Lesson 4 Linear Algebra

Transcription

1 Lesso Lier Algebr A fmily of vectors is lierly idepedet if oe of them c be writte s lier combitio of fiitely my other vectors i the collectio. Cosider m lierly idepedet equtios i ukows:, +, , +, m, + m, ,, m, = = = = = = = b = b = b There re, (clled osigulr system), or ifiitely my solutios. Defiitios:. If m =, the we cll the mtri squre mtri. b. Mtri A = mtri B if d oly if they ech hve the sme umber of rows d the sme umber of colums d ech elemet of A equls ech elemet of B. c. The zero mtri is where every elemet is zero. m = d. A mtri cosistig of oe row d severl colums is clled row mtri or row vector. [ ],, L, e. A mtri cosistig of oe colum d severl rows is clled colum mtri or colum vector. M m.,, f. If m =, the the idetity mtri is the mtri with s log mi digol d zeroes everwhere else.

2 g. If A hs m rows d colums the A is m mtri. A. Row Reductio Betty d Bob fid tresure t osis. They will tke home gold, wter d fuel. Ech cotier of gold weighs lbs, tkes hrs to pck d occupies cu.ft. Ech cotier of wter weighs lbs, tkes hrs to pck, d occupies cu.ft. Ech cotier of fuel weighs lbs, tkes hrs to pck, d occupies cu.ft. Their vehicle must tke lbs, they must use hrs d cu.ft. How my cotiers of gold, wter, d fuel must they tke? Let = the umber of cotiers of gold = the umber of cotiers of wter = the umber of cotiers of fuel Cosider the followig system of lier equtios: + + = + + = + + = We c replce y equtio i the system by multiple of itself or multiple of itself ± multiple of other equtio i the system. This is clled row reductio. For emple, the equtio + + = is give s true. Therefore, by multiplyig both sides by : or, dividig both sides by : + + = = re lso true equtios. Also, we c subtrct oe true equtio from other so tht: + + = ( + + = ) = is lso true equtio. Let s begi usig these rules to solve the system of lier equtios.

3 R R 5 5 R R R R R R R R 5 Now, So, =, d therefore, So, = 5, d therefore, So, = 5. 5 R 8 8 R R 5 = 9 + = = B. Crmer s Rule Let s strt gi with the previous system of lier equtios, ugmetig the mtri of coefficiets to get: Tkig the dow digols d the up digols, we get: = = = = = 8 = Sum: 8 Sum: Now, 8 = 8, the mtri determit, D, or A. Strt g ugmetig the mtri of coefficiets s show: Tkig the dow digols d the up digols, we get:

4 = = 8 = = = 9 = Sum: Sum: Now, =, D. Strt g ugmetig the mtri of coefficiets s show: Tkig the dow digols d the up digols, we get: = = 8 = = = = 8 Sum: 5 Sum: 58 Now, 5 58 =, D. Strt g ugmetig the mtri of coefficiets s show: Tkig the dow digols d the up digols, we get: = = 8 = = 8 = = Sum: 88 Sum: 8 Now, 88 8 = 8, D. Usig the vlues we hve foud we c clculte,, d. C. Mtri Mth = D / D = / 8 = 5 = D / D = / 8 = 5 = D / D = 8 / 8 = A m mtr tht is with dimesios m by, is rry of m rows d colums of umbers, clled elemets.,,..., =,,..., A,,..., m, m,... m, If m = the we cll the mtri squre mtri. A = B, if d oly if they ech hve the sme umber of rows d the sme umber of colums d every elemet of A is equl to every

5 elemet of B; tht is i = b i for ll. A mtri is sid to equl, if d oly if every elemet is equl to zero. Tht is, if A =, the i =. A = = A mtri tht cosists of oe row d severl colums is clled row mtri or row vector. [,,..., ] A mtri tht cosists of oe colum d severl rows is clled colum mtri or colum vector.,,... m, If m = the I, the idetity mtr hs oes log the mi digol. I = D. Additio d Subtrctio To dd two mtrices or to subtrct oe mtri from other, the two must hve the sme dimesios d: For emple, E. Sclr Multiplictio [ ] ± [ b ] = [ b ] ± = To multiply mtri by sclr, tht is oe by oe mtr multiply ech elemet i the mtri by the sclr. For emple, F. Mtri Multiplictio [ ] = [ c ] c = I order to multiply two mtrices together, sy A times B, the umber of colums i A must equl the umber of rows i B. The solutio mtri will hve dimesios equl to the umber of rows i A d the umber of colums i B. The, tke the i th row of A d multiply by the th colum of B. For emple, 8 8 9

6 A = B = 8 AB = 8 + = = Note tht mtri multiplictio is o commuttive. So, BA AB. G. Mtri Iversio Cosider squre mtr A. If there eists squre mtri,b, such tht AB = BA = I, the B is clled the iverse of A. Tht is, B = A -. Also the, A is sid to be ivertible or osigulr. Mtri A is osigulr if d oly if D( A ). Cosider system of lier equtios such tht AX = B. If A is osigulr, the A - AX = IA - B. The, X = A - B is the solutio. To fid A -, where: A = To fid A -, ugmet A to get [ A I ]. [ A I ] = The, pply row reductio to fid [ I A - ]..5 A =.5 Give tht AX = B, the system c be solved by computig A - B. H. Eigevlues d Eigevectors A eigevector of give lier trsformtio is o-zero vector which is multiplied by costt clled the eigevlue s result of tht trsformtio. A importt tool for describig eigevlues of squre mtrices is the chrcteristic polyomils, where λ is eigevlue of A is equivlet to ( A λi ) v = hs o-zero solutio v, eigevector, d so it is equivlet to the mtri determit, det( A λi ), beig zero. It follows tht we c compute ll the eigevlues of mtri A by solvig the equtio p A (λ) =. If A is -by- mtr the p A hs degree d A c therefore hve t most eigevlues.

7 Proect. Use the Row Reductio Method to solve y z = y + z = + y z =. Use the Row Reductio Method to solve y + z = y + z = + y + z = 8. Use the Row Reductio Method to solve + 5y + z = + y z =. Use Crmer s Rule (determits) to solve + y + z = + y z = + y + z = 5. Solve the o lier system + y = + y = Iterpret your results geometriclly.. Fid the dimesios of the followig mtrices ( m ). b. c. ( ) d. 5 5 e. / / f. ( ).. Fid

8 5 b. Fid 5 i.e the product of the mtri with itself. Ay Commets? b 8. Let A =. If A +A = (the zero mtri) the fid d b. b 9. Use the iverse mtri method to solve + y + z = + 5y + z = + y + 8z = 5. (Portfolio Problem) A portfolio of stocks hve tble A B C Portfolio Proportio X Y Z Retur Risk(bet) 5.8 Use the iverse mtri method to solve the system of equtios d ukows tht is result of this iformtio. Wht re your proportios?