This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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here are the solutions to questions 8a and 8b. 8. a) Solve the equations for x and y. Given the equation: (x - i2y) + (y - i3x) = 2 + i3 Step 1: Group the real and imaginary parts on the left side of the equation. The real parts are x and y. The imaginary parts are -2y (from -i2y) and -3x (from -i3x). So, the equation becomes: (x+y) + i(-2y-3x) = 2 + i3 Step 2: Equate the real parts and the imaginary parts from both sides of the equation. Equating the real parts: x + y = 2 (Equation 1) Equating the imaginary parts: -2y - 3x = 3 (Equation 2) Step 3: Solve the system of linear equations. From Equation 1, express y in terms of x: y = 2 - x Substitute this expression for y into Equation 2: -2(2 - x) - 3x = 3 -4 + 2x - 3x = 3 -4 - x = 3 -x = 3 + 4 -x = 7 x = -7 Now substitute the value of x back into the expression for y: y = 2 - (-7) y = 2 + 7 y = 9 The solutions are x = -7, y = 9. 8. b) Find the eigenvalues and eigenvectors for the given matrix A. Given the matrix: A = 2 & 2 & -2 \\ 1 & 3 & 2 \\ -1 & 1 & 1 Step 1: Find the eigenvalues by solving the characteristic equation (A - I) = 0. A - I = 2- & 2 & -2 \\ 1 & 3- & 2 \\ -1 & 1 & 1- Calculate the determinant: (A - I) = (2-)[(3-)(1-) - (2)(1)] - 2[(1)(1-) - (2)(-1)] + (-2)[(1)(1) - (3-)(-1)] = (2-)[^2 - 4 + 3 - 2] - 2[1-+2] - 2[1+3-] = (2-)[^2 - 4 + 1] - 2[3-] - 2[4-] = (2^2 - 8 + 2 - ^3 + 4^2 - ) - (6 - 2) - (8 - 2) = -^3 + 6^2 - 5 - 12 Set the determinant to zero: -^3 + 6^2 - 5 - 12 = 0 ^3 - 6^2 + 5 + 12 = 0 By testing integer factors of 12, we find that = -1 is a root: (-1)^3 - 6(-1)^2 + 5(-1) + 12 = -1 - 6 - 5 + 12 = 0. So, (+1) is a factor. Dividing the polynomial by (+1) gives ^2 - 7 + 12. Factoring the quadratic equation ^2 - 7 + 12 = 0: (-3)(-4) = 0 The eigenvalues are _1 = -1, _2 = 3, _3 = 4. Step 2: Find the eigenvectors for each eigenvalue. For _1 = -1: Solve (A - (-1)I)v = 0 (A+I)v = 0: 3 & 2 & -2 \\ 1 & 4 & 2 \\ -1 & 1 & 2 v_1 \\ v_2 \\ v_3 = 0 \\ 0 \\ 0 From the second and third rows, we can perform row operations: R_2 3R_2 - R_1: 3(1)v_1 + 3(4)v_2 + 3(2)v_3 - (3v_1 + 2v_2 - 2v_3) = 0 10v_2 + 8v_3 = 0 5v_2 + 4v_3 = 0. R_3 3R_3 + R_1: 3(-1)v_1 + 3(1)v_2 + 3(2)v_3 + (3v_1 + 2v_2 - 2v_3) = 0 5v_2 + 4v_3 = 0. Let v_2 = 4k. Then 5(4k) + 4v_3 = 0 20k + 4v_3 = 0 v_3 = -5k. Substitute v_2 and v_3 into the first equation: 3v_1 + 2(4k) - 2(-5k) = 0 3v_1 + 8k + 10k = 0 3v_1 + 18k = 0 v_1 = -6k. Choosing k=1, the eigenvector is v_1 = -6 \\ 4 \\ -5 . For _2 = 3: Solve (A - 3I)v = 0: -1 & 2 & -2 \\ 1 & 0 & 2 \\ -1 & 1 & -2 v_1 \\ v_2 \\ v_3 = 0 \\ 0 \\ 0 From the second equation: v_1 + 2v_3 = 0 v_1 = -2v_3. Substitute v_1 into the first equation: -(-2v_3) + 2v_2 - 2v_3 = 0 2v_3 + 2v_2 - 2v_3 = 0 2v_2 = 0 v_2 = 0. Let v_3 = k. Then v_1 = -2k and v_2 = 0. Choosing k=1, the eigenvector is v_2 = -2 \\ 0 \\ 1 . For _3 = 4: Solve (A - 4I)v = 0: -2 & 2 & -2 \\ 1 & -1 & 2 \\ -1 & 1 & -3 v_1 \\ v_2 \\ v_3 = 0 \\ 0 \\ 0 From the first equation, divide by -2: v_1 - v_2 + v_3 = 0 v_1 = v_2 - v_3. Substitute v_1 into the second equation: (v_2 - v_3) - v_2 + 2v_3 = 0 v_3 = 0. Substitute v_3 = 0 into v_1 = v_2 - v_3: v_1 = v_2 - 0 v_1 = v_2. Let v_2 = k. Then v_1 = k and v_3 = 0. Choosing k=1, the eigenvector is v_3 = 1 \\ 1 \\ 0 . Drop the next question.