Math Computation Test 1 September 26 th, 2016 Debate: Computation vs. Theory Whatever wins, it ll be Huuuge!

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1 Math 5- Computation Test September 6 th, 6 Debate: Computation vs. Theory Whatever wins, it ll be Huuuge! Name: Answer Key: Making Math Great Again Be sure to show your work!. (8 points) Consider the following system of equations: x x + x = x x + x x = x + x + x + x = Write down the associated augmented matrix and perform Gauss-Jordan elimination (as defined in class). SHOW EACH STEP! Then, using the reduced row echelon form, write the (new) associated system of equations. Finally, write the general solution in parametric form (label the free variables t, s, etc.) or if the system has no solution, simply write inconsistent. : : : : : : R+R : : : : : : R R+R R+R : : : : : : R+R Writing down the equivalent system of equations and letting x = s, x = t (the second and fourth columns are non-pivot columns and thus correspond to free variable), we get the general solution... x = + t + s x x x x = x = t x x + x = x = s x + t + s x = s. ( points) Consider the following system of equations: x + y z = x y + z = x + y + z = Find a PLU-decomposition for the coefficient matrix of this system. Then use the PLU-decomposition to solve this system. [You must use the decomposition to get credit for solving the system.] A = R+R ( ) R+R ( ) R R () () ( ) P = L = U = Letting b =, we are to solve Ax = P LUx = b using our decomposition. Thus we should solve LUx = P b =. First, we solve Ly = P b... u = u + v = u + w = Now solve Ux = y... x + y z = y + z = z = u = () + v = so v = + w = so w = z = so z = y + () = so y = x + ( ) = so x = 5 x y = x 5 y z I don t mean to be braggadocious, but I did this test myself in minutes. It was a yuuuge success. I did fantastic. Not like my competition. I m sure if the FBI did their job, my competitors would be in jail. They are the worst. Like the worst ever.

2 [ ]. (8 points) Let A = and B = [ ]. Compute each of AB, BA, A B T, and B T A or explain why such a product does not exist. First, A is so A is. B is so B T is. Since A has columns and B has rows, the product AB exists ( match = = ). On the other hand, BA does not exist since B has columns and A has rows. Next, A B T does not exist since A has columns and B T has rows. On the other hand, B T A does exist since B T has columns and A has rows ( match = = ). [ ] [ ] [ ] 6 AB = = B T A = [ ] = 7 7 [ ] [ a b d b I used (and recommend knowing) the inverse formula: = c d ad bc c a (Withdrawn Problem) All matrices in this problem are real matrices. Let E ij be a standard unit matrix with in the (i, j)-entry and s otherwise. Also, let I denote the identity matrix. (a) Let E = I E + 9E. First, in terms of row and column operations on some matrix A, explain what EA and AE are. Then, given det(a) = 5, compute det(ea). Finally, if E exists, give a formula for it. E = I E + 9E = I E E + E. This means EA removes the fourth row of A and then scales in the first row by. AE removes the fourth column of A and scales the first column by. Since E has a row of zeros (i.e. its fourth row), det(e) = and so det(ea) = det(e) det(a) = (5) = and E does not exist. (b) Let E = I E. First, in terms of row and column operations on some matrix A, explain what EA and AE are. Then, given det(a) = 5, compute det(ea). Finally, if E exists, give a formula for it. EA adds times the fourth row of A to the second row of A (this is a type operation so E is an elementary matrix). AE adds times the second column of A to the fourth column of A. Type operations do not effect determinants so det(ea) = det(a) = 5. To undo this operation (add times row to row ) we should add times row to row. Thus E = I + E. : 7 : : : : : [ ]. ( points) Let A : a : b : c = : : : G.E. : : : : : : : : : : 6 : : : : : (a) Is the system of equations Ax = a consistent? If not, why? If so, give its general solution in vector form. Yes, this system is consistent since a s column doesn t introduce a new pivot. Reading off the RREF we get: x + x =, x + x = and x =. Thus letting x = t (it s free), we get the general solution (where t R)... x = t x = t x = t x = = x x x x + t (b) Is the system of equations Ax = b consistent? If not, why? If so, give its general solution in vector form. No, this system is inconsistent. To see this notice that the last row gives us the equation x + x + x + x = (so = ). [Alternatively, we have that this system is inconsistent because b introduces a new pivot column.] (Withdrawn part) Is the system of equations Ax = c consistent? If not, why? If so, give its general solution in vector form. No, this system is also inconsistent. Notice that the last row gives the equation = in this case. [Alternatively, this is inconsistent because the final column involves a pivot not found in A.] ].

3 (c) Let S be the set of columns of A. Do a, b, or c belong to Span(S)? If not, why not? If so, write that vector as a linear combination of the vectors in S. By the linear correspondence we can see that b and c do not belong to Span(S) since the corresponding columns in the RREF are not linear combinations of columns corresponding to A in the RREF. In particular, both involve a new (fourth) pivot (i.e. b itself). On the other hand, a Span(S). In fact, the linear correspondence tells us that a = a + a + a where a, a, and a are the first, second, and fourth columns of A (i.e. the first, second, and third pivot columns). Explicitly, 7 a = (6 points) Straight Up Linear Correspondence Problem! Fill in the boxes! G.E The second column is times the first column. The fourth column is times the first column plus times the third column (i.e. second pivot column). The fifth column is times the first column plus times the third column (i.e. second pivot column). Finally the final column is the first, third, and sixth columns added together (i.e. the sum of the pivot columns). 6. (9 points) Let A =. Compute det(a) using... (I) the trick, (II) expansion along the second column, and (III) the forward pass of Gaussian Elimination. = 7 (I) Bottom Top = 5 7 = 9 = 5 (II) ( ) + () + ( )+ () + ( )+ () = ( )[6 ] + ()[ ] + ( )[ ] = (III) R+R 5 R+R 5 5 R+R 5 ( 5)(/5) = /5 Note: For (III), we only did type operations so no adjustment was needed. If we had done any row swaps, we would have needed to adjust the product of the diagonal elements accordingly (6 points) Let A =. det(a) = [( )+ ()] = [( ) + ()] [( ) + ()] ( = [( ) + ()] [( ) + ()] ( ) + () + ( )+ () ) = ( )() (( )( ) + ( )( 6)) = ( +) = I expanded along the third column then first column then second column then used formulas.

4 8. (6 points) Let A, B, and C be real matrices such that det(a) =, det(b) = 5, and det(c) =. Also, let I be the identity matrix. Let E be the identity matrix with rows and 6 swapped. Let F be the identity with row scaled by 7. (a) Compute det(a B T C) = det(a ) det(b T ) det(c) = det(a) det(b) det(c) = 5 Here we used the fact that the determinant of a product is the product of determinants, the determinant of the inverse of a matrix is the reciprocal of the determinant, and the determinant of a transpose is just the determinant (transposing does nothing to a determinant). (b) Compute det( EF A ) = ( ) det(e) det(f ) det(a) = (+)( )( 7)( ) = 8 Again we expanded the product. We used the fact that det(cx) = c n det(x) for an n n matrix X and scalar c (this is because c scales all n rows of X, so it multiplies the determinant of X by c a total of n times). Next, a row swap flips the sign of the determinant (i.e. det(e) = ). And scaling a row by 7, scales the determinant by 7 (i.e. det(f ) = 7). 9. ( points) Find the eigenvalues and bases for eigenspaces for each of the following matrices. State if the matrix is diagonalizable. (a) A = When a matrix is upper (or lower) triangular, you can read its eigenvalues off of the diagonal. This means that the only eigenvalue of A is λ = (its algebraic multiplicity is ). Alternatively, we could compute the characteristic polynomial (and learn the same thing)... t f(t) = det(ti A) = det t = (t ) t To find a basis for the eigenspace associated with λ = we need to solve (A I)x =. A I = is already in RREF. This corresponds to the equations: x =, =, =. So x and x are x s free, say x = s and x = t. Then x s + t t. A basis for the λ = eigenspace:,. x t Notice that the geometric multiplicity of λ = is whereas the algebraic multiplicity is. Thus A is not diagonalizable. (b) B = f(t) = det(ti B) = det t ([ ]) t t = (t ) det = (t )(t + ) = (t )(t i)(t + i) t t Note: It would be foolish to try to expand along anything other than the second row (or column). The trick would be a particularly poor choice. When computing characteristic polynomials, take care not to destroy pre-existing factorization (if possible). Since B has distinct eigenvalues (all with algebraic multiplicity ), B is diagonalizable (over C). Of course, B is not diagonalizable over fields like R or Q since B has complex eigenvalues. Next, let s find bases for eigenspaces. First, λ = (so we should solve (B I)x = ). B I = (use the linear correspondence and avoid actually doing Gaussian elimination). We get the basis.

5 Now let s find the basis for the λ = i eigenspaces. B ii = i i i (again use the linear i correspondence to avoid doing Gaussian elimination). This time we have equations: x + ix = and x = thus x = t is free. We get x x it t i. We get a x t i i basis (for the λ = i eigenspace). Likewise we get a basis for the eigenspace associated with λ = i.. (8 points) Given A = G.E. Find bases for the row, column, and null spaces of A. The non-zero rows of the RREF give a basis for the row space: {[ - -], [ -], [ ]}. Be careful! The first rows of the original matrix do not give a basis for the row space. In fact, notice that the second row is times the first row the set of the first rows is linearly dependent! The pivot columns (in the original matrix) provide us with a basis for the column space:, 6, 6. To find a basis for the null space we covert the RREF back to homogeneous equations: x x x 5 =, x x 5 =, and x + x 5 =. We have x = s and x 5 = t free. This gives us... x x x x x 5 s + t s t t s + t = Basis: t,. (9 points) Suppose that A is a square matrix whose characteristic polynomial is f(t) = (t + ) (t )(t ). Answer the following questions as much as is possible given just the characteristic polynomial. If an answer is unknown, give the possible range of answers. Notice that the characteristic polynomial is a fifth degree polynomial. This means that our mystery matrix must be a 5 5 matrix. Next, we can read off the polynomial s roots:,,. Recall that the determinant is the product of the eigenvalues (taking into account repeats) so the determinant is ( ) ( ) ( ) =. Since the determinant is non-zero, A is invertible. Next, geometric multiplicities are bounded below by and above by the algebraic multiplicity. This forces the geometric multiplicity of λ = and λ = to be. Finally, we cannot pin down the geometric multiplicity of λ =. All we can say is that it s,, or. Since we cannot tell if this geometric multiplicity matches its algebraic multiplicity, we cannot tell if A is diagonalizable or not. Circle your answer: The matrix A is / is not / may or may not be diagonalizable. The matrix A has size 5 5 and its determinant is ( ) ()() =. Circle your answer: A is Invertible / Singular / Need more Information. The eigenvalues of A are,,. λ = has algebraic multiplicity and geometric multiplicity,, or. λ = has algebraic multiplicity and geometric multiplicity. λ = has algebraic multiplicity and geometric multiplicity. 5

6 . ( points) Run the Gram-Schmidt algorithm on the columns of A =. We need to run Gram-Schmidt on v =, v =, v =. So first, w = v =. Next, w = v v w w w w = 6 + Finally, w = v v w w w w v w w w w =. Thus {w, w, w } =,, is an orthogonal basis for Span({v, v, v }) = Col(A). 6