1. For each of the following matrices, determine whether it is in row echelon form, reduced row echelon form, or neither.


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1 Math Exam  Practice Problem Solutions. For each of the following matrices, determine whether it is in row echelon form, reduced row echelon form, or neither. (a) 5 (c) Since each row has a leading that is down and to the right of the leading in the previous row, this matrix is in row echelon form. Since some of the columns with a leading have other nonzero entries, it is not in reduced row echelon form. 8 Since each nonzero row has a leading that is down and to the right of the leading in the previous row, each column with a leading has no other nonzero entries, and the zero rows is at the bottom of the matrix, this matrix is in reduced row echelon form. Since the last row is not a zero row but does not have a leading, this matrix is in neither row echelon form nor reduced row echelon form.. Put each of the following matrices into row echelon form. (a) 8 8 r r r r r r (c) r r r r r 8 8 r r r r 8 sinθ sinθ sinθ sinθ = sinθ sin θ+cos θ r r = tanθ r r r 8 r r 8 r +r r sinθ sinθ r r r r r r r +sinθr r sinθ sin θ r r tanθ r r 9 + = sinθ sin θ + cos θ
2 . Put each of the matrices from the previous problem into reduced row echelon form. (a) Continuing from above: 8 r +r r 8 r+ r r r r r 8 Continuing from above: r r r r +r r (c) Continuing from above: tanθ r tanθr r. Use Gaussian Elimination to find all solutions to the following systems of linear equations. x+y +z = 9 (a) x z = x+y +z = We begin by finding the augmented matrix associated with this system of equations and then carry out row operations: r r r r r r 8 r r r r r 5 8 Then y +z = 5, so y = 5 z, and x+y +z = 9, or, substituting, x+(5 z)+z = 9. Therefore, x = z +z = 9, so x = +z Hence the solutions to this system are all of the form ( +t,5 t,t) for some t R. { x+y +z = 9 x+y +z = We begin by finding the augmented matrix associated with this system of equations and then carry out row operations: r r r 8 r r 5 Then, exactly as above, y +z = 5, so y = 5 z, and x+y +z = 9, or, substituting, x+(5 z)+z = 9. Therefore, x = z +z = 9, so x = +z Hence the solutions to this system are all of the form ( +t,5 t,t) for some t R. 5. Use GaussJordan reduction to find all solutions to the following systems of linear equations. x y +z = 6 (a) x y +z = x+y z = 9 We begin by finding the augmented matrix associated with this system of equations and then carry out row operations: 6 r r r r r r r r r 5 r r
3 r +r r r +r r r r r r r r r r +5r r 5 Therefore, the unique solution is x =, y = 5, and z = x y +z = x+y z = x y +z = We begin by finding the augmented matrix associated with this system of equations and then carry out row operations: r r r r r r r r 8 r r r Therefore, this system has no solutions (the equation = is inconsistent). 6. Use quadratic interpolation to find a quadratic function passing through the points (, ), (, ), and (, ). If we start with the general quadratic p(x) = ax +bx+c and the three conditions given above, we have the following system of equations: c = a+b+c = a+b+c = In matrix form, this is: r r r r r r r r r 5 r r r r r r r r r 5 Therefore, we have a =, b = 5, and c =, so p(x) = x 5x+. Once can quickly verify that this quadratic passes through all three desired points.. Construct a quadratic polynomial p(x) = ax + bx + c satisfying the conditions: p() = f(), p () = f (), and p () = f () if f(x) = xlnx+x. First, notice that since f(x) = xlnx+x, then f (x) = lnx+x ( ) x +x = lnx+x+ and f (x) = ( x) +. Then f() =, f () =, and f () =. Next, notice that since p(x) = ax +bx+c, then p (x) = ax+b and p (x) = a. Then p() = a+b+c, p () = a+b, and p (x) = a a+b+c = This gives the system of equations: a+b = a = The matrix form of this equation is: r r r r r r r r r r Thus, we have a =, b =, and c =. Hence p(x) = x. One can verify this this quadratic satisfies all of the given conditions.
4 8. Find the elementary matrices corresponding to carrying out each of the following elementary row operations on a matrix: (a) r r E = r r E = (c) r +r r E = 9. Find the inverse of each of the elementary matrices you found in the previous problem. (a) E = E = (c) E =. Write the matrix A = as a product of elementary matrices. Hint: what elementary operations are needed to put A into reduced row echelon form We begin by putting A into r.r.e.f.: r r r r +r r r r Next, we observe that the elementary matrices used in this reduction process are: E = and E =. Then E E E A = I, so A = E E E. Hence A = ( A ) = E E E. Thus A =., E =,. For each of the following matrices, either find the inverse or prove that the matrix is singular. (a) r r r r r r r r r +5r r Thus A = r r 5 r r 5 r r r 5 Since we encountered a row of zeros, A is a singular matrix, so A does not exist. (c) 5 The standard row reduction algorithm yields: A = 5
5 (d) 5 The standard row reduction algorithm yields: A = Find all values of a for which the matrix A = the cases where it exists. a r r r r r a r r r r + a r r Thus A = a a a a a a a a a a. Given the matrix A = a a (a ) a a a a a a a (a ) a 5 has an inverse. Find the form of A (in terms of a) for r r a r r. Notice that we must have a for the inverse to exist. : (a) Find matrix B equivalent to A having the form described by Theorem. a a a We will solve both parts of this question at the same time by utilizing matrices of the form A I n r r r r r r r +r r r r r 5 Restarting our right matrix: c c c Then the reduced form of A is: c +c c Find matrices P and Q such that B = PAQ (for the matrices A and B from above). Using the right hand sides of the reductions performed above (the row and column reductions respectively) we see that: P = and Q =. For each permutation given below, first find the number of inversions. Then, classify the permutation as either even or odd.
6 (a) 5 5 (c) 5 Notice that this permutation has inversions, so it is an odd permutation. Notice that this permutation has inversions, so it is an odd permutation. Notice that this permutation has 5 inversions, so it is an odd permutation. 5. Find the determinant of each of the following matrices: (a) 5 5 = ()() ( )(5) = + =. (c) (d) = ( 5)+()+( ) () (96) =. = a A +++ = t t+ t t+ = t(t+) 6 = t +t 6 = (t+)(t ) = ++ 6 = 6. For which values of t is the matrix in part (d) above singular? Justify your answer. RecallthatamatrixAissingularifandonlyifdet(A) =. Forthismatrix, det(a) = t(t+) 6 = t +t 6 = (t+)(t ). Hence this matrix is singular when t = and t =.. Evaluate each of the following: 5 (a) = ()()( )() = 8 t t t t t = (t )(t )(t ) = t 6t +t 6 t
7 (c) 6 6 = 6 = (Notice that the second matrix has a repeated column). 8. Compute the determinant of the following matrix by putting it into triangular form: A = 8 8 r +r r = ( ) r r r r r r = 6 8 r r = ( ) Notice that since we encountered a row of zeros, then det(a) = ( )() =. 9. If det(ab) =, must det(a) or det(b) =? Justify your answer Suppose that det(ab) =. Since det(ab) = det(a)det(b), then we have det(a)det(b) =. Recall that a product of two real numbers is zero if and only if at least one of the numbers is zero. Thus we must have either det(a) = or det(b) = (including the possibility that both of them are zero).. Show that if k is a scalar and A is an n n matrix, then det(ka) = k n det(a). Let A = a ij be an n n matrix. Let B = ka ij be the matrix obtained by multiplying A by a constant k. If k = then B has a row of zeros and hence det(b) = = n det(a). Otherwise, using the definition, det(b) = (±)bj b j b ljell b nj n = (±)ka j ka j ka ljell ka nj n (since the entries have all been scaled by k) Then, factoring out the k from each term and pulling it outside the sum, det(b) = k n (±)a j a j a ljell a njn = k n det(a).. Prove Theorem.5 Theorem.5: If B is obtained from A by multiplying a row (column) of A by a real number k, then det(b) = kdet(a). Let A = a ij be an n n matrix. Let B = b ij be the matrix obtained by multiplying the lth row of A by a constant k. If k = then B has a row of zeros and hence det(b) = = det(a). Otherwise, using the definition, det(b) = (±)b j b j b ljell b njn = (±)a j a j ka ljell a njn (since the entries for all other rows remain the same, and the entry drawn from row l has been scaled by k) Then, factoring out the k from each term and pulling it outside the sum, det(b) = k (±)a j a j a ljell a njn = kdet(a) The column case follows by observing that a column operation on A is a row operation on A T and using the fact that det(a) = det(a T )... Prove Theorem.8 Theorem.8: If A is an n n matrix, then A is nonsingular if and only if det(a).
8 Proof: First, suppose that A is nonsingular. Then, by Theorem.8, A = E E E n where each E i is an elementary matrix. Then, using Lemma. n times, det(a) = det(e E E n ) = det(e )det(e ) det(e n ) (as discussed in the sketch of the proof of Lemma. above, for each i, det(e i ) ). Conversely, suppose A is singular. Then, using Theorem., A is row equivalent to a matrix B with a row of zeros. Then, by Theorem., det(b) =. Therefore, since A = F F F l B, where each F i is an elementary matrix, det(a) = det(f F F l B) = det(f )det(f ) det(f l )det(b) =.. Let A = (a) M = 5 M = 5 (c) det(m ) = 5 5. Find: = ( )+()+( 6) () () (9) = 5 (d) A ( ) + det(m ) = ( )( 5) = 5 (e) A = ( ) + det(m ) = ( ) 5 = ()+()+() (5) () ( ) = ( )( ) =. Find the determinant of the matrix A from the previous problem by using the expansion of det(a) along the last row. det(a) = ()A +( )A +()A +()A = +( ) 5 +() +() 5 = +( )()+()+(8) () () ( )+()()+()+( ) () () ( )+()()+( )+() () ( 8) ( 5) = +( )()+()( )+()() = 9+6 =
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