In this section make precise the idea of a matrix inverse and develop a method to find the inverse of a given square matrix when it exists.


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1 Mth 52 Sec S060/S0602 Notes Mtrices IV 5 Inverse Mtrices 5 Introduction In our erlier work on mtrix multipliction, we sw the ide of the inverse of mtrix Tht is, for squre mtrix A, there my exist mtrix B with the property tht AB = BA = I This is useful concept, nd gives us yet nother method for solving systems of equtions To illustrte, consider the simple system 2x 5y = 6 x + 3y = Insted of writing this s n ugmented mtrix, write this s mtrix eqution using product: 2 5 x 6 = 3 y 2 5 x 6 If we let A =, X =, nd C =, then the eqution we wish to solve is 3 y AX = C If we knew A, we could solve this esily for the unknown X: (left) multiply both sides of the eqution by A to find A (AX) = A C (A A)X = A C IX = A C X = A C We see this is much like solving the simple eqution x = c for the unknown x where nd c re rel numbers In this section mke precise the ide of mtrix inverse nd develop method to find the inverse of given squre mtrix when it exists 52 Definition Suppose A is squre mtrix of order n A mtrix B with the property tht BA = I is clled n inverse of A If A hs n inverse, it is clled invertible, nd we write A to denote the inverse Some notes concerning this definition: If A is invertible, then AA = A A = I 2 If mtrix A hs n inverse, then the inverse is unique, so we my spek of the inverse A 3 Not ll squre mtrices hve inverses p of 6
2 Mth 52 Sec S060/S0602 Notes Mtrices IV 53 Procedure for Finding the Inverse of Mtrix Here we give method for finding the inverse of squre mtrix We will see tht this involves nothing more thn row reduction tht we hve seen before For the purposes of the explntion 2 2 mtrices re used, but the method extends to squre mtrices of ny size Suppose A is invertible, where A = 2, 2 22 b b nd we wish to find mtrix B = 2 such tht AB = I Tht is, we wnt b 2 b 22 2 b b 2 0 = 2 22 b 2 b 22 0 This mtrix multipliction my be expressed s two systems of equtions: b + 2 b 2 = 2 b + 22 b 2 = 0 nd b b 22 = 0 2 b b 22 = If A is invertible, then there re vlues of b, b 2, b 2, b 22 which solve this system In ugmented mtrix form these two systems of equtions become 2 nd Now, if A is invertible, gin mening tht these two systems hve unique solutions, then fter reduction by elementry row opertions the result would be 0 b 0 b2 nd 0 b 2 0 b 22 Here s the key observtion: the elementry row opertions used to reduce A re the sme for both systems! Therefore, we cn do both reductions simultneously using n ugmented mtrix of the form 2 0 reduce 0 b b b 2 b 22 Notice wht this sys: if A exists, then A reduces to I nd produces A in the ugmented mtrix bove It lso tells us something more: if A fils to reduce to I with this procedure, then A does not exist So this procedure not only gives the inverse when it exists, it lso tells us with certinty when A does not exist The procedure cn be summrized very concisely: to find the inverse of the mtrix A: A I reduce I A If the originl mtrix A does not reduce to I in this procedure, then A does not exist p 2 of 6
3 Mth 52 Sec S060/S0602 Notes Mtrices IV 54 Exmples Exmple: Bck to our problem from the beginning of this section: solve the system using mtrix inverses Solution: Letting A = x, X = y 2x 5y = 6 x + 3y = 6, nd C = AX = C To find A, first set up Now reduce: Therefore, A = 3/ 5/ / 2/ 3 0 R R 2 : ( 2)R + R 2 : ( /)R 2 : 0 / 2/ 0 3/ 5/ ( 3)R 2 + R : 0 / 2/, nd so, we wish to solve X = A C 3/ 5/ = / 2/ 23/ = 4/ 6 Exmple: Let A = p 3 of 6
4 Mth 52 Sec S060/S0602 Notes Mtrices IV Find A Solution: Set up Now reduce: (2)R + R 2 : ( 5)R + R 3 : ( )R 2 : (2)R 2 + R : ( 3)R 2 + R 3 : (5)R 3 + R : (3)R 3 + R 2 : Since A reduced to I in the left hnd side of the ugmented mtrix, the right hnd side is A : A = A check shows tht indeed, AA = A A = I Exmple: Let Find A Solution: Set up A = p 4 of 6
5 Mth 52 Sec S060/S0602 Notes Mtrices IV Now reduce: ( 2)R + R 2 : ( )R + R 3 : ( /5)R 2 : ( 3)R 2 + R : (2)R 2 + R 3 : /5 / /5 3/ /5 / /5 2/5 Notice: the left hnd side of the ugmented mtrix is now reduced, but it is not the 3 3 identity mtrix Therefore, A does not exist b Exmple: Find A if A = c d Solution: b 0 c d 0 b (/) R : 0 c d 0 ssuming 0 b 0 ( c)r + R 2 : 0 d bc c ( ) b R 2 : d bc 0 c d bc 0 d bc ssuming d bc 0 ( b ) R 2 + R : 0 d d bc b 0 c d bc d bc d bc b The conclusion is tht if A = where d bc 0, then A d b = Note c d d bc c tht even though we stted tht 0 in the first row reduction step, the finl result is vlid even if = 0 This form for A is very convenient in prctice p 5 of 6
6 Mth 52 Sec S060/S0602 Notes Mtrices IV Problems for Section 5 Find the inverses (if they exist) of the following mtrices: Use your nswers to nd 3 to verify tht (AB) = B A Solutions to Problems for Section Does not exist 3 2/5 3/5 /5 4/5 4 5 Does not exist 3/8 /4 /8 /8 3/4 3/8 /4 /2 /4 p 6 of 6
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