September 13 Homework Solutions


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1 College of Engineering nd Computer Science Mechnicl Engineering Deprtment Mechnicl Engineering 5A Seminr in Engineering Anlysis Fll Ticket: 5966 Instructor: Lrry Cretto Septemer Homework Solutions. Are ll ordered polynomils in, of degree not eceeding three, vector spce or not? If they re find the dimension nd sis. Any possile polynomil of degree three or less cn e epressed y the eqution p = If we define sis for these polynomils s the four functions f =, f =, f =, nd f =, we see tht these four functions llow us to epress ny polynomil (of degree three or less) s the weighted sum of these four functions. (I.e., we cn write p =f + f + f + f.) Thus, these four functions form the sis for vector spce nd consequently, the dimension of the spce, which is the numer of sis functions, is four. This vector spce stisfies ll the requirements of commuttivity nd ssocitivity for vector ddition nd distriutivity nd ssocitivity for sclr multipliction. The sclr zero,, is the null element nd one is the unit element.. Show tht the vectors [ 8] T nd [ 4 ] T re orthogonl. Two vectors re orthogonl if their inner product is zero. (The inner product is the sum of the products of ech component.) For the two vectors shown, the inner product is ()() + ()() +()() + ()(4) + (8)() = =. Thus, the two vectors re orthogonl.. Find ll vectors, v, orthogonl to = [ ] T. Do they form vector spce? A vector, v, with components [v v v] T will hve n inner product with = [ ] T tht is equl to v + v + v = v + v. All vectors for which this inner product is zero will e orthogonl to. Since v is not prt of this formul, v cn hve ny vlue, sy. If v = y, then v = y will give zero innerproduct. Thus ny vector of the form [y y ] will e orthogonl to. These vectors stisfy ll the requirements of commuttivity nd ssocitivity for vector ddition nd distriutivity nd ssocitivity for sclr multipliction. The sclr zero,, is the null element nd one is the unit element. They re not sis for n ritrry threedimensionl vector spce, ut they do form vector spce for this limited cse. 4. Solve the following set of equtions using Guss elimintion or indicte if no solution eists. + 4y  z = 4 w  + y + z = w + + y = 6 8w y  z = 4 We first rerrnge the equtions s shown elow to simplify the elimintion process. w + + y = 6 [I] + 4y  z = 4 [II] w  + y + z = [III] 8w y  z = 4 [IV] Net we replce eqution [III] y [III] + [I] nd eqution [IV] y [IV] 8[I] to eliminte the w coefficients in ll equtions elow the first.
2 Septemer homework solutions ME5A, L. S. Cretto, Fll Pge w + + y = 6 [I] + 4y  z = 4 [II] [ +] + [ + ]y + z = [ + (6)] [III] [4 8] + [6 8]y  z = [4 8(6)] [IV] Crrying out the indicted multiplictions nd sutrctions gives. W + + y = 6 [I] + 4y  z = 4 [II] y + z = [III] y z = 44 [IV] Now we dd (4/) times eqution [II] to eqution three nd (4/) times eqution [II] to eqution [IV] to eliminte the coefficient in the equtions [III] nd [IV]. (4)() (4)(4) (4)( ) (4)(6) 4 4 [ ] y z III (4)() (4)(4) (4)( ) (4)(6) 4 8 y z 44 Crrying out the indicted rithmetic in these two equtions, the foureqution set ecomes W + + y = 6 [I] + 4y  z = 4 [II] + 9.6y.8z = 4.4 [III] + 4.8y 8.4z = 6.8 [IV] [ IV ] Finlly we sutrct 4.8/9.6 times eqution [III] from eqution [IV] to complete the upper tringulr form. (4.8)(9.6) (4.8)(.8) (4.8)(4.4) 4.8 y 8.4 z 6.8 [ IV ] The upper tringulr form of the equtions otined fter completing the indicted rithmetic is shown elow. W + + y = 6 [I] + 4y  z = 4 [II] + 9.6y.8z = 4.4 [III] (49/)z = 98 [IV] We now complete the solution y ck sustitution. First we solve eqution [IV] for z = 98/( 49/) or z = 6. Net, we use this vlue of z in eqution [III] to solve for y z 4.4.8(6) y Now we use y = nd z = 6 to solve eqution [II] for. 4 z 4y 4 (6) 4() Finlly, with, y, nd z known, we cn solve eqution [] for w. 6 y 6 w 4
3 Septemer homework solutions ME5A, L. S. Cretto, Fll Pge 5. Are the vectors [ 4], [ 9], nd [ 5] linerly dependent or independent? Since these re twodimensionl vectors, they must e linerly dependent. (The mimum numer of linerly independent vectors equls the dimension of the spce.) In this cse, it is esy to see tht [ 5] = [ 4] + [ 9]. 6. Find the rnk of the mtri t the right. We see tht this mtri hs two columns, which is less thn the numer of rows (three). Thus, its rnk cn e t most two, the numer of columns. Are the columns linerly independent? By inspection we cn see tht the first column is times the second column. Thus, this mtri hs only one linerly independent column so its rnk is one. Notice tht the first row is 4 times the middle row nd the lst row is times the middle row. Thus, the mtri hs only one linerly independent row. We epect this since either the numer of linerly independent rows or the numer of linerly independent columns determines the rnk of mtri. Both re the sme.. Find the rnk of the mtri shown t the right. We see tht this mtri hs three columns, which is less thn the numer of rows (four). Thus, its rnk cn e t most three, the numer of columns. Are the columns linerly independent? One pproch to determining the rnk of the mtri, which equls the numer of linerly independent rows, which is the sme s the numer of linerly independent columns, is to convert the mtri to rowechelon form. To convert the mtri to row echelon form, we use process similr to Guss elimintion in which we seek to set ll the mtri elements elow the min digonl to zero. Since the second row lredy hs zero in the first column, we cn tke the first step of getting ll coefficients in the first column, in the second row nd elow, equl to zero s follows. Replce the third row y the sum of the third nd first row nd replce the fourth row y the sum of the first row minus three times the fourth row When we do this, we get the result t the right ove, which is seen to hve only two linerly independent rows. Hence the rnk of the mtri is two. (See the net prolem for dditionl informtion out this mtri.) 8. Find the sis for the row spce nd the column spce for the mtri in prolem, immeditely ove. Since the rnk of the mtri is two, ny two linerly independent rows form sis for the row spce. We see tht the two rows [ 4] nd [ 5 8] re linerly independent (ecuse they re not multiples of ech other) nd cn form one possile sis. Similrly, ny two columns cn form the sis for the column spce of rnk two mtri. Here we cn pick the first two columns (gin since they re not multiples of ech other) s one possile sis. Hence the sis we choose for the column spce is [  ] T nd [ 5 4 ] T. Note tht other nswers re possile. You cn multiply ny of these nswers y constnt nd hve correct sis. You could lso tke the sum of two sis vectors nd sustitute tht sum for one of the sis vectors shown here. If you re not sure tht these two column vectors re n pproprite sis you cn trnspose the mtri nd pply Guss elimintion to show tht the rnk of the trnsposed mtri is lso two.
4 Septemer homework solutions ME5A, L. S. Cretto, Fll Pge 4 9. Solve the system of equtions shown elow y Guss elimintion nd check the result using Crmer s rule. The originl set of equtions, in the center, is rerrnged s shown on the right. + y + 8z =  + 9z = y 6z = If we sutrct times the first eqution from the third eqution nd use the result to replce the third eqution, we hve the following result for our set of three equtions Finlly, sutrcting / times the second eqution from the third eqution, nd replcing the third eqution with the result gives the system of equtions shown t the right. y  4 6z = [I] + 9z = 5 [II] y + + 8z =  [III] y  4 6z = [I] + 9z = 5 [II] + 9z =  [III] y  4 6z = [I] + 9z = 5 [II] 5.5z = 5.5 [III] We now find the results y cksustitution. First we see tht z = 5.5/5.5 or z =. We use this vlue of z to solve for. 5 9z 5 9() Finlly, we use =  nd z = to find y from eqution [I]. 6z 4 6() 4( ) y The solution using Crmer s rule for three simultneous equtions is given on slide 4 of the Septemer lecture. If the generl eqution is written s i + i + i = i, i =,,, the solution is The determinnt in the denomintor is the sme for ll unknowns. For this system of equtions (with =, = y, nd = z), we cn evlute the denomintor s follows ()()( 6) ()()(8) ( 4)()(9) ( 4)()(8) ()()( 6) ()()(9) With this determinnt for the denomintor, we cn then find the vlues of,, nd, s follows. 5 8 ( )()( 6) ( 5)()(8) 4 9 ()()(9) ()()(8) 6 6 ( 5)()( 6) ( )()(9) 9
5 Septemer homework solutions ME5A, L. S. Cretto, Fll Pge ()( 5)( 6) ()()(8) 9 9 ( 4)( )(9) ( 4)( 5)(8) ()( )( 6) ()()(9) ()()() ()()( ) 6 5 ( 4)()( 5) ( 4)()( ) 4 ()()() ()()( 5) 8 5 Thus, s epected, Crmer s rule nd Guss elimintion give the sme results in this emple.
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