Mtrix Algebr ECON 33 Lecture Notes: Ch 4 nd Ch 5. Gives us shorthnd wy of writing lrge system of equtions.. Allows us to test for the existnce of solutions to simultneous systems. 3. Allows us to solve simultneous system. DRAWBACK: Only works for liner systems. However, we cn often covert non-liner to liner systems. Exmple y x b Mtrices nd Vectors Given In mtrix form 3 ln y ln + b ln x y 0 x x + y 0 y +3x 3x + y x y 0 In generl Mtrix of Coefficients Vector of Unknows Vector of Constnts x + x +...+ n x n d x + x +...+ n x n d... m x + m x +...+ mn x n d m n-unknowns (x, x,...x n ) Mtrix form Mtrix shorthnd Where: A coefficient mrtrix or n rry x vector of unknowns or n rry d vector of constnts or n rry... n... n........ m m... mn Ax d x x. x n d d. d m Subscript nottion is the coefficient found in the i-th row (i,...,m) nd the j-th column (j,...,n) ij
. Vectors s specil mtrices The number of rows nd the number of columns define the DIMENSION of mtrix. A is m rows nd n is columns or mxn. A mtrix contining column is clled column VECTOR xisn columnvector d is m column vector If x were rrnged in horizontl rry we would hve row vector. Row vectors re denoted by prime x x,x,...,x n A vector is known s sclr. Mtrix Opertors If we hve two mtrices, A nd B, then x 4 is sclr A B iff ij b ij Addition nd Subtrction of Mtrices Suppose A is n m nmtrixndbisp q mtrixthena nd B is possible only if mp nd nq. Mtrices must hve the sme dimensions. b b + b b Subtrction is identicl to ddition 9 4 7 3 6 ( + b ) ( + b ) ( + b ) ( + b ) (9 7) (4 ) (3 ) ( 6) 5 c c c c Sclr Multipliction Suppose we wnt to multiply mtrix by sclr We multiply every element in A by the sclr k Exmple 6 Let k3 nd A 4 5 then ka ka ka k A m n k k... k n k k... k n. k m k m... k mn 3 6 3 3 4 3 5 8 6 5 Multipliction of Mtrices To multiply two mtrices, A nd B, together it must be true tht for A B C m n n q m q Tht A must hve the sme number of columns (n) s B hs rows (n). Theproductmtrix,C,willhvethesmenumberofrowssAndthesmenumberofcolumnssB.
Exmple A B C ( 3) (3 4) ( 4) row 3rows row 3cols 4cols 4cols In generl A B C D E (3 ) ( 5) (5 4) (4 ) (3 ) To multiply two mtrices: () Multiply ech element in given row by ech element in given column () Sum up their products Exmple b b c c b b c c Where: c b + b (sum of row times column ) c b + b (sum of row times column ) c b + b (sum of row times column ) c b + b (sum of row times column ) Exmple Exmple 3 3 3 4 3 4 is the inner product of two vectors. (3 ) +( 3) (3 ) +( 4) 9 4 (3 ) + ( ) + ( 4) Suppose therefore However Exmple 4 x x x then x x x x x x x x x x + x xx bymtrix x x x x x x x x x x Ab A 3 8 4 0 b 5 9 ( 5) + (3 9) ( 5) + (8 9) (4 5) + (0 9) 3 8 0 3
Exmple This produces 6 3 4 4 5 Ax d x x A x d x 3 0 (3 3) (3 ) (3 ) 6x +3x + x 3 x +4x x 3 4x x +5x 3 0.. Ntionl Income Model Arrnge s Mtrix form b y c + I 0 + G 0 C + by y C I 0 + G 0 by + C Y C A x d Io + G o.. Division in Mtrix Algebr In ordinry lgebr b c is well defined iff b 0. Now b cn be rewritten s b, therefore b c, lso b c. But in mtrix lgebr is not defined. However, is well defined. BUT B is clled the inverse of B A B C AB C AB B A B B In some wys B hs the sme properties s b but in other wys it differs. We will explore these differences lter. 4
. Liner Dependnce Suppose we hve two equtions x +x 3x +6x 3 To solve 3 x + 6x 3 6x +3 6x 3 33 There is no solution. These two equtions re linerly dependent. Eqution is equl to two times eqution one. x 3 6 x 3 where A is two column vectors OrAistworowvector U 3 Ax d U 6 V V 3 6 Where column two is twice column one nd/or row two is three times row one U U or 3V V Liner Dependence Generlly: A set of vectors is sid to be linerly dependent iff ny one of them cn be expressed s liner combintion of the remining vectors. Exmple: Three vectors, re linerly dependent since V 7 V 8 3V V V 3 6 6 or expressed s 3V V V 3 0 Generl Rule A set of vectors, V, V,..., V n re linerly dependent if there exsists set of sclrs (i,...,n). Not ll equl to zero, such tht Note n i n i k i V i 0 V 3 4 5 4 5 k i V i k V + k V +...+ k n V n 5
.3 Commuttive, Associtive, nd Distributive Lws From Highschool lgebr we know commuttive lw of ddition, + b b + commuttive lw of multipliction, b b Associtive lw of ddition, ( + b) +c +(b + c) ssocitive lw of multipliction, (b)c (bc) Distributive lw (b + c) b + c In mtrix lgebr most, but not ll, of these lws re true..3. Communictive Lw of Addition A + B B + A Since we re dding individul elements nd ij + b ij b ij + ij forllindj..3. Similrly Associtive Lw of Addition for the sme resons. A +(B + C) (A + B)+C.3.3 Mtrix Multipliction Mtrix multipliction in not communttive Exmple Let A be 3 nd B be 3 AB BA Exmple Let A 3 4 But A B C wheres B A C ( 3) (3 ) ( ) (3 ) ( 3) (3 3) 0 nd B 6 7 ( 0) + ( 6) ( ) + ( 7) AB (3 0) + (4 6) (3 ) + (4 7) (0)() ()(3) (0)() ()(4) BA (6)() + (7)(3) (6)() + (7)(4) 3 4 5 3 4 7 40 Therefore, we relize the distinction of post multiply nd pre multiply. In the cse AB C B is pre multiplied by A, A is post multiplied by B. 6
.3.4 Associtive Lw Mtrix multipliction is ssocitive (AB)C A(BC) ABC s long s their dimensions conform to our erlier rules of multipliction..3.5 Distributive Lw Mtrix multipliction is distributive A B C (m n) (n p) (p q) A(B + C) AB + AC Pre multipliction (B + C)A BA + CA Post multipliction.4 Identity Mtrices nd Null Mtrices.4. Identity mtrix: is squre mtrix with ones on its principl digonls nd zeros everywhere else. I 0 0 I 3 0 0 0 0 0 0 In 0... n 0..... 0 0... 0 Identity Mtrix in sclr lgebr we know In mtrix lgebr the identity mtrix plys the sme role Exmple 3 Let A 4 IA 0 0 3 4 IA AI A ( ) + (0 ) ( 3) + (0 4) (0 ) + ( ) (0 3) + ( 4) 3 4 Exmple 3 Let A 0 3 Furthermore, IA AI 0 0 3 0 3 3 0 3 0 0 0 0 0 0 3 0 3 3 0 3 A {I Cse} A {I 3 Cse} AIB (AI)B A(IB) AB (m n)(n p) (m n)(n p) 7
.4. Null Mtrices A null mtrix is simply mtrix where ll elements equl zero. 0 0 0 0 0 0 0 0 0 0 0 0 ( ) ( 3) The rules of sclr lgebr pply to mtrix lgebr in this cse. Exmple A +0 A 0 +0 {sclr} 0 0 + 0 0 3 3 0 0 0 A {mtrix} 0 0 0.5 Idiosyncrcies of mtrix lgebr ) We know AB BA )b0 implies or b0 In mtrix AB 4 4.5. Trnsposes nd Inverses )Trnspose: is when the rows nd columns re interchnged. Trnspose of AA or A T Exmple 3 8 9 If A 0 4 A 3 8 0 9 4 Symmetrix Mtrix 3 4 nd B 7 nd B 3 4 7 If A 0 4 0 3 7 4 7 A is symmetric mtrix. then A 0 4 0 3 7 4 7 0 0 0 0 Properties of Trnsposes ) (A+B) A +B 3) (AB) B A ) (A ) A Inverses nd their Properties In sclr lgebr if x b 8
then x b or b In mtrix lgebr if then where A is the inverse of A. Ax d x A d Properties of Inverses ) Not ll mtrices hve inverses non-singulr: if there is n inverse singulr: if there is no inverse ) A mtrix must be squre in order to hve n inverse. (Necessry but not sifficient) 3) In sclr lgebr, in mtrix lgebr AA A A I 4)Ifninverseexiststhenitmustbeunique. Exmple 3 Let A nd A 3 6 0 0 { A by fctoring is sclr} 6 0 3 6 Post Multipliction 3 AA 0 0 3 6 0 6 0 6 0 6 0 Pre Multipliction A A 6 0 3 3 0 6 6 0 0 6 0 0 Further properties If A nd B re squre nd non-singulr then: ) (A ) A ) (AB) B A 3) (A ) (A ) Solving liner system Suppose then A x d (3 3) (3 ) (3 ) A A x A d (3 3) (3 3) (3 ) (3 3) (3 ) I x A d (3 3) (3 ) (3 3) (3 ) Exmple x A d Ax d 9
then A 6 3 4 4 5 x x x 3 x x x x 3 5 d 0 8 6 0 3 6 3 7 8 A 5 0 3 8 6 0 3 6 3 7 8 x x 3 x 3 Liner Dependence nd Determinnts Suppose we hve the following. x +x. x +4x where eqution two is twice eqution one. Therefore, there is no solution for x,x. In mtrix form: Ax d A 4 The determinnt of the coefficient mtrix is x x x d A ()(4) ()() 0 determinnt of zero tells us tht the equtions re linerly dependent. Sometimes clled vnishing determinnt. In generl, the determinnt of squre mtrix, A is written s A or deta. For two by two cse A k where k is unique ny k 0implies liner independence Exmple 3 A 5 Exmple 6 B 8 4 A (3 5) ( ) 3 B ( 4) (6 8) 0 {Non-singulr} {Singulr} Three by three cse 0
Given A 3 b b b 3 c c c 3 then A ( b c 3 )+( b 3 c )+(b c 3 ) ( 3 b c ) ( b c 3 ) (b 3 c ) Cross-digonls Use viso to disply cross digonls 3 b b b 3 c c c 3 Multiple long the digonls nd dd up their products The product long the BLUE lines re given positive sign The product of the RED lines re negtive.. Using Lplce expnsion The cross digonl method does not work for mtrices greter thn three by three Lplce expnsion evlutes the determinnt of mtrix, A, by mens of subdeterminnts of A. Subdeterminnts or Minors Given A 3 b b b 3 c c c 3 By deleting the first row nd first column, we get M b b 3 c c 3 The determinnt of this mtrix is the minor element. M ij is the subdeterminnt from deleting the i-th row nd the j-th column. Given A 3 b b b 3 c c c 3 then M 3 3 33.. Cofctors A cofctor is minor with specific lgebric sign. C ij ( ) i+j M ij 3 M 3 3 therefore C ( ) M M The determinnt by Lplce Expnding down the first column C ( ) 3 M M A 3 3 3 3 33
Note: minus sign (-) (+) A C + C + 3 C 3 3 i i C i A 3 3 3 3 + 33 3 3 33 3 A 33 3 3 33 3 3 + 3 3 3 Lplce expnsion cn be used to expnd long ny row or ny column. Exmple Third row Exmple A 8 3 4 0 6 0 3 ()Expnd the first column A 3 3 3 3 + 3 33 3 A 8 0 0 3 4 3 +6 3 0 3 0 ()Expnd the second column A (8 0) (4 3) + (6 ) 6 A 4 6 3 +0 8 3 0 8 3 6 3 4 A ( 6) + (0) (0) 6 Suggestion: Try to choose n esy row or column to expnd. (i.e. the ones with zero s in it.). Rnk of Mtrix Definition The rnk of mtrix is the mximum number linerly independent rows in the mtrix. If A is n m n mtrix, then the rnk of A is r(a) min m, n Red s: the rnk of A is less thn or equl to the minimum of m or n. Using Determinnts to Find the Rnk () If A is n m nd A 0 () Then delete one row nd one column, nd find the determinnt of this new (n-) (n-) mtrix. (3) Continue this process until you hve non-zero determinnt.
3 Mtrix Inversion Given n n n mtrix, A, the inverse of A is A A AdjA where AdjA is the djoint mtrix of A. AdjA is the trnspose of mtrix A s cofctor mtrix. It is lso the djoint, which is n n n mtrix Cofctor Mtrix (denoted C) The cofctor mtrix of A is mtrix who s elements re the cofctors of the elements of A C C If A then C C C Exmple 3 Let A A - 0 Step : Find the cofctor mtrix C C C C C 0 3 Step : Trnspose the cofctor mtrix C T 0 AdjA 3 Step 3: Multiply ll the elements of AdjA by to find A A A A AdjA ( ) 0 3 0 3 Step 4: Check by AA I 3 0 0 0 0 3 (3)(0) + ()( ) (3)() + ()( 3 ) ()(0) + (0)( ) ()() + (0)( 3 ) 4 Crmer s Rule Suppose: Eqution x + x d or where Solve for x by substitution From eqution Eqution b x + b x d A x d d d x b b x A b b 0 x d x 3
nd eqution therefore: Cross multiply Collect terms x d b x b d x d b x b d b b x d b x d b d ( b b )x x d b d b b The denomintor is the determinnt of A The numertor is the sme s the denomintor except d d replces b. Crmer s Rule d b d x b b Where the d vector replces column in the A mtrix To find x replce column with the d vector b d d x b b d b d b b d d b b b Generlly: to find x i,replce column i with vector d; find the determinnt. x i the rtio of two determinnts x i Ai A 4.0. Exmple: The Mrket Model Eqution Q d 0 P Or Q + P 0 Eqution Q s P Or Q + P Mtrix form A x d Q P 0 A ()() ( )() Find Q e Q e 0 0 4 4
Find P e 0 P e ( 0) 6 Substitute P nd Q into either eqution or eqution to verify 4.0. Exmple: Ntionl Income Model Q d 0 P 0 64 Y C + I 0 + G 0 Or Y C I 0 + G 0 C + by Or by + c In mtrix form Solve for Y e b Y C Y e I 0 + G 0 I 0 + G 0 b I 0 + G 0 + b Solve for C e C e I 0 + G 0 b b + b(i 0 + G 0 ) b 5