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 non-zero entries, it is not in reduced row echelon form. 8 Since each non-zero 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 non-zero 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 Gauss-Jordan 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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