AfterMidterm2-PracticeProblems-Solutions

AfterMidterm2-PracticeProblems-Solutions - Math 331...

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Math 331 practice problems for material covered after Midterm 2 – Solutions This is a list of practice problems for Math 331-Wylie covering material after the second midterm. 1. True of False: (a) If A is a 4 × 4 matrix with two eigenvalues and one eigenspace is three dimensional, then A is diagonalizable. True (b) Every diagonalizable matrix is invertible. False (c) Every list of 5 orthogonal non-zero vectors in R 5 is a basis of R 5 . True (d) If A is a diagonalizable n × n matrix, then so is A 7 . True (e) If A is an n × n matrix and v is an eigenvector for A then it is also an eigenvector for A T . False 2. (a) Find the area of the parrallelogram whose vertices are (0 , 0), (1 , 2), (3 , 3) and (4 , 6). Solution: The area of the parallelogram is the absolute value of det ±² 1 3 2 3 ³´ = 3 - 6 = - 3 So the area is 3. (b) Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices (1 , - 1 , 2), (3 , 2 , 1), and ( - 1 , - 1 , 1). Solution: The volume of the parallelepiped is the absolute value of det 1 3 - 1 - 1 2 - 1 2 1 1 = 1 · det ² 2 - 1 1 1 ³ - ( - 1) · det ² 3 - 1 1 1 ³ + 2 · det ² 3 - 1 2 - 1 ³ = (2 + 1) + (3 + 1) + 2( - 3 + 2) = 7 - 6 + 4 = 5 So the volume is 5.

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3. Determine the values of the parameter s for which the system has a unique solution and describe the solution. 2 sx + sy = 2 3 sx + 3 y = 1 Solution: The system has a unique solution when det ± 2 s s 3 s 3 ² = 6 s - 3 s 2 = - 3 s ( s - 2) 6 = 0 . So the system has a unique solution when s 6 = 0 and s 6 = 2. To ﬁnd the formula for the solution, we can use Cramer’s rule by computing det( A 1 ) = det ± 2 s 1 3 ² = 6 - s det( A 2 ) = det ± 2 s 2 3 s 1 ² = - 4 s So the solution is x = det( A 1 ) det( A ) = 6 - s - 3 s ( s - 2) y = det( A 2 ) det( A ) = - 4 s - 3 s ( s - 2) Check: This solution can be checked to be correct by plugging the formulas back into the original system of equations. 4. Determine which of the following matrices are diagonalizable (over the real numbers) (a) ± 3 - 1 - 1 3 ² Solution: The characteristic polynomial is det ± 3 - λ - 1 - 1 3 - λ ² = (3 - λ ) 2 - 1 = λ 2 - 6 λ + 9 - 1 = λ 2 - 6 λ + 8 = ( λ - 4)( λ - 2) So the eigenvalues are 4 and 2. Since the 2 × 2 matrix has 2 distinct eigenvalues, it must be diagonalizable. (b) 1 3 7 9 4 0 2 1 5 8 0 0 - 2 3 7 0 0 0 5 1 0 0 0 0 3 Solution: Since the matrix is upper triangular, its eigenvalues are 1, 2, - 2, 5, and 3. Since the 5 × 5 matrix has 5 distinct eigenvalues, it is diagonalizable.
(c) 1 0 0 0 0 1 0 - 2 3 Solution: The characteristic polynomial is det 1 - λ 0 0 0 - λ 1 0 - 2 3 - λ = (1 - λ )det ± - λ 1 - 2 3 - λ ² = (1 - λ )( - λ (3 - λ ) + 2) = (1 - λ )( λ 2 - 3 λ + 2) = (1 - λ )( λ - 1)( λ - 2) = - ( λ - 1) 2 ( λ - 2) So the eigenvalues are 1 and 2. Since we do not have 3 distinct eigenvalues, we must investigate the 1-eigenspace to see if the matrix is diagonalizable. 1

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AfterMidterm2-PracticeProblems-Solutions - Math 331...

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