MATH 2030: ASSIGNMENT 3 SOLUTIONS

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1 MATH : ASSIGNMENT SOLUTIONS Matrix Operations Q.: pg 9, q. Write the system of linear equations as a matrix equation of the form Ax = b. x + x =, x x =, x + x = A.. x x =. x Q.: pg 9, q. Compute AB by block multiplication using the indicated partitioning: A =, B = 4 4. A.. We may write A as [A I ] where I is the identity matrix and A =. 4 I Similarly B = u B where is the -component column vector, and u = Then the product AB becomes, and B = 4 [A + I u A I + I B ] Computing the product of these individual matrices separately we find that the product is simply Matrix Algebra Q.: pg 67, q. Solve the equation for X from X = A B where A and B are matrices: A =, B =. 4

2 MATH : ASSIGNMENT SOLUTIONS A.. Using the properties of matrix addition and scalar multiplication we may solve for the simplest expression on the right hand side first ( ) () A B = = () 4 () dividing by gives the expression since X = (A B) X =. Q.4: pg 67, q 4. Determine whether the three matrices are linearly independent, 6,, A.4. Multiply each matrix by c, c and c respectively, [ ] 6 c c + c 4 + c 7 8 = + c + c c + 6c + c c + 7c + c = 4c + 8c + c we equate the resulting matrix with the zero matrix to determine linear independence, which gives us four linear equations with augmented matrix Applying row reduction from top to bottom and left to right, the row echelon form of A is 4 it is clear that the system is consistent and that the reduced row echelon matrix of the coefficient matrix gives c = c 4, c = c 4 and c is arbitrary - so the three matrices are linearly dependent. Q.: pg 68, q 44. The trace of a n n matrix A = [a ij ] is the sum of the entries on its main diagonal and is denoted by tr(a), i.e. tr(a) = a + a a nn. If A and B are n n matrices, prove that tr(a + B) = tr(a) + tr(b) tr(ka) = ktr(a), where k is a scalar A.. Using the component description of matrices, the trace can be seen as Σ n i= a ii, where A = a[a ij ] is a n n matrix. If A and B=[b ij ] are the same size we may add them together and this would be described as A + B = [a ij + b ij ] and so the trace of A + B is then tr(a + B) = Σ n i=(a ii + b ii ) = Σ n i=a ii + Σ n i=b ii = tr(a) + tr(b)

3 MATH : ASSIGNMENT SOLUTIONS this can always be done as the components of A and B are finite and real-valued, so we have proven the first identity. For the second we use the distributivity of the real numbers to note that ka = [ka ij ] and hence the trace of ka is tr(ka) = Σ n i=ka ii = k (Σ n i=a ii ) = ktr(a). Matrix Inverse Q.6: pg 84 q. Solve the given matrix equation for X, you may assume that the matrices A, B and X are invertible: (A X) = (AB ) (AB ). Simplify the expression for X as much as possible. A.6. To start we apply the inverse operator on both sides (A X) ) = [(AB ) (AB )] A X = (AB ) AB Next we simplify the right hand side by noting (AB ) = B A so that this becomes A X = B A AB = B IB = B. Finally left-multiplying by A on both sides we find that X = AB. Q.7: pg 8, q 6,8. Find the inverse of the two elementary matrices:, c. A.7. As the first matrix is a row interchange R R the inverse will be the transpose which is which is actually the same matrix, it is its own inverse! The second elementary matrix corresponds to the row operation R +cr, this has an inverse row operation, namely R cr with corresponding elementary matrix c this new elementary matrix is the inverse we are looking for.

4 4 MATH : ASSIGNMENT SOLUTIONS The LU factorization. Q.8: pg 9, q 4. Solve the system Ax = b using the given LU factorization of A, A = 4 4 = 4 4, b =. A.8. To start we denote U x = y and consider the simpler problem Ly = b. Writing y t = [y, y, y ] we find the following system of linear equations y =, y + y =, y + y = by substituting y into y and y we find y = and y = 4, so that y =. 4 Next we solve the problem Ux = y, with the system of linear equations x 4x =, x + 4x =, x = 4 Solving for x = and substituting into the remaining two equations we find that x 4x = and x =. One more substitution gives the last value x =. Checking we find that x = is indeed a solution to the original problem Ax = b Q.9: pg 96, q. Find the LU factorization for the matrix Q.9. Applying the row operations, R R, R R, then R + 4 R we find the upper triangular matrix Noting the row operations this implies the lower triangular matrix L has entries L =, L = and L = 4 as entries below the diagonal, thus L = ; 4 and so A = LU. Q.: pg 96 q 6. For an invertible matrix with an LU factorization A = LU both L and U will be invertible and so A = U L. Find L and U and then calculate A for the matrix given in Q.8.

5 MATH : ASSIGNMENT SOLUTIONS Q.. To compute the inverse of L we could apply the Gauss-Jordan method and row reduced the super augmented matrix [L I until this matrix is of the form [I L ]. Instead we note that if we apply the row operations R R, R + R, we may transform L into I. By recording these operations as elementary matrices and computing their product E = E E E we find that EL = I and so L = To compute the inverse of U we row reduce the super augmented matrix [U I ] until U is in reduced row echelon form (the identity matrix since it s invertible), i.e. [I n U ] U = 4 Computing the inverse A = U L we find A = 4 4 References [] D. Poole, Linear Algebra: A modern introduction - rd Edition, Brooks/Cole ().

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