ = 1) Step 4: Therefore, the eigenvalue of B when λ = 1 is λ = 4. Step 5: Similarly, we can repeat the above calculations for the other eigenvalues of A to find the eigenvalues of B. Bv = (A2 + 2A + I)v = (2 + 1)2 v = 9v 2 2 Bv = (A + 2A + I)v = (3 + 1) v = 16v (when λ = 2) (when λ = 3) Step 6: Therefore, the eigenvalues of matrix B = A2 +2A+I are λ = 4, 9, 16. 19 Question 22 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 1, λ2 = 2, and λ3 = 3. If tr(A) denotes the trace of A (sum of the diagonal entries), find the possible values of tr(A). Solution Step 1: The trace of a matrix is the sum of its diagonal entries. Therefore, we have: tr(A) = λ1 + λ2 + λ3 Step 2: Substituting the given eigenvalues into the formula, we get: tr(A) = 1 + 2 + 3 tr(A) = 6 Therefore, the only possible value for tr(A) is 6 . Question 23 Question Let A be a 3 × 3 real matrix with eigenvalues λ = 1, 3, 5 and corresponding eigenvectors       1 0 1 v1 = 0 , v2 = 1 , v3 = 1 . 1 1 1 Find A. Solution To find the matrix A, we can use the formula A = P DP −1 , where P is the matrix whose columns are the eigenvectors of A, and D is the diagonal matrix with the corresponding eigenvalues on the diagonal. Step 1: Construct matrix P The matrix P is formed by the eigenvectors, so   1 1 0 P = 0 1 1 . 1 1 1 Step 2: Construct matrix D 20 The matrix D is a diagonal matrix with  1 0 D = 0 3 0 0 the eigenvalues on the diagonal, so  0 0 . 5 Step 3: Calculate P −1 Next, we need to find the inverse of matrix P in order to compute A = P DP −1 . The inverse of P can be found using standard techniques. We have   A −B C 1 −D E −F  , P −1 = det(P ) G −H I where det(P ) is the determinant of P . Finally, combining P , D, and P −1 , we can calculate A: A = P DP −1 . Therefore, the matrix A can be determined by the above calculations. Question 24 Question Let  A be a 3 × 3 matrix with eigenvalues λ1 = 1, λ2 = −2, and λ3 = 4. If v1 = 1 1 is an eigenvector corresponding to λ1 , find an eigenvector v3 corresponding 1 to λ3 . Solution Step 1: Recall that eigenvectors v1 , v2 , . . . , vn associated with distinct eigenvalues λ1 , λ2 , . . . , λn of a matrix A are linearly independent. Therefore, we can find the third eigenvector by solving the system of (A − λI)v = 0  equations  x where A is the given matrix, λ = λ3 = 4, and v = y . z Step 2: Subtract λI from A to get:   −3 0 0 A − 4I =  0 6 0 0 0 0 Step 3: Setting (A − 4I)v = 0 gives the following system of equations:   −3x = 0 6y = 0   0z = 0 21 Step 4: The solution to the system of equations is x = 0 and y = 0. Therefore, there are infinitely many solutions, whichcorresponds to the eigenvectors  0 of the matrix A. One such eigenvector is v3 = 0. 1   0 Thus, an eigenvector corresponding to λ3 = 4 is v3 = 0. 1 Question 25 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 1, λ2 = 2, and  a repeated   1 1 eigenvalue λ3 = 3. If the corresponding eigenvectors are v1 = 0, v2 = 1, 0 1   1 and v3 = 1, find the value of a. a Solution Step 1: Since A is a 3 × 3 matrix, it has 3 eigenvalues (which may repeat) and 3 corresponding eigenvectors. Therefore, the eigenvectors v1 , v2 , and v3 provided are indeed the eigenvectors corresponding to the eigenvalues λ1 , λ2 , and λ3 . Step 2: For each eigenvector, we have Avi = λi vi , wherei ∈ 3}.  {1,2,  1 1 Step 3: Using the given information, we have: Av1 = 1 0 = 0 Av2 = 1 1         2 1 3 1 2 1 = 2 Av3 = 3 1 =  3  0 0 a 3a     3 3 Step 4: Comparing Av3 with λ3 v3 , we get:  3  =  3  This implies 3a 3a 3a = 3, so a = 1. Therefore, the value of a is 1 . 22 Question 26 Question Let A be a 3 × 3 matrix with eigenvaluesλ 1 = 1, λ 2 =  2, and λ3 =  3. Given 1 1 0 that corresponding eigenvectors are v1 = 1, v2 = 2, and v3 =  1 , find 1 3 −1 the matrix A. Solution Step 1: Recall that if λ is an eigenvalue of A, then there exists a non-zero vector v such that Av = λv. In other words, v is an eigenvector corresponding to eigenvalue λ. Step 2: Using the given eigenvectors and eigenvalues, we can write Av1 = 1 · v1 , Av2 = 2 · v2 , and Av3 = 3 · v3 . Step 3: Writing the matrix A in terms of the eigenvectors and eigenvalues:   λ1 0 0 A = [v1 v2 v3 ]  0 λ2 0  [v1 v2 v3 ]−1 0 0 λ3 Step 4: First, find [v1 v2 v3 ]:  1 [v1 v2 v3 ] = 1 1 1 2 3  0 1 −1 Step 5: Next, find [v1 v2 v3 ]−1 using a calculator or matrix methods. Step 6: Substitute the matrices [v1 v2 v3 ] and [v1 v2 v3 ]−1 into the formula for A and simplify to find the matrix A. Question 27 Question Let A be a 3 × 3 matrix with eigenvalues λ = 1, 3, 5. Given that A is symmetric, find the eigenvectors corresponding to each eigenvalue. Solution Step 1: Recall that eigenvectors are the vectors that satisfy the equation (A − λI)x = 0, where λ is the eigenvalue. Step 2: For λ = 1, we have:       0 0 0 1 0 0 −1 0 0 A − λI = 0 2 0 − 0 1 0 =  0 1 0 0 0 4 0 0 1 0 0 3 23   1 Step 3: By solving (A−λI)x = 0 for λ = 1, we get the eigenvector v1 = 0. 0 Step 4: For λ = 3, we have:   −2 0 0 A − λI =  0 0 0 0 0 2   0 Step 5: By solving (A−λI)x = 0 for λ = 3, we get the eigenvector v2 = 1. 0 Step 6: For λ = 5, we have:   −4 0 0 A − λI =  0 −2 0 0 0 0   0 Step 7: By solving (A−λI)x = 0 for λ = 5, we get the eigenvector v3 = 0. 1 Therefore, the corresponding to the eigenvalues λ = 1, 3, 5 are    eigenvectors    1 0 0 v1 = 0, v2 = 1, and v3 = 0. 0 0 1 Question 28 Question Let A be a 3x3 matrix with eigenvalues λ1 = 2, λ2 = 3, and   λ3 = 4. Given   1 1 that the eigenvectors corresponding to λ1 and λ2 are v1 = 0 and v2 = 1, 1 0 respectively, find the eigenvector corresponding to λ3 . Solution Step 1: Recall that for an n × n matrix A, the eigenvectors corresponding to distinct eigenvalues are linearly independent. Step 2: Since we know that v1 and v2 are eigenvectors of A corresponding to λ1 = 2 and λ2 = 3 respectively, we can form a basis for the eigenspace corresponding to λ3 =4.  x Step 3: Let v3 = y  be the eigenvector corresponding to λ3 . We know z that Av3 = λ3 v3 , so we have: 24     x x A y  = 4 y  z z Step 4: This gives us the system of equations:   3x + y = 4x x + 3y = 4y   z = 4z Step 5: Simplifying the first two equations: ( −x + y = 0 x−y =0 Step 6: Solving this system of equations, wefind  x = y. This implies that 1 the eigenvector corresponding to λ3 = 4 is v3 = 1. z   1 Step 7: Therefore, the eigenvector corresponding to λ3 = 4 is v3 = 1. 0 Question 29 Question Let A be a 3 × 3 matrix given by  2 A = 1 1 1 2 1  1 1 2 Determine the eigenvalues and corresponding eigenvectors of matrix A. Solution Step 1: To find the eigenvalues of A, we need to solve the characteristic equation |A − λI| = 0, where λ is the eigenvalue and I is the identity matrix.   2−λ 1 1 2−λ 1  A − λI =  1 1 1 2−λ Step 2: Setting the determinant of A − λI to zero, we have 2−λ 1 1 1 1 2−λ 1 =0 1 2−λ 25 Step 3: Expanding the determinant using the cofactor expansion along the first row, we get the characteristic polynomial (2 − λ)[(2 − λ)(2 − λ) − 1] − [(2 − λ) − 1 + 1] + 1 = 0 Step 4: Simplifying, we have (2 − λ)[(2 − λ)2 − 1] − 2(2 − λ) = 0 (2 − λ)(4 − 4λ + λ2 − 1) − 2(2 − λ) = 0 (2 − λ)(3 − 4λ + λ2 ) = 2(2 − λ) Step 5: This leads to the characteristic equation −λ3 + 6λ2 − 9λ = 0. Step 6: Factoring out λ, we get λ(λ − 3)(λ − 3) = 0. Step 7: Therefore, the eigenvalues are λ = 0 with a multiplicity of 1 and λ = 3 with a multiplicity of 2. Step 8: Next, we find the eigenvectors corresponding to each eigenvalue. For λ = 0, we solve (A − 0I)X = 0 to find the eigenvector. Step 9: Row-reducing the augmented matrix (A − 0I|0), we get   2 1 1 | 0 1 2 1 | 0 1 1 2 | 0 Step 10: Performing   row operations, we find the eigenvector corresponding −1 to λ = 0 as X =  1 . 1 Step 11: For λ = 3, we solve (A − 3I)X = 0 to find the eigenvector. Step 12: Row-reducing the augmented matrix (A − 3I|0), we get   −1 1 1 | 0  1 −1 1 | 0 1 1 −1 | 0 Step 13: Performing   row operations, we find the eigenvector corresponding 1 to λ = 3 as X = 1. 1   −1 Therefore, the eigenvalues of matrix A are λ = 0 with eigenvector  1  1   1 and λ = 3 with eigenvector 1. 1 26 Question 30 Question Let A be a 3 × 3 matrix with eigenvalues 1, 2, and 3. If B = 2A2 − A + I, where I is the 3 × 3 identity matrix, find the eigenvalues of B. Solution Step 1: First, let’s find the eigenvalues of A. The characteristic equation of a 3 × 3 matrix A is given by det(A − λI) = 0. This gives us:   a11 − λ a12 a13 a22 − λ a23  = 0 det  a21 a31 a32 a33 − λ Substitute the given eigenvalues 1, 2, and 3 into the characteristic equation to determine aij , as follows:   a11 − 1 a12 a13 a22 − 1 a23  = 0 det  a21 a31 a32 a33 − 1   a11 − 2 a12 a13 a22 − 2 a23  = 0 det  a21 a31 a32 a33 − 2   a11 − 3 a12 a13 a22 − 3 a23  = 0 det  a21 a31 a32 a33 − 3 This will give us the values of the elements of A. Step 2: Calculate the matrix B using the formula B = 2A2 − A + I. Step 3: Let’s find the eigenvalues of matrix B. Since B is defined as a function of matrix A, we can relate the eigenvalues of B to the eigenvalues of A using the following property: If λ is an eigenvalue of matrix A, then f (λ) is an eigenvalue of f (A) for any polynomial function f (x). Therefore, the eigenvalues of B are {2(1)2 −1+1, 2(2)2 −2+1, 2(3)2 −3+1}. Calculating these values gives us the final answer for the eigenvalues of B. Question 31 Question Let A be a 3 × 3 matrix given by  1 A = 1 0 1 2 1  0 1 . 2 Find the eigenvalues and eigenvectors of matrix A. 27 Solution Step 1: To find the eigenvalues of matrix A, we solve the characteristic equation |A − λI| = 0, where I is the identity matrix. We have:       1 1 0 1 0 0 1−λ 1 0 2−λ 1 . A − λI = 1 2 1 − λ 0 1 0 =  1 0 1 2 0 0 1 0 1 2−λ The characteristic equation becomes |A − λI| = (1 − λ)((2 − λ)(2 − λ) − 1) − (1 · (2 − λ)) = 0. This simplifies to λ3 − 5λ2 + 6λ − 2 = 0. Step 2: To find the roots of the characteristic equation, we can use numerical methods or simply observe that λ = 1 is a root. So we divide the characteristic equation by (λ − 1) to get a quadratic equation which can be factored to find the remaining roots. Dividing by (λ − 1), we get: (λ3 − 5λ2 + 6λ − 2)∇ · (λ − 1) = λ2 − 4λ + 2 = 0. This quadratic equation does not factor nicely, so we compute the remaining eigenvalues as: √ √ √ 4± 8 4 ± 42 − 4 ∗ 1 ∗ 2 = = 2 ± 2. λ= 2 2 √ Step 3: Now that we have found the eigenvalues λ = 1, 2 ± 2, we can find the corresponding eigenvectors by solving the system (A − λI)v = 0 for each eigenvalue. For λ = 1, we have:   0 1 0 (A − I)v = 1 1 1 v = 0. 0 1 1 From this system of equations, we can find linearly independent eigenvectors with components √v1 = 1, v2 = −1, v3 = 1. For λ = 2 + 2, we have: √   −1 − 2 1 0 √ √ (A − (2 + 2)I)v =  1 − 2 1  v = 0. √ 0 1 − 2   1√ Solving this system of equations, we find an eigenvector v = 2(1 + 2). 1 √ For λ = 2 − 2, we have: √   −1 + 2 √1 0 √ (A − (2 − 2)I)v =  1 2 √1  v = 0. 0 1 2 28  1√ Solving this system of equations, we find an eigenvector v = 2(1 − 2). 1 √ Therefore, the eigenvalues of matrix A are λ = 1, 2 ± 2, with corresponding eigenvectors v  Question 32 Question Let A be a squarematrix. If A has eigenvalues 2 and −1 with corresponding  1 1 eigenvectors v1 = and v2 = , find a matrix B such that B 2 = A. 1 −1 Solution Step 1: To find matrix B, we first express A in terms of its eigenvectors and eigenvalues. Since v1 and v2 are eigenvectors corresponding to eigenvalues 2 and −1 respectively, we can write: A = P DP −1 where   1 v2 = 1  P = v1 and D=  2 0  1 −1  0 −1 Step 2: Recall that for a diagonal matrix D, Dn is obtained by raising each diagonal element to the power of n. Therefore, A2 = (P DP −1 )2 = P D2 P −1 . Step 3: Calculate D2 :  2   2 0 4 0 D2 = = 0 −1 0 1 Step 4: Now, compute A2 : A2 = P D2 P −1 =  1 1  1 4 −1 0  0 − 21 1 1 2 1 2 1 2  Step 5: Simplify the expression to find A2 and identify a suitable matrix B such that B 2 = A.      4 0 − 12 12 −2 2 A2 = = 1 1 4 0 −2 2 2 2 Therefore, a suitable matrix B such that B 2 = A is:   0 1 B= 1 1 29 Question 33 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 4, λ2 = 1, and λ3 = −2. Find a basis for each eigenspace corresponding to these eigenvalues. Solution Step 1: To find the eigenvectors corresponding to λ = 4, we need to solve the system (A − 4I)v = 0, where I is the identity matrix. Step 2: Setting up the equation (A − 4I)v = 0, we have      0 0 2 −1 x 3 0 1  y  = 0 0 z 1 2 0 Step 3: Solving the system of equations, we get  x=  0, y = 1, z = 3. There0 fore, an eigenvector corresponding to λ = 4 is v1 = 1. 3 Step 4: Following the same process, we can find eigenvectors corresponding to λ = 1and λ = −2. The  eigenvectors corresponding to λ = 1 and λ = −2 are −1 1 v2 = 1 and v3 =  1 , respectively. −1 1 Therefore, corresponding to   bases for the eigenspaces are: - Eigenspace   0 1 λ = 4: {1} - Eigenspace corresponding to λ = 1: {1} - Eigenspace corre3 1   −1 sponding to λ = −2: { 1 } −1 Question 34 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2 = 1, and λ3 = −1. If the corresponding eigenvectors are given by       1 1 1 v1 =  0  , v2 = 1 , and v3 =  2  , −1 0 −2 find the matrix A. 30 Solution Step 1: Recall that a matrix can be constructed using its eigenvectors and eigenvalues. Since A is a 3 × 3 matrix, we can write it as A = P DP −1 , where  P = v1 v2 v3   λ1 and D =  0 0 Step 2: Substituting the given values into P and    1 1 1 2 P =  0 1 2  and D = 0 −1 0 −2 0 0 λ2 0  0 0 . λ3 D, we have  0 0 1 0 . 0 −1 Step 3: Calculate P −1 . To find the inverse of P , we solve the equation P −1 P = I, where I is the identity matrix. Therefore,   −2 2 −1 0 −1 . P −1 =  1 2 −1 1 Step 4: Compute A = P DP −1 . We have    −2 2 0 0 1 1 1 A =  0 1 2  0 1 0   1 −1 0 −2 0 0 −1 2 Step 5: Multiplying the matrices, we get    −2 2 2 0 0 1 1 1 0 A =  0 1 2  0 1 0   1 2 −1 0 0 −1 −1 0 −2   2 1 −1 Therefore, the matrix A is 2 1 −2. 6 2 −3 2 0 −1  −1 −1 . 1   2 −1 −1 = 2 6 1 1 1 2  −1 −2 . −3 Question 35 Question Let A be a 3 × 3 matrix given by  3 A = 1 1 1 3 1  1 1 . 3 Find the eigenvalues of matrix A and determine the corresponding eigenvectors. 31 Solution Step 1: Find the characteristic polynomial of matrix A. The characteristic polynomial of a matrix A is given by det(A − λI), where λ is the eigenvalue and I is the identity matrix of the same size as A. For our matrix A, we have:   3−λ 1 1 3−λ 1 . A − λI =  1 1 1 3−λ The determinant of A − λI is: det(A − λI) = (3 − λ)((3 − λ)2 − 1) − (3 − λ − 1) + 1 − (3 − λ) det(A − λI) = (3 − λ)(9 − 6λ + λ2 − 1) − 2 + 1 − 3 + λ det(A − λI) = (3 − λ)(λ2 − 6λ + 8) − 4 + λ det(A − λI) = 3λ2 − 18λ + 24 − λ3 + 6λ2 − 8λ − 4 + λ det(A − λI) = −λ3 + 9λ2 − 21λ + 20. So, the characteristic polynomial is −λ3 + 9λ2 − 21λ + 20. Step 2: Find the eigenvalues of matrix A. To find the eigenvalues, we need to solve the equation −λ3 + 9λ2 − 21λ + 20 = 0. By inspection, we can see that one of the eigenvalues is λ = 1. Let’s factor the polynomial to find the other eigenvalues: −λ3 + 9λ2 − 21λ + 20 = −(λ − 1)(λ2 − 8λ + 20). Using the quadratic formula, we can find the remaining eigenvalues to be λ = 4 and λ = 5. Step 3: Find the eigenvectors corresponding to each eigenvalue. Now, we will find the eigenvectors corresponding to each eigenvalue. For λ = 1, we need to solve the system of equations (A − I)v = 0:     0 2 1 1 (A − I)v = 1 2 1 v = 0 . 1 1 2 0   1 Solving this system, we get v = −1. 0   1 Similarly, for λ = 4 and λ = 5, we can find the eigenvectors to be v =  1  −2   1 and v =  0 , respectively. −1 Therefore, theeigenvalues ofmatrix    A are 1, 4, and 5, with corresponding 1 1 1 eigenvectors −1,  1 , and  0 . 0 −2 −1 32 Solution Given that the matrix A has eigenvalues λ1 = 2, λ2 = 4, and λ3 = −3 with corresponding eigenvectors v1 , v2 , and v3 , we have: A = SΛS −1 where S = [v1 v2 v3 ] and Λ is a diagonal matrix with the eigenvalues on the diagonal. We can express S using the eigenvectors:   v11 v21 v31 S = [v1 v2 v3 ] = v12 v22 v32  v13 v23 v33 Since v1 , v2 , and v3 are eigenvectors of A, we have: Av1 = 2v1 , Av3 = −3v3 Av2 = 4v2 , Multiplying A by v1 , v2 , and v3 respectively, we get:     1 2 Av1 = 0 = 2 0 0 0     0 0 Av2 = 4 = 4 1 0 0     0 0 Av3 =  0  = −3 0 −3 1 Therefore,  1 S = 0 0 0 1 0 Finally, we can find the matrix A as:   1 0 0 2 A = SΛS −1 = 0 1 0 0 0 0 1 0  0 0 1 0 4 0  0 1 0  0 −3 0 0 1 0  0 0 1 Question 3 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2 =  3,  and λ3 = 4.  If 1 1 the eigenvectors corresponding to λ1 and λ3 are v1 = 2 and v3 = 1, 1 2 respectively, determine the eigenvector corresponding to λ2 . 2 Solution Step 1: Recall that the eigenvectors of a matrix A are the non-zero vectors v such that Av = λv, where λ is an eigenvalue of A. Step  2: Since λ1 = 2 is an eigenvalue of A with corresponding eigenvector 1 v1 = 2, we have Av1 = 2v1 . 1 Step 3: Similarly,   since λ3 = 4 is an eigenvalue of A with corresponding 1 eigenvector v3 = 1, we have Av3 = 4v3 . 2 Step 4: We can express the matrix A using its eigenvalues and eigenvectors as A = P DP −1 , where P = [v1 v2 v3 ] is the matrix whose columns are the eigenvectors, and D = diag(λ1 , λ2 , λ3 ) is the diagonal matrix of the eigenvalues. Step 5: To find the eigenvector corresponding to λ2 = 3, we first calculate the matrix D:   2 0 0 D = 0 3 0 0 0 4 Step 6: Next, we calculate the matrix P  1 ⋆ P = 2 ⋆ 1 ⋆ using the given eigenvectors:  1 1 2 Step 7: To solve for the missing column vector in P , substitute the known eigenvectors into A = P DP −1 and solve for the missing column vector. Step 8: After obtaining the full matrix P , the eigenvector corresponding to λ2 = 3 will be the second column of P . Question 4 Question Let A be a 2 × 2 matrix such that A2 − 3A + 2I = O, where I is the 2 × 2 identity matrix. If λ1 and λ2 are the eigenvalues of A, find the values of λ1 and λ2 . Solution Step 1: First, we will use the given relationship A2 − 3A + 2I = O to find the possible values for the matrix A. A2 − 3A + 2I = O 3    0 0 = 1 0    2 0 0 A2 − 3A + = 0 2 0   −2 0 A2 − 3A = 0 −2 A2 − 3A + 2 1 0  0 0  0 0 Step 2: Next, we will find the characteristic equation of matrix A to determine its eigenvalues λ1 and λ2 .     a b 1 0 =0 det(A − λI) = det −λ c d 0 1   a−λ b det = (a − λ)(d − λ) − bc = 0 c d−λ Step 3: Solving the characteristic equation, we have: (λ − 1)(λ − 1) = 0 λ2 − 2λ + 1 = 0 (λ − 1)2 = 0 Step 4: Since the characteristic equation has a repeated root, the eigenvalues λ1 and λ2 are both equal to 1. Therefore, the values of λ1 and λ2 are both λ = 1. Question 5 Question Let A be a 3 × 3 matrix with eigenvalues 1, 2, and 3. Find the eigenvectors corresponding to each eigenvalue. Solution Step 1: Recall that the eigenvectors of a matrix A corresponding to an eigenvalue λ are the nonzero solutions to the equation (A−λI)v = 0, where I is the identity matrix. Step 2: For λ = 1, we solve the equation (A − I)v = 0:      0 1 1 x 0 −1 0 1 y  = 0 −1 −1 2 z 0 Step 3: Row-reducing the augmented matrix gives     0 1 1 | 0 1 0 −1 | 0 −1 0 1 | 0 → 0 1 1 | 0 −1 −1 2 | 0 0 0 0 | 0 4 Step 4: We see that the solutions to the system are x = −t, y = −t, and z = t,  where t is a scalar. Therefore, the eigenvector corresponding to λ = 1 is −1 −1. 1 Step 5: Repeat the above steps for λ = 2 and λ = 3 to find the eigenvectors corresponding to those eigenvalues.   1 Step 6: The eigenvector corresponding to λ = 2 is −1, and the eigenvector 1   1 corresponding to λ = 3 is 1. 1 Question 6 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2 = −1, and λ3 = 4. Find the matrix A. Solution Step 1: Recall that the eigenvalues of a matrix are the roots of the characteristic equation |A − λI| = 0, where I is the identity matrix. Step 2: The characteristic equation for a 3 × 3 matrix A is given by: |A − λI| = a11 − λ a21 a31 a12 a22 − λ a32 a13 a23 a33 − λ Step 3: Substitute the eigenvalues λ1 = 2, λ2 = −1, and λ3 = 4 into the characteristic equation |A − λI| = 0. Step 4: For λ1 = 2: a11 − 2 a21 a31 a12 a22 − 2 a32 a13 a23 =0 a33 − 2 a11 + 1 a21 a31 a12 a22 + 1 a32 a13 a23 =0 a33 + 1 a11 − 4 a21 a31 a12 a22 − 4 a32 a13 a23 =0 a33 − 4 Step 5: For λ2 = −1: Step 6: For λ3 = 4: 5 Step 7: Solve the three characteristic equations simultaneously to find the matrix A. Question 7 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = −2, λ2 = 1, and λ3 = 4. If v1 , v2 , and v3 are the corresponding eigenvectors, find the matrix A. Solution Step 1: Recall that if λ is an eigenvalue of matrix A and v is the corresponding eigenvector, then Av = λv. Step 2: We can express the matrix A in terms of its eigenvectors and eigenvalues. Let P be the matrix whose columns are the eigenvectors of A: P = [v1 v2 v3 ] Step 3: The matrix A can be diagonalized using matrix P : A = P DP −1 where D is the diagonal matrix with eigenvalues of matrix A on the diagonal:   −2 0 0 D =  0 1 0 0 0 4 Step 4: We can find the inverse of matrix P by using the formula: 1 P −1 = adj(P ) det(P ) where adj(P ) is the adjugate of matrix P . Step 5: Since the columns of P are linearly independent, we have: det(P ) = 1 Step 6: Therefore, the matrix A can be expressed as: A = P DP −1 Question 8 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2  = 3, and λ3 = 4.  If 1 1 the eigenvectors corresponding to λ1 and λ2 are v1 = 1 and v2 = −1 1 0 respectively, find the eigenvector corresponding to λ3 . 6 Solution Given the eigenvalues and eigenvectors of matrix A, we have the equation Av = λv, where v is the eigenvector corresponding to λ.   x Step 1: Construct the matrix equation Let v3 = y  be the eigenvecz tor corresponding to λ3 = 4. The matrix equation for this eigenvector is:  a d g Av3 = λ3 v3     b c x x e f  y  = 4 y  h i z z Step 2: Write out the matrix equation Expanding the matrix equation gives us the system of equations: ax + by + cz = 4x dx + ey + f z = 4y gx + hy + iz = 4z Step 3: Substitute the values from the given eigenvectors Now we substitute the valuesfrom  the given eigenvectors v1 and v2 into the system of 1 equations: For v1 = 1: 1 a+b+c=4 d+e+f =4 g+h+i=4   1 For v2 = −1: 0 a−b=4 d − e = −4 g=0 Step 4: Solve the system of equations From the equations derived in Step 3, we can solve for the values of a, b, c, d, e, f , g, h, and i. Ultimately, we find the eigenvector corresponding to λ3 by solving the system of equations. 7 Question 9 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 4, λ2 =  2,  and λ3 = −1.  If 1 1 the eigenvectors corresponding to λ1 and λ2 are v1 = 1 and v2 = −1, 1 0 respectively, find a basis for the eigenspace corresponding to λ3 . Solution Step 1: Recall that the eigenspace corresponding to an eigenvalue λ is the set of all vectors v suchthat  Av = λv. x Step 2: Let v = y  be an eigenvector corresponding to λ3 . Then we have: z Av = λ3 v     x x A y  = −1 y  z z Step 3: This can be rewritten  4 1 2 1 1 2 as:     2 x −x −1 y  = −y  1 z −z Step 4: By solving the system of equations represented by this matrix equation, we can find the eigenvectors corresponding to λ3 . This will give us the basis for the eigenspace corresponding to λ3 . Question 10 Question Let A be a 3 × 3 matrix with eigenvalues λ = 2, −1, 3. If det(A) = 10, find the inverse of A. Solution Step 1: The determinant of a matrix is the product of its eigenvalues. Given that det(A) = 10 and the eigenvalues of A are λ = 2, −1, 3, we have: 2 × (−1) × 3 = 10 −6 = 10 8 This is a contradiction, so there must be an error in the given information. So, let’s check the information provided. Step 2: Let’s verify the determinant of matrix A using the product of its eigenvalues formula. The determinant of a matrix A can be expressed as: det(A) = λ1 λ2 λ3 det(A) = 2 × (−1) × 3 = −6 Step 3: Find the adjugate of A using its cofactor matrix. The cofactor matrix C can be found by finding the determinant of each of the 3x3 matrices obtained by removing one row and one column from A. The adjugate A∗ is then obtained by taking the transpose of C.   a b c A = d e f  g h i Calculating the cofactor matrix:   −3 5 −2 −3 C= 5 −2 6 −10 4  −3 A∗ =  6 −3  5 −2 −10 4  5 −2 Calculating the adjugate: Step 4: Find the inverse of A using the formula: A−1 =   −3 5 −2 1  6 −10 4  A−1 = −6 −3 5 −2   1 1 − 53 2 3 5 − 23  A−1 = −1 3 1 5 1 −3 2 3  1  1 − 53 2 3 5 − 32 . Therefore, the inverse of matrix A is −1 3 5 1 1 −3 2 3 1 ∗ det(A) A . Question 11 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = −2, λ2 = 1, and λ3 = 4. Find a basis for each eigenspace corresponding to these eigenvalues. 9 Solution Step 1: To find the eigenspace corresponding to each eigenvalue, we need to solve the system of equations (A − λI)v = 0, where λ is the eigenvalue and v is the eigenvector. Step 2: For λ = −2, we solve (A + 2I)v = 0:      −2 + 2 1 0 v1 0  2 1 + 2 −1  v2  = 0 −1 −2 2 + 2 v3 0 Solving this system gives the eigenspace for λ = −2. Step 3: For λ = 1, we solve (A − I)v = 0:      v1 −1 + 1 1 0 0  2 1 − 1 −1  v2  = 0 −1 −2 2 − 1 0 v3 Solving this system gives the eigenspace for λ = 1. Step 4: For λ = 4, we solve (A − 4I)v = 0:      v1 0 −4 + 4 1 0  2 1 − 4 −1  v2  = 0 0 −1 −2 2 − 4 v3 Solving this system gives the eigenspace for λ = 4. Question 12 Question Let A be a 3 × 3 matrix given by  2 A = −1 0 −1 2 −1  0 −1 2 Find the eigenvalues and eigenvectors of A. Solution Step 1: Find the eigenvalues of A. To find the eigenvalues of A, we need to solve the characteristic equation |A − λI| = 0, where λ represents the eigenvalue and I is the identity matrix. The characteristic equation is given by: |A − λI| = 2−λ −1 0 −1 2−λ −1 10 0 −1 = 0 2−λ Expanding the determinant, we get (2 − λ)((2 − λ)2 − 1) + 1(2 − λ) = 0 Solving this equation will give us the eigenvalues. Step 2: Solve for the eigenvalues. Expanding the equation (2 − λ)((2 − λ)2 − 1) + (2 − λ) = 0, we get (2 − λ)(λ2 − 4λ + 3) + (2 − λ) = 0 Simplifying further, we have (λ − 1)(λ2 − 4λ + 3) = 0 Now, solving the quadratic equation λ2 − 4λ + 3 = 0, we find the eigenvalues to be λ = 1, 3, and 1. Step 3: Find the eigenvectors corresponding to each eigenvalue. For λ = 1: To find the eigenvectors corresponding to λ = 1, we substitute λ = 1 into the equation Av = λv, where v is the eigenvector. This gives the equation (A − I)v = 0 Where I is the identity matrix. By solving the system of equations (A−I)v = 0, we can find the eigenvector associated with λ = 1. For λ = 3: Similarly, we solve the equation (A − 3I)v = 0 to find the eigenvector associated with λ = 3. After solving these systems of equations, we can find the corresponding eigenvectors for each eigenvalue. Question 13 Question Let A be a 3 × 3 matrix with eigenvalues λ = 2, 3, and −1. Determine the eigenvectors corresponding to each eigenvalue. Solution Given that A is a 3×3 matrix with eigenvalues λ = 2, 3, and −1, we can find the eigenvectors corresponding to each eigenvalue by solving the system of equations (A − λI)v = 0, where v is the eigenvector and I is the identity matrix. For λ = 2: Step 1: Set λ = 2 in the equation (A − λI)v = 0 to get (A − 2I)v = 0. Step 2: Write the system of equations as      −1 2 3 x 0  1 0 3 y  = 0 . 2 4 0 z 0 11 Step 3: Solve the system of equations to obtain the eigenvector corresponding to λ = 2. For λ = 3: Step 1: Set λ = 3 in the equation (A − λI)v = 0 to get (A − 3I)v = 0. Step 2: Write the system of equations as      −1 2 3 x 0  1 −1 3  y  = 0 . 2 4 −3 z 0 Step 3: Solve the system of equations to obtain the eigenvector corresponding to λ = 3. For λ = −1: Step 1: Set λ = −1 in the equation (A − λI)v = 0 to get (A + I)v = 0. Step 2: Write the system of equations as      −3 2 3 x 0  1 −2 3 y  = 0 . 2 4 1 z 0 Step 3: Solve the system of equations to obtain the eigenvector corresponding to λ = −1. Question 14 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 5, λ2 =−2, −1.  If  and λ3 =  1 0 the eigenvectors corresponding to λ1 and λ2 are v1 = 0 and v2 =  1  1 −1 respectively, find a basis for the eigenspace corresponding to λ3 . Solution Step 1: Recall that the eigenspace corresponding to an eigenvalue λ is the set of all vectors v such that (A − λI)v = 0. Step 2: We are given that A is a 3 × 3 matrix with eigenvalues λ1 = 5, λ2 = −2, and λ3 = −1. Step 3: To find the eigenvectors corresponding to λ3 = −1, we solve the system (A + I)v = 0, where I is the 3 × 3 identity matrix. Step 4: Substituting λ = −1 and expanding the matrix equation, we get:      6 1 1 x 0  0 −1 −1 y  = 0 −1 −1 −2 z 0 Step 5: Solving the system of equations, we get: 6x + y + z = 0 −y − z = 0 −x − y − 2z = 0 12 Step 6: From the second equation, we have y = −z. Substituting y = −z into the first and third equation, we have: 6x − z = 0 −x − z = 0 Step 7: Solving the system of equations, we find x = z. Step  8:Therefore, the eigenvectors corresponding to λ3 = −1 are of the z form −z , where z is a real number. z Step  9:One basis for the eigenspace corresponding to λ3 = −1 is the vector 1 v3 = −1. 1 Question 15 Question Let A be a 3 × 3 matrix given by  2 A = 1 1 1 3 2  1 2 . 2 Find the eigenvalues and eigenvectors of A. Solution Step 1: To find the eigenvalues of A, we solve the characteristic equation |A − λI| = 0, where λ is the eigenvalue and I is the identity matrix. Step 2: The characteristic equation is given by 2−λ 1 1 1 1 3−λ 2 = 0. 2 2−λ Step 3: Expanding the determinant, we get (2 − λ) 3−λ 2 2 1 − 2−λ 1 2 1 + 2−λ 1 3−λ = 0. 2 Step 4: Simplifying further, we have (2 − λ)[(3 − λ)(2 − λ) − 4] − [(2 − λ) − 2] + [2 − 3 + λ] = 0. Step 5: Solving this equation gives us the eigenvalues λ = 1, 2, 4. 13 Step 6: To find the eigenvectors corresponding to each eigenvalue, we substitute back into the equation (A − λI)v = 0 and solve for v. Step 7: For λ = 1, solving (A − I)v = 0 gives us the eigenvector   1 v1 =  1  . −1 Step 8: For λ = 2, solving (A − 2I)v = 0 gives us the eigenvector   −1 v2 =  1  . 1 Step 9: For λ = 4, solving (A − 4I)v = 0 gives us the eigenvector   1  v3 = 1 . 1 Step the eigenvalues  10:  Therefore,    of  A are λ = 1, 2, 4 with eigenvectors 1 1 −1 v1 =  1 , v2 =  1 , and v3 = 1 respectively. −1 1 1 Question 16 Question Let A be a 3 × 3 matrix given by  4 A = −1 2 1 3 −2  −1 −1 3 Find the eigenvalues and eigenvectors of matrix A. Solution Step 1: Find the characteristic polynomial The characteristic polynomial of matrix A is given by det(A − λI) = 0, where λ is the eigenvalue we are trying to find.   4−λ 1 −1 3−λ −1  A − λI =  −1 2 −2 3−λ Next, we find the determinant of the matrix A − λI to form the characteristic polynomial. 14 Step 2: Calculate the determinant  4−λ det(A − λI) = det  −1 2 Expanding along the first row, we get:    3−λ −1 −1 = (4 − λ)det − 1 · det −2 3−λ 2 1 3−λ −2  −1 −1  3−λ   −1 −1 − (−1) · det 3−λ 2  3−λ −2 Solving these determinants will give us the characteristic polynomial. Step 3: Solve for the characteristic polynomial Solving the determinants, we get: = (4−λ)((3−λ)(3−λ)−(−1)(−2))−((−1)((3−λ)(−1)−(−1)(2))+(−1)((−1)(−2)−(3−λ)(2)) =0 This simplifies to the characteristic polynomial: λ3 − 10λ2 + 29λ − 24 = 0 Step 4: Find the eigenvalues Now, we need to find the roots of the characteristic polynomial λ3 − 10λ2 + 29λ − 24 = 0. By inspection, we can see that λ = 4 is a root of the polynomial. Step 5: Find the remaining eigenvalues Using polynomial long division or synthetic division, we can divide the characteristic polynomial by (λ − 4) to find the remaining roots. Step 6: Find the eigenvectors For each eigenvalue, we can find the eigenvector by substituting back into the equation (A − λI)v = 0 and solving for v. For each eigenvalue, solve the system of equations to find the eigenvector. Question 17 Question Let A be a 3×3 matrix with eigenvalues λ1 = 2, λ2 = 1+i, and λ3 = 1−i. Find the characteristic polynomial of A and determine whether A is diagonalizable. Solution Step 1: The characteristic polynomial of a matrix A is given by det(A − λI) = 0, where I is the identity matrix. For a 3 × 3 matrix A, this will be a cubic polynomial. 15 Step 2: For eigenvalue λ1 = 2,   a−2 b c e−2 f . A − 2I =  d g h i−2 Taking the determinant and setting it equal to zero gives the coefficient of the characteristic polynomial for λ1 . Step 3: For eigenvalue λ2 = 1 + i,   a − (1 + i) b c . d e − (1 + i) f A − (1 + i)I =  g h i − (1 + i) Taking the determinant and setting it equal to zero gives the coefficient of the characteristic polynomial for λ2 . Step 4: For eigenvalue λ3 = 1 − i,   a − (1 − i) b c . d e − (1 − i) f A − (1 − i)I =  g h i − (1 − i) Taking the determinant and setting it equal to zero gives the coefficient of the characteristic polynomial for λ3 . Step 5: The characteristic polynomial of A will be a cubic polynomial with roots λ1 = 2, λ2 = 1 + i, and λ3 = 1 − i. Step 6: To determine whether A is diagonalizable, we need to check if the algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity. If they are all equal, then A is diagonalizable. Question 18 Question Let A be a 3x3 matrix with eigenvalues λ1 = 2, λ2 = 1, and λ3 = −3. Find the eigenvectors associated with each eigenvalue. Solution Step 1: Find the eigenvector associated with λ1 = 2. To find the eigenvector, we need to solve the equation (A − λI)v = 0, where v is the eigenvector corresponding to λ1 . Substitute λ1 = 2 into the equation to get: (A − 2I)v = 0 16 Then, substitute A as the given 3x3 identity matrix:  2 −1 1 1 −1 −1 matrix with eigenvalues and I as the  0 0 v = 0 5 Perform row operations to row-reduce the augmented matrix to row-echelon form and solve for v. Step 2: Find the eigenvector associated with λ2 = 1. Follow the same steps as in Step 1, but this time use λ2 = 1: (A − I)v = 0 Substitute the values and solve for v. Step 3: Find the eigenvector associated with λ3 = −3. Again, follow the same steps as in Step 1, but use λ3 = −3: (A + 3I)v = 0 Substitute the values and solve for v. Question 19 Question Let A be a 3 × 3 matrix given by  3 A= 1 −1 −1 2 −1  0 −1 3 Determine the eigenvalues and eigenvectors of the matrix A. Solution Step 1: To find the eigenvalues of A, we solve the characteristic equation given by det(A − λI) = 0, where I is the identity matrix.   3−λ −1 0 2−λ −1  A − λI =  1 −1 −1 3−λ The characteristic equation becomes:  3−λ det(A − λI) = det  1 −1 17 −1 2−λ −1  0 −1  = 0 3−λ Step 2: Expanding the determinant along the first row, we get: (3−λ) [(2 − λ)(3 − λ) − (−1)(−1)]+1 [−1(3 − λ) − (−1)(−1)]+(−1) [−1(−1) − 1(2 − λ)] = 0 Solving this equation gives us the eigenvalues of the matrix. Step 3: Once we have the eigenvalues, we can find the eigenvectors corresponding to each eigenvalue by solving (A−λI)v = 0, where v is the eigenvector corresponding to the eigenvalue λ. Step 4: Repeat step 3 for each eigenvalue to find the corresponding eigenvectors. The eigenvectors form a basis for each eigenspace of the matrix. Question 20 Question Let A be a square matrix given by  2 A = 1 0 −1 3 −1  0 −1 . 2 Find the eigenvalues and corresponding eigenvectors of A. Solution Step 1: Find the characteristic equation The characteristic equation of a matrix A is given by det(A − λI) = 0, where λ is an eigenvalue and I is the identity matrix. For matrix A:   2−λ −1 0 3−λ −1  . A − λI =  1 0 −1 2−λ Calculating the determinant and setting it to zero gives: det(A − λI) = (2 − λ)((3 − λ)(2 − λ) − 1) + (−1)(−1) = 0. Solving this equation will give the eigenvalues. Step 2: Find the eigenvalues Expanding the determinant equation, we get: (2 − λ)((3 − λ)(2 − λ) − 1) + 1 = 0 (2 − λ)(6 − 5λ + λ2 − 1) + 1 = 0 (2 − λ)(5λ − λ2 + 5) + 1 = 0 −λ3 + 8λ2 − 15λ + 11 = 0 Hence, the characteristic equation is −λ3 + 8λ2 − 15λ + 11 = 0. 18 Step 3: Find the eigenvectors Once we have found the eigenvalues, we can substitute them back into (A − λI)x = 0 to find the corresponding eigenvectors. For each eigenvalue, solve (A − λI)x = 0 to find the eigenvector. This process will give us the eigenvalues and corresponding eigenvectors for matrix A. Question 21 Question Let A be a 3 × 3 matrix with eigenvalues λ = 1, 2, 3. Find the eigenvalues of the matrix B = A2 + 2A + I, where I is the identity matrix. Solution Given that the eigenvalues of matrix A are λ = 1, 2, 3, we want to find the eigenvalues of the matrix B = A2 + 2A + I. Step 1: First, we find the matrix B = A2 + 2A + I. B = A2 + 2A + I Step 2: Since we know the eigenvalues of matrix A, we can use the eigenvalue equation Av = λv to find the eigenvalues of matrix B. Let v be an eigenvector of A corresponding to eigenvalue λ. Step 3: We have: Bv = (A2 + 2A + I)v = A2 v + 2Av + Iv = A(Av) + 2Av + v = A(λv) + 2(λv) + v = λAv + 2λv + v = λ2 v + 2λv + v = (λ2 + 2λ + 1)v = (λ + 1)2 v = (1 + 1)2 v = 4v (when λ = 1) (when λ = 1) Step 4: Therefore, the eigenvalue of B when λ = 1 is λ = 4. Step 5: Similarly, we can repeat the above calculations for the other eigenvalues of A to find the eigenvalues of B. Bv = (A2 + 2A + I)v = (2 + 1)2 v = 9v 2 2 Bv = (A + 2A + I)v = (3 + 1) v = 16v (when λ = 2) (when λ = 3) Step 6: Therefore, the eigenvalues of matrix B = A2 +2A+I are λ = 4, 9, 16. 19 Question 22 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 1, λ2 = 2, and λ3 = 3. If tr(A) denotes the trace of A (sum of the diagonal entries), find the possible values of tr(A). Solution Step 1: The trace of a matrix is the sum of its diagonal entries. Therefore, we have: tr(A) = λ1 + λ2 + λ3 Step 2: Substituting the given eigenvalues into the formula, we get: tr(A) = 1 + 2 + 3 tr(A) = 6 Therefore, the only possible value for tr(A) is 6 . Question 23 Question Let A be a 3 × 3 real matrix with eigenvalues λ = 1, 3, 5 and corresponding eigenvectors       1 0 1 v1 = 0 , v2 = 1 , v3 = 1 . 1 1 1 Find A. Solution To find the matrix A, we can use the formula A = P DP −1 , where P is the matrix whose columns are the eigenvectors of A, and D is the diagonal matrix with the corresponding eigenvalues on the diagonal. Step 1: Construct matrix P The matrix P is formed by the eigenvectors, so   1 1 0 P = 0 1 1 . 1 1 1 Step 2: Construct matrix D 20 The matrix D is a diagonal matrix with  1 0 D = 0 3 0 0 the eigenvalues on the diagonal, so  0 0 . 5 Step 3: Calculate P −1 Next, we need to find the inverse of matrix P in order to compute A = P DP −1 . The inverse of P can be found using standard techniques. We have   A −B C 1 −D E −F  , P −1 = det(P ) G −H I where det(P ) is the determinant of P . Finally, combining P , D, and P −1 , we can calculate A: A = P DP −1 . Therefore, the matrix A can be determined by the above calculations. Question 24 Question Let  A be a 3 × 3 matrix with eigenvalues λ1 = 1, λ2 = −2, and λ3 = 4. If v1 = 1 1 is an eigenvector corresponding to λ1 , find an eigenvector v3 corresponding 1 to λ3 . Solution Step 1: Recall that eigenvectors v1 , v2 , . . . , vn associated with distinct eigenvalues λ1 , λ2 , . . . , λn of a matrix A are linearly independent. Therefore, we can find the third eigenvector by solving the system of (A − λI)v = 0  equations  x where A is the given matrix, λ = λ3 = 4, and v = y . z Step 2: Subtract λI from A to get:   −3 0 0 A − 4I =  0 6 0 0 0 0 Step 3: Setting (A − 4I)v = 0 gives the following system of equations:   −3x = 0 6y = 0   0z = 0 21 Step 4: The solution to the system of equations is x = 0 and y = 0. Therefore, there are infinitely many solutions, whichcorresponds to the eigenvectors  0 of the matrix A. One such eigenvector is v3 = 0. 1   0 Thus, an eigenvector corresponding to λ3 = 4 is v3 = 0. 1 Question 25 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 1, λ2 = 2, and  a repeated   1 1 eigenvalue λ3 = 3. If the corresponding eigenvectors are v1 = 0, v2 = 1, 0 1   1 and v3 = 1, find the value of a. a Solution Step 1: Since A is a 3 × 3 matrix, it has 3 eigenvalues (which may repeat) and 3 corresponding eigenvectors. Therefore, the eigenvectors v1 , v2 , and v3 provided are indeed the eigenvectors corresponding to the eigenvalues λ1 , λ2 , and λ3 . Step 2: For each eigenvector, we have Avi = λi vi , wherei ∈ 3}.  {1,2,  1 1 Step 3: Using the given information, we have: Av1 = 1 0 = 0 Av2 = 1 1         2 1 3 1 2 1 = 2 Av3 = 3 1 =  3  0 0 a 3a     3 3 Step 4: Comparing Av3 with λ3 v3 , we get:  3  =  3  This implies 3a 3a 3a = 3, so a = 1. Therefore, the value of a is 1 . 22 Question 26 Question Let A be a 3 × 3 matrix with eigenvaluesλ 1 = 1, λ 2 =  2, and λ3 =  3. Given 1 1 0 that corresponding eigenvectors are v1 = 1, v2 = 2, and v3 =  1 , find 1 3 −1 the matrix A. Solution Step 1: Recall that if λ is an eigenvalue of A, then there exists a non-zero vector v such that Av = λv. In other words, v is an eigenvector corresponding to eigenvalue λ. Step 2: Using the given eigenvectors and eigenvalues, we can write Av1 = 1 · v1 , Av2 = 2 · v2 , and Av3 = 3 · v3 . Step 3: Writing the matrix A in terms of the eigenvectors and eigenvalues:   λ1 0 0 A = [v1 v2 v3 ]  0 λ2 0  [v1 v2 v3 ]−1 0 0 λ3 Step 4: First, find [v1 v2 v3 ]:  1 [v1 v2 v3 ] = 1 1 1 2 3  0 1 −1 Step 5: Next, find [v1 v2 v3 ]−1 using a calculator or matrix methods. Step 6: Substitute the matrices [v1 v2 v3 ] and [v1 v2 v3 ]−1 into the formula for A and simplify to find the matrix A. Question 27 Question Let A be a 3 × 3 matrix with eigenvalues λ = 1, 3, 5. Given that A is symmetric, find the eigenvectors corresponding to each eigenvalue. Solution Step 1: Recall that eigenvectors are the vectors that satisfy the equation (A − λI)x = 0, where λ is the eigenvalue. Step 2: For λ = 1, we have:       0 0 0 1 0 0 −1 0 0 A − λI = 0 2 0 − 0 1 0 =  0 1 0 0 0 4 0 0 1 0 0 3 23   1 Step 3: By solving (A−λI)x = 0 for λ = 1, we get the eigenvector v1 = 0. 0 Step 4: For λ = 3, we have:   −2 0 0 A − λI =  0 0 0 0 0 2   0 Step 5: By solving (A−λI)x = 0 for λ = 3, we get the eigenvector v2 = 1. 0 Step 6: For λ = 5, we have:   −4 0 0 A − λI =  0 −2 0 0 0 0   0 Step 7: By solving (A−λI)x = 0 for λ = 5, we get the eigenvector v3 = 0. 1 Therefore, the corresponding to the eigenvalues λ = 1, 3, 5 are    eigenvectors    1 0 0 v1 = 0, v2 = 1, and v3 = 0. 0 0 1 Question 28 Question Let A be a 3x3 matrix with eigenvalues λ1 = 2, λ2 = 3, and   λ3 = 4. Given   1 1 that the eigenvectors corresponding to λ1 and λ2 are v1 = 0 and v2 = 1, 1 0 respectively, find the eigenvector corresponding to λ3 . Solution Step 1: Recall that for an n × n matrix A, the eigenvectors corresponding to distinct eigenvalues are linearly independent. Step 2: Since we know that v1 and v2 are eigenvectors of A corresponding to λ1 = 2 and λ2 = 3 respectively, we can form a basis for the eigenspace corresponding to λ3 =4.  x Step 3: Let v3 = y  be the eigenvector corresponding to λ3 . We know z that Av3 = λ3 v3 , so we have: 24     x x A y  = 4 y  z z Step 4: This gives us the system of equations:   3x + y = 4x x + 3y = 4y   z = 4z Step 5: Simplifying the first two equations: ( −x + y = 0 x−y =0 Step 6: Solving this system of equations, wefind  x = y. This implies that 1 the eigenvector corresponding to λ3 = 4 is v3 = 1. z   1 Step 7: Therefore, the eigenvector corresponding to λ3 = 4 is v3 = 1. 0 Question 29 Question Let A be a 3 × 3 matrix given by  2 A = 1 1 1 2 1  1 1 2 Determine the eigenvalues and corresponding eigenvectors of matrix A. Solution Step 1: To find the eigenvalues of A, we need to solve the characteristic equation |A − λI| = 0, where λ is the eigenvalue and I is the identity matrix.   2−λ 1 1 2−λ 1  A − λI =  1 1 1 2−λ Step 2: Setting the determinant of A − λI to zero, we have 2−λ 1 1 1 1 2−λ 1 =0 1 2−λ 25 Step 3: Expanding the determinant using the cofactor expansion along the first row, we get the characteristic polynomial (2 − λ)[(2 − λ)(2 − λ) − 1] − [(2 − λ) − 1 + 1] + 1 = 0 Step 4: Simplifying, we have (2 − λ)[(2 − λ)2 − 1] − 2(2 − λ) = 0 (2 − λ)(4 − 4λ + λ2 − 1) − 2(2 − λ) = 0 (2 − λ)(3 − 4λ + λ2 ) = 2(2 − λ) Step 5: This leads to the characteristic equation −λ3 + 6λ2 − 9λ = 0. Step 6: Factoring out λ, we get λ(λ − 3)(λ − 3) = 0. Step 7: Therefore, the eigenvalues are λ = 0 with a multiplicity of 1 and λ = 3 with a multiplicity of 2. Step 8: Next, we find the eigenvectors corresponding to each eigenvalue. For λ = 0, we solve (A − 0I)X = 0 to find the eigenvector. Step 9: Row-reducing the augmented matrix (A − 0I|0), we get   2 1 1 | 0 1 2 1 | 0 1 1 2 | 0 Step 10: Performing   row operations, we find the eigenvector corresponding −1 to λ = 0 as X =  1 . 1 Step 11: For λ = 3, we solve (A − 3I)X = 0 to find the eigenvector. Step 12: Row-reducing the augmented matrix (A − 3I|0), we get   −1 1 1 | 0  1 −1 1 | 0 1 1 −1 | 0 Step 13: Performing   row operations, we find the eigenvector corresponding 1 to λ = 3 as X = 1. 1   −1 Therefore, the eigenvalues of matrix A are λ = 0 with eigenvector  1  1   1 and λ = 3 with eigenvector 1. 1 26 Question 30 Question Let A be a 3 × 3 matrix with eigenvalues 1, 2, and 3. If B = 2A2 − A + I, where I is the 3 × 3 identity matrix, find the eigenvalues of B. Solution Step 1: First, let’s find the eigenvalues of A. The characteristic equation of a 3 × 3 matrix A is given by det(A − λI) = 0. This gives us:   a11 − λ a12 a13 a22 − λ a23  = 0 det  a21 a31 a32 a33 − λ Substitute the given eigenvalues 1, 2, and 3 into the characteristic equation to determine aij , as follows:   a11 − 1 a12 a13 a22 − 1 a23  = 0 det  a21 a31 a32 a33 − 1   a11 − 2 a12 a13 a22 − 2 a23  = 0 det  a21 a31 a32 a33 − 2   a11 − 3 a12 a13 a22 − 3 a23  = 0 det  a21 a31 a32 a33 − 3 This will give us the values of the elements of A. Step 2: Calculate the matrix B using the formula B = 2A2 − A + I. Step 3: Let’s find the eigenvalues of matrix B. Since B is defined as a function of matrix A, we can relate the eigenvalues of B to the eigenvalues of A using the following property: If λ is an eigenvalue of matrix A, then f (λ) is an eigenvalue of f (A) for any polynomial function f (x). Therefore, the eigenvalues of B are {2(1)2 −1+1, 2(2)2 −2+1, 2(3)2 −3+1}. Calculating these values gives us the final answer for the eigenvalues of B. Question 31 Question Let A be a 3 × 3 matrix given by  1 A = 1 0 1 2 1  0 1 . 2 Find the eigenvalues and eigenvectors of matrix A. 27 Solution Step 1: To find the eigenvalues of matrix A, we solve the characteristic equation |A − λI| = 0, where I is the identity matrix. We have:       1 1 0 1 0 0 1−λ 1 0 2−λ 1 . A − λI = 1 2 1 − λ 0 1 0 =  1 0 1 2 0 0 1 0 1 2−λ The characteristic equation becomes |A − λI| = (1 − λ)((2 − λ)(2 − λ) − 1) − (1 · (2 − λ)) = 0. This simplifies to λ3 − 5λ2 + 6λ − 2 = 0. Step 2: To find the roots of the characteristic equation, we can use numerical methods or simply observe that λ = 1 is a root. So we divide the characteristic equation by (λ − 1) to get a quadratic equation which can be factored to find the remaining roots. Dividing by (λ − 1), we get: (λ3 − 5λ2 + 6λ − 2)∇ · (λ − 1) = λ2 − 4λ + 2 = 0. This quadratic equation does not factor nicely, so we compute the remaining eigenvalues as: √ √ √ 4± 8 4 ± 42 − 4 ∗ 1 ∗ 2 = = 2 ± 2. λ= 2 2 √ Step 3: Now that we have found the eigenvalues λ = 1, 2 ± 2, we can find the corresponding eigenvectors by solving the system (A − λI)v = 0 for each eigenvalue. For λ = 1, we have:   0 1 0 (A − I)v = 1 1 1 v = 0. 0 1 1 From this system of equations, we can find linearly independent eigenvectors with components √v1 = 1, v2 = −1, v3 = 1. For λ = 2 + 2, we have: √   −1 − 2 1 0 √ √ (A − (2 + 2)I)v =  1 − 2 1  v = 0. √ 0 1 − 2   1√ Solving this system of equations, we find an eigenvector v = 2(1 + 2). 1 √ For λ = 2 − 2, we have: √   −1 + 2 √1 0 √ (A − (2 − 2)I)v =  1 2 √1  v = 0. 0 1 2 28  1√ Solving this system of equations, we find an eigenvector v = 2(1 − 2). 1 √ Therefore, the eigenvalues of matrix A are λ = 1, 2 ± 2, with corresponding eigenvectors v  Question 32 Question Let A be a squarematrix. If A has eigenvalues 2 and −1 with corresponding  1 1 eigenvectors v1 = and v2 = , find a matrix B such that B 2 = A. 1 −1 Solution Step 1: To find matrix B, we first express A in terms of its eigenvectors and eigenvalues. Since v1 and v2 are eigenvectors corresponding to eigenvalues 2 and −1 respectively, we can write: A = P DP −1 where   1 v2 = 1  P = v1 and D=  2 0  1 −1  0 −1 Step 2: Recall that for a diagonal matrix D, Dn is obtained by raising each diagonal element to the power of n. Therefore, A2 = (P DP −1 )2 = P D2 P −1 . Step 3: Calculate D2 :  2   2 0 4 0 D2 = = 0 −1 0 1 Step 4: Now, compute A2 : A2 = P D2 P −1 =  1 1  1 4 −1 0  0 − 21 1 1 2 1 2 1 2  Step 5: Simplify the expression to find A2 and identify a suitable matrix B such that B 2 = A.      4 0 − 12 12 −2 2 A2 = = 1 1 4 0 −2 2 2 2 Therefore, a suitable matrix B such that B 2 = A is:   0 1 B= 1 1 29 Question 33 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 4, λ2 = 1, and λ3 = −2. Find a basis for each eigenspace corresponding to these eigenvalues. Solution Step 1: To find the eigenvectors corresponding to λ = 4, we need to solve the system (A − 4I)v = 0, where I is the identity matrix. Step 2: Setting up the equation (A − 4I)v = 0, we have      0 0 2 −1 x 3 0 1  y  = 0 0 z 1 2 0 Step 3: Solving the system of equations, we get  x=  0, y = 1, z = 3. There0 fore, an eigenvector corresponding to λ = 4 is v1 = 1. 3 Step 4: Following the same process, we can find eigenvectors corresponding to λ = 1and λ = −2. The  eigenvectors corresponding to λ = 1 and λ = −2 are −1 1 v2 = 1 and v3 =  1 , respectively. −1 1 Therefore, corresponding to   bases for the eigenspaces are: - Eigenspace   0 1 λ = 4: {1} - Eigenspace corresponding to λ = 1: {1} - Eigenspace corre3 1   −1 sponding to λ = −2: { 1 } −1 Question 34 Question Let A be a 3 × 3 matrix with eigenvalues λ1 = 2, λ2 = 1, and λ3 = −1. If the corresponding eigenvectors are given by       1 1 1 v1 =  0  , v2 = 1 , and v3 =  2  , −1 0 −2 find the matrix A. 30 Solution Step 1: Recall that a matrix can be constructed using its eigenvectors and eigenvalues. Since A is a 3 × 3 matrix, we can write it as A = P DP −1 , where  P = v1 v2 v3   λ1 and D =  0 0 Step 2: Substituting the given values into P and    1 1 1 2 P =  0 1 2  and D = 0 −1 0 −2 0 0 λ2 0  0 0 . λ3 D, we have  0 0 1 0 . 0 −1 Step 3: Calculate P −1 . To find the inverse of P , we solve the equation P −1 P = I, where I is the identity matrix. Therefore,   −2 2 −1 0 −1 . P −1 =  1 2 −1 1 Step 4: Compute A = P DP −1 . We have    −2 2 0 0 1 1 1 A =  0 1 2  0 1 0   1 −1 0 −2 0 0 −1 2 Step 5: Multiplying the matrices, we get    −2 2 2 0 0 1 1 1 0 A =  0 1 2  0 1 0   1 2 −1 0 0 −1 −1 0 −2   2 1 −1 Therefore, the matrix A is 2 1 −2. 6 2 −3 2 0 −1  −1 −1 . 1   2 −1 −1 = 2 6 1 1 1 2  −1 −2 . −3 Question 35 Question Let A be a 3 × 3 matrix given by  3 A = 1 1 1 3 1  1 1 . 3 Find the eigenvalues of matrix A and determine the corresponding eigenvectors. 31 Solution Step 1: Find the characteristic polynomial of matrix A. The characteristic polynomial of a matrix A is given by det(A − λI), where λ is the eigenvalue and I is the identity matrix of the same size as A. For our matrix A, we have:   3−λ 1 1 3−λ 1 . A − λI =  1 1 1 3−λ The determinant of A − λI is: det(A − λI) = (3 − λ)((3 − λ)2 − 1) − (3 − λ − 1) + 1 − (3 − λ) det(A − λI) = (3 − λ)(9 − 6λ + λ2 − 1) − 2 + 1 − 3 + λ det(A − λI) = (3 − λ)(λ2 − 6λ + 8) − 4 + λ det(A − λI) = 3λ2 − 18λ + 24 − λ3 + 6λ2 − 8λ − 4 + λ det(A − λI) = −λ3 + 9λ2 − 21λ + 20. So, the characteristic polynomial is −λ3 + 9λ2 − 21λ + 20. Step 2: Find the eigenvalues of matrix A. To find the eigenvalues, we need to solve the equation −λ3 + 9λ2 − 21λ + 20 = 0. By inspection, we can see that one of the eigenvalues is λ = 1. Let’s factor the polynomial to find the other eigenvalues: −λ3 + 9λ2 − 21λ + 20 = −(λ − 1)(λ2 − 8λ + 20). Using the quadratic formula, we can find the remaining eigenvalues to be λ = 4 and λ = 5. Step 3: Find the eigenvectors corresponding to each eigenvalue. Now, we will find the eigenvectors corresponding to each eigenvalue. For λ = 1, we need to solve the system of equations (A − I)v = 0:     0 2 1 1 (A − I)v = 1 2 1 v = 0 . 1 1 2 0   1 Solving this system, we get v = −1. 0   1 Similarly, for λ = 4 and λ = 5, we can find the eigenvectors to be v =  1  −2   1 and v =  0 , respectively. −1 Therefore, theeigenvalues ofmatrix    A are 1, 4, and 5, with corresponding 1 1 1 eigenvectors −1,  1 , and  0 . 0 −2 −1 32