Math 250B Final Exam Review Session Spring 2015 SOLUTIONS

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1 Math 5B Final Exam Review Session Spring 5 SOLUTIONS Problem Solve x x + y + 54te 3t and y x + 4y + 9e 3t λ SOLUTION: We have det(a λi) if and only if if and 4 λ only if λ 3λ This means that the eigenvalues of A are λ and λ 3 [ Eigenvectors for λ : We have A, which gives rise to corresponding 4 eigenvectors of the form v r(, ), where [ r R Hence, one solution to the associated homogeneous system is x (t) [ 4 Eigenvectors for λ 3: We have A 3I, which gives rise to corresponding eigenvectors of the form v r(, ), where r[ R Hence, a second solution to the associated homogeneous system is x (t) e 3t Therefore, the general solution to the associated homogeneous system is x c (t) c [ + c e 3t [ and the fundamental matrix is [ e 3t X(t) e 3t We have Therefore, u(t) X (t) 3 t X (s)b(s)ds t [ 3 [ e 3t e 3t e 3s e 3s [ 54se 3s 9e 3s ds

2 t [ 36se 3s 3e 3s 8s + 6 [ e ds 3t (t 5) 3t( 3t) Hence, we obtain the particular solution [ [ e 3t e x p (t) X(t)u(t) 3t (t 5) e 3t 3t( 3t) e 3t [ 9t 3t + 8t 4t + 5 Thus, the general solution to the nonhomogeneous system of differential equations is [ [ [ 9t x(t) x c (t) + x p (t) c + c e 3t e 3t 3t + 8t 4t + 5 Problem Let S : {a + bx + cx + dx 3 : a + c 3d } (a): Show that S is a subspace of P 3 (R) SOLUTION: We must show that S is closed under addition and scalar multiplication Closure under Addition: Suppose p (x) and p (x) belong to S Write p (x) a + b x + c x + d x 3 and p (x) a + b x + c x + d x 3 We must have Now a + c 3d and a + c 3d p (x) + p (x) (a + b x + c x + d x 3 ) + (a + b x + c x + d x 3 ) Note that (a + a ) + (b + b )x + (c + c )x + (d + d )x 3 (a + a ) + (c + c ) 3(d + d ) (a + c 3d ) + (a + c 3d ) +, which shows that p (x) + p (x) belongs to S

3 Closure under Scalar Multiplication: Let p (x) be as above and let k be a scalar Then k p (x) k(a + b x + c x + d x 3 ) (ka ) + (kb )x + (kc )x + (kd )x 3, and since (ka ) + (kc ) 3(kd ) k(a + c 3d ) k, we know that k p (x) belongs to S (b): Find a basis and dimension for S SOLUTION: Using a + c 3d, we can write p(x) a + bx + cx + dx 3 as p(x) (3d c) + bx + cx + dx 3 bx + c( + x ) + d(3 + x 3 ) So this collection of polynomials is spanned by {x, + x, 3 + x 3 }, which is a basis for S The dimension is therefore three: dim[s 3 Problem 3 Show that the vectors {x, +x, 3+x 3 } is linearly independent by two different methods SOLUTION: One approach is by defiinition of linear independence Assume that c (x) + c ( + x ) + c 3 (3 + x 3 ) Then ( c + 3c 3 ) + c x + c x + c 3 x 3 For a polynomial to be zero, each coefficient must be zero Therefore, we conclude that c c c 3, as required Alternatively, we can use the Wronskian to verify this We have x + x 3 + x 3 W (x) x 3x 6x x3 ++(3+x 3 ) 6x 3 6x( +x ) x 3 +6x+6 If we can find even one value of x such that W (x), then the functions given are linearly independent In this case, we could choose x to get W () 6, as needed Problem 4 Let S : {[ a b b 3b } : a, b R (a): Show that S is a subspace of M (R) SOLUTION: We must show that S is closed under addition and scalar multiplication [ [ a b b c d d Closure under Addition: Suppose A and B 3b 3d

4 belong to S Therefore, [ (a b) + (c d) b + d A + B 3b 3d [ (a + c) (b + d) b + d 3(b + d), which has the right form for membership in S Closure under Scalar Multiplication: Let A be as above and let k be a scalar Then [ [ [ a b b k(a b) kb ka kb kb ka k, 3b 3(kb) 3(kb) which again has the right form for membership in S, as needed (b): Find a basis and dimension for S SOLUTION: We can write [ a b b 3b [ a [ + b 3, so the basis is {[ The dimension of S is [, 3 } Problem 5 Define T : M (R) R via ( ) a b T (a b 3c, b + d) c d (a): Show that T is a linear transformation SOLUTION: We must show that T respects addition and scalar multiplication Check T respects addition: We have (( ) ( )) ( ) a b T a b + a + a T b + b c d c d c + c d + d ((a + a ) (b + b ) 3(c + c ), (b + b ) + (d + d )) ( ) ( a b (a b 3c, b + d ) + (a b 3c, b + d ) T a b + T c d c d )

5 Check T respects scalar multiplication: Let k be a scalar We have ( ( )) ( ) ( a b ka kb a b T k T (ka kb 3kc, kb+kd) k(a b 3c, b+d) kt c d kc kd c d ), as needed (b): Is T one-to-one, onto, both, or neither? SOLUTION: ONTO only We see immediately that T cannot be one-to-one since dim[v > dim[w The only question is whether or not T is onto (ie Does Rng(T ) W?) We have Rng(T ) span{t Thus, T is onto ([ ) ([, T ) ([, T span{(, ), (, ), ( 3, ), (, )} R W ) ([, T ) } (c): Find a basis for Ker(T ) and Rng(T ) ( ) a b SOLUTION: To find Ker(T ), we set T (, ) and solve for a, b, c, d We c d have a b 3c and b + d We see that c and d are free variables Let us say that d t and c r Then b t/, and a t/ + 3r Thus, {[ } {[ [ } t/ + 3r t/ / / 3 Ker(T ) span,, r t so these two matrices form a basis for Ker(T ) As for Rng(T ), we have already observed that Rng(T ) R, so it is natural to use the basis {(, ), (, )} for Rng(T ) Problem 6 Let V : P (R) and define a + a x + a x, b + b x + b x : a b + a b + a b Show that this is NOT a valid inner product SOLUTION: Consider the vectors v + x V We have v, v <, which violates the first axiom of an inner product

6 Problem 7 Let V : P (R) and define a + a x + a x, b + b x + b x : a b + a b + a b (a): Show that this is a valid inner product on V SOLUTION: We must check all of the axioms (i) - (iv) of an inner product space Check (i): We have a + a x + a x, a + a x + a x a + a + a, and moreover, the only way for a + a + a is if a + a x + a x Check (ii): We have a + a x + a x, b + b x + b x a b + a b + a b b a + b a + b a Check (iii): We have b + b x + b x, a + a x + a x k(a + a x + a x ), b + b x + b x ka b + ka b + ka b k(a b + a b + a b ) Check (iv): We have k a + a x + a x, b + b x + b x (a +a x+a x )+(c +c x+c x ), b +b x+b x (a +c )b +(a +c )b +(a +c )b (a b + a b + a b ) + (c b + c b + c b ) a + a x + a x, b + b x + b x + c + c x + c x, b + b x + b x (b): Use this inner product to find the angle between the vectors p(x) x + x and q(x) x x SOLUTION: We use the formula p, q p q cos θ We know that p p, p 9 3 and q 6 and p, q Thus, cos θ 3 6

7 (c): Find an orthogonal basis for S : span{ x + x, x x } SOLUTION: We let v x+x and v x x Using the Gram-Schmidt Process, we set w v Next, ( ) v, w w v w w ( x x ) 9 ( x + x ) Problem 8 (a): Give an example of two diagonalizable matrices A and B such that A + B is not diagonalizable [ SOLUTION: Many examples are possible For instance, we could take A [ and B Both A and B are diagonalizable since they have distinct eigenvalues, but A + B is not diagonalizable (the repeated eigenvalue λ, [ does not pull its weight) (b): Consider the system x ( 4 ) x As t, do the functions x (t) and x (t) tend to the origin? SOLUTION: It is easy to verify that the eigenvalues of the coefficient matrix are λ 3 ± i With this information alone, we can already say that the solutions to the system do NOT tend towards the origin as t, so all solutions with be multiplied by the scalar function e 3t, which makes the solutions swell to infinity, not shrink to the origin (c): Assume A and B are 4 4 matrices with det(a) and det(b) 3 Find det((a B) T (B) )

8 SOLUTION: Note that det(a B) T det(a B) 3 Also, det(b) Thus, det(b) Hence, our final answer is (d): True or False: If A is a 7 9 matrix with nullity(a), then rowspace(a) R 7 SOLUTION: FALSE Even though the Rank-Nullity Theorem tells us that rank(a) 7 (which implies that the dimension of the rowspace is 7), we must recognize that rowspace(a) is a subspace of R 9, not R 7 Thus, we should say that rowspace(a) is a 7-dimensional subspace of R 9 Problem 9 Solve: SOLUTION: dy dy dx 3 y /3 dy dx y 3 x y 6y/3 x ln x 3 dx x 6y/3 x ln x or x y/3 6x ln x Let u y /3 Substituting these results into dy y /3 dx x y/3 6x ln x yields du dy 3 y /3 dx dx so d dx (x u) u x 4x ln x An integrating factor for this equation is I(x) x 4x ln x x u 4 x ln xdx+c x u x ln x x +c u(x) x(x ln x x + c) y /3 x(x ln x x + c) Problem Solve y + 9y 8 sec 3 (3x) SOLUTION: Setting y + 9y r + 9 r {3i, 3i} y c (x) c cos 3x + c sin 3x Let y (x) cos 3x and y (x) sin 3x Then, W [y, y (x) (cos 3x)(3 cos 3x) (sin 3x)( 3 sin 3x) 3 so that y p (x) y u + y u, where x (sin 3t)(8 sec 3 3t) x u dt 6 (sin 3t)(sec 3 3t)dt tan 3x 3 and Consequently, x (cos 3t)(8 sec 3 3t) x u dt 6 sec 3tdt tan 3x 3 y p (x) cos 3x tan 3x + sin 3x tan 3x

9 Hence, y(x) c cos 3x + c sin 3x cos 3x tan 3x + sin 3x tan 3x Problem By seeking solutions to x 3 y + x y xy + y of the form y(x) x r, find three linearly independent solutions (and verify that your solutions are linearly independent with the Wronskian) to this differential equation SOLUTION: We have y (x) rx r, y (x) r(r )x r, y (x) r(r )(r )x r 3 Plugging these values into the differential equation, it reads or That is, Setting x 3 [r(r )(r )x r 3 + x [r(r )x r x[rx r + [x r, we factor to obtain or [r(r )(r ) + r(r ) r + x r (r 3 r r + )x r r 3 r r +, r (r ) (r ), (r )(r ) The roots are r,, Thus, we obtain the solutions y (x) x, y (x) x, y 3 (x) x To check the linear independence of these solutions, we use the Wronskian: x x x W [y, y, y 3 det x x x 3

10 If we perform cofactor expansion along the bottom row, we obtain W [y, y, y 3 [ x 3 [x x + x + x x x + 4 x 6 x, which is never zero Thus, the functions y, y, y 3 are linearly independent Problem (a): State any two of the axioms of an inner product and verify that on the vector space V C[,, the formula satisfies your two axioms f, g tf(t)g(t)dt SOLUTION: The axioms are: f, f f, g g, f 3 kf, g k f, g 4 f + f, g f, g + f, g To verify these, we observe that f, f tf(t) dt, since tf(t) throughout the interval t of integration Next, f, g tf(t)g(t)dt tg(t)f(t)dt g, f Next kf, g t(kf)(t)g(t)dt k tf(t)g(t)dt k f, g

11 Finally, f + f, g t(f + f )(t)g(t)dt tf (t)g(t)dt + [tf (t)g(t) + tf (t)g(t)dt tf (t)g(t)dt f, g + f, g (b): Using the inner product given in part (a), if θ denotes the angle between the vectors f(x) x and g(x) x, compute cos θ SOLUTION: We have cos θ f, g f g x(x )(x)dx x(x ) dx x(x) dx