Solutions to Homework 3, Introduction to Differential Equations, 3450: , Dr. Montero, Spring 2009 M = 3 3 Det(M) = = 24.

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1 Solutions to Homework 3, Introduction to Differential Equations, 3450: , Dr. Montero, Spring 2009 Problem 1. Find the determinant of the matrix M = Solution: The determinant is easy to compute along the third row. We obtain 3 3 Det(M) = = 24. Problem 2. Find the determinant of the matrix M = Solution: We compute this determinant along the third row. We obtain Det(M) = = 1 ( 3) 2 ( 2) 6 = 18. Note that both the 3 3 matrices we obtained have ether a column or a row with all but one entry equal to zero. Hence I took advantage of this to find their determinants. Problem 3. Find the determinant of the n n matrix A with 5 s on the diagonal, 1 s above the diagonal, and 0 s below the diagonal. Solution: The elements of the diagonal are those with equal indices, like a 11, a 2 2 and so on. For example the 2 2 matrix with 5s in the diagonal and 1s above it is the matrix ( ) 5 1 A =. 0 5

2 The determinant of this matrix is 25. The 3 3 matrix with 5s in the diagonal, 1s above it and 0s below the diagonal is the matrix A = and the determinant of this matrix is 125. The 4 4 matrix with 5s in the diagonal, 1s , above it and 0s below the diagonal is the matrix A = , and the determinant of this matrix is 625. Hence, for the n n matrix A with 5 s on the diagonal, 1 s above the diagonal, and 0 s below the diagonal the determinant is 5 n. Problem 4. If the determinant of a 7 7 matrix A is det (A) = 10, and the matrix B is obtained from A by multiplying the fifth row by by 9 then det (B) = Solution: We recall that multiplying a whole row (or a whole column) by a constant has the effect of multiplying the determinant by the same constant. Hence det(b) = 9 det(a) = 90. Problem 5. If the determinant of a matrix A is det (A) = 23, and the matrix C is obtained from A by swapping the third ad twelfth columns, then det (C) = Solution: Recall that swapping two rows changes the sign of the Determinant. The same happens if we swap two columns. Hence det(c) = 23. Problem 6. Find the determinant of the matrix M = Solution: This is very similar to Problem 1; just longer. We compute the determinant along the third row. We obtain det(m) = M =

3 We compute the 3 3 determinants separately. First = = 1 (7) (11) = 16. Second, = ( 1) = ( 7) + ( 12) = 22. We conclude that det(m) = 3 ( 16) 2 ( 22) = 4. Problem 7. Determine if the vectors (0, 4, 2), ( 2, 1, 4) and ( 2, 3, 3) are linearly dependent or independent. The vectors are written horizontally to save space. The problem is equivalent if written with column vectors. Solution: We compute the determinant, for example, along the first column ( ) ( ) det = ( 2) det + ( 2) det = = 0. The vectors are dependent. In fact we have = Problem 8. Let us consider the functions f(t) = e t, g(t) = t and h(t) = 2 + 3t. Determine if these functions are dependent or not.

4 Solution: We compute the Wronskian f g h W(t) = det f g h = det f g h e t t 2 + 3t e t 1 3 e t 0 0 = e t t 2 + 3t 1 3 = 2et. The functions are independent. Problem 9. Determine if the following pairs of functions are linearly independent or not. 1. f(θ) = cos(3θ) and g(θ) = 3 cos 3 (θ) 3 cos(θ) 2. f(t) = 3t t and g(t) = 3t 2 21t. 3. f(t) = 3t and g(t) = t. 4. f(t) = e λt cos(µt) and g(t) = e λt sin(µt) Solution: For the first pair we try to prove that if αf(θ) + βg(θ) = 0 for all θ then necessarily α = 0 and β = 0. This means that f(θ) and g(θ) must be linearly independent. To prove this we notice that if αf(θ) + βg(θ) = 0 (0.1) for all θ, then this equation is valid for both θ = 0 and also for θ = π. Note that g(0) = 0 and 3 f(0) = 1 so replacing θ = 0 in (0.1) gives directly α = 0. Now we recall that cos(π/e) = 1/2. Hence g(π/3) = 9/8. Recalling that we already know that α = 0, plugging θ = π/3 in (0.1) gives β = 0. Hence f and g are independent. For the second pair we compute the determinant ( ) f g W(t) = det f g ( ) 3t t 3t 2 21t = det 6t t 21 The functions are independent. = (3t t)(6t 21) (3t t)(6t + 21) = 126t 2. We considered a problem pretty much like the third in class.

5 For the last pair we note that, if αe λt cos(µt) + βe λt cos(µt) = e λt (α cos(µt) + β cos(µt)) = 0 for all t, then α cos(µt) + β cos(µt) = 0 for all t. But for h 1 (t) = cos(µt) and h 2 (t) = sin(µt) we have ( ) ( h 1 h 2 cos(µt) sin(µt) W(t) = det = det h 1 h 2 µ sin(µt) µ cos(µt) ) = µ, so these functions are independent whenever µ 0. This means that cos(µt) and sin(µt) are independent, so if α cos(µt) + β cos(µt) = 0 for all t, then α = β = 0. This means that f(t) = e λt cos(µt) and g(t) = e λt sin(µt) are independent. One could also compute ( ) ( ) f g e λt cos(µt) e λt sin(µt) det = det = µe 2λt. f g λe λt cos(µt) µe λt sin(µt) λe λt sin(µt) + µe λt cos(µt) we conclude, as before, that these functions are independent unless µ = 0. Problem 10. Determine if the following pairs of functions are independent. 1. f(θ) = 5 cos(3θ) and g(θ) = 20 cos 3 (θ) 15 cos(θ) 2. f(t) = t 2 + 5t and g(t) = t 2 5t. 3. f(t) = t and g(t) = t. Solution: Recall the formulas cos(α + β) = cos(α) cos(β) sin(α) sin(β) and sin(α + β) = sin(α) cos(β) + cos(α) sin(β). In particular cos(2α) = cos 2 (α) sin 2 (α) = 2 cos 2 (α) 1 and sin(2α) = 2 sin(α) cos(α). For the first pair we use these formulas to obtain cos(3θ) = cos(2θ) cos(θ) sin(2θ) sin(θ) = (2 cos 2 (θ) 1) cos(θ) 2 cos(θ) sin 2 (θ) = 2 cos 3 (θ) cos(θ) 2 cos(θ)(1 cos 2 (θ)) = 4 cos 3 (θ) 3 cos(θ).

6 This means that 5 cos(3θ) = 20 cos 3 (θ) 15 cos(θ), so the functions are in fact equal. We conclude that they are linearly dependent. The second pair of functions is pretty much the same as one in the last question. The third pair of functions we looked at in class. Problem 11. Determine the value of k for which the vectors below are linearly independent , 9 and k 1 Solution: We compute the determinant of the matrix M = k 1 along the third column and set it equal to zero. This gives us 9 9 det(m) = k = 9(9 + k) 63 + ( ) = 135 9k = This means that k = = 15.