Solutions to practice questions for the final

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1 Math A UC Davis, Winter Prof. Dan Romik Solutions to practice questions for the final. You are given the linear system of equations x + 4x + x 3 + x 4 = 8 x + x + x 3 = 5 x x + x 3 x 4 = x + x + x 4 = 3 (a) Write an augmented matrix representing the system (b) Find a reduced row echelon form (RREF) matrix that is row-equivalent to the augmented matrix. 3 (c) Find the general solution of the system. x x x 3 x 4 = 3 + λ + λ (d) Write the homogeneous system of equations associated with the above (nonhomogeneous) system and find its general solution.

2 matrix The homogeneous system is represented by the augmented 4 This is the same as the original system except that the rightmost column is the zero vector. The equivalent RREF is therefore and the general solution is x x x 3 = λ x 4. Define M = 3. + λ (a) Find the inverse matrix of M using elementary row operations. M = (b) Compute the adjoint matrix adj(m) using the definition of the adjoint matrix. (c) Compute det(m) and verify using the results of the above computations that adj(m) = det(m)m. matrix. det(m) = ( ) =, since M is a lower-triangular

3 3. Define M = 3 3. (a) Name an easily observable property of the matrix M that guarantees that it is diagonalizable. M is a symmetric matrix. (b) Compute the characteristic polynomial P M (λ) of M and find all its zeros. P M (λ) = det(λi M) = det λ λ 3 λ 3 = λ(λ 6λ + 8) = λ(λ )(λ 4) = λ[(λ 3) ] The zeros of P M (λ) (which are the eigenvalues of M) are λ =, λ = and λ 3 = 4. (c) Find a basis of R 3 consisting of eigenvectors of M. A basis of eigenvectors corresponding to the eigenvalues λ, λ and λ 3 is given by v =, v =, v 3 =. (d) Find a 3 3 invertible matrix P and a 3 3 diagonal matrix D such that M = P DP. The matrix P is constructed by putting the vectors v, v, v 3 in the columns of a matrix and the matrix D is the diagonal matrix having the eigenvalues λ, λ, λ 3 in the diagonal (they need to appear in the same order as the order of the eigenvectors in the columns of P ). Thus: D = 4, P =. 3

4 4. (a) Let {u, v} be an orthonormal basis for R. Let a, b be two real numbers such that a + b =. Show that the vectors {w, z} given by w = au + bv z = bu + av also form an orthonormal basis for R. Compute the dot products w w, w z and z z, using the information that u u = v v = and u v = : w w = (au + bv) (au + bv) = a (u u) + ab(u v) + ab(v u) + b (v v) = a + b =, w z = (au + bv) ( bu + av) = ab(u u) + ab(v v) + a (u v) b (u v) = ab + ab =, and similarly z z =. So w, z are orthonormal. Note in particular that they are linearly independent (otherwise one is a scalar multiple of the other, in contradiction to the orthogonality), and since R is two-dimensional this implies that they are a basis. (b) Define the linear subspace U = span{(,, )} of R 3. Find a basis for its orthogonal complement U. U consists of all vectors (x, y, z) such that (x, y, z) (,, ) = x+y+z =. This is a system of one linear equation in three variables. Its solution set is found (using the standard method) to be span{v = (,, ), v = (,, )}. These two vectors are linearly independent so they form a basis for U. (c) Find an orthogonal basis for the space U defined above. Apply the Gram-Schmidt orthogonalization process to the vectors v, v to get an orthogonal basis u = v = (,, ), u = v v u u = (,, ) (,, ) = (/, /, ). u u 5. (a) If u, v are two linearly independent vectors in a vector space V, prove that u + v, u v are also linearly independent. 4

5 Denote p = u + v and q = u v. If a, b are scalars such that ap + bq =, then = a(u + v) + b(u v) = (a + b)u + (a b)v. This is a linear combination of u and v that gives the zero vector. Since u, v are linearly independent this implies that the coefficients of the linear combination are both zero. So: a + b =, a b = and this clearly implies (by considering the sum and the difference of the two equations) that a = b =, proving linear independence. (b) If u,..., u k are vectors that span R 5, what are the possible values of k? dim(r 5 ) = 5, so a spanning set (which can be diluted to get a basis) must contain at least 5 vectors. Hence k 5. (c) If M is a 5 5 matrix such that the vectors v =, v = form a basis for the kernel ker(m), find the rank of M. The data implies in particular that dim(ker(m)) = (the nullity of M). By the dimension formula, dim(ker(m)) + rank(m) = 5, so rank(m) = 3. 5

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