Iverse Matrix Two square matrices A ad B of dimesios are called iverses to oe aother if the followig holds, AB BA I (11) The otio is dual but we ofte write 1 B A meaig that matrix B is a iverse of matrix A 1 Obviously because of the duality of defiitio (11) we have 1 the iverse of a matrix is uique ie, there are o two iverses for the same matrix A A This also implies that Example 1 Matrices 1 3 A 1 3 B 1 1 are iverses to oe aother Verify (atted the class)! Example Matrices 0 1 1 0 A 3 0 4 1 B 1 0 3 0 are iverses to oe aother Verify (atted the class)! Obviously oe should ot expect that iverse of a matrix with iteger etries is a matrix with iteger etries, for example are iverses to oe aother Verify! 1 3 / 5 / 5 A 1 3 B 1/ 5 1/ 5 Not every square matrix has a iverse, for example 0 0 0 0 1
clearly caot have ad iverse There are may more matrices that caot have a iverse For example a product of 0 0 0 4 with ay other matrix would always have the first etry 0 Iverse is ot defied for o-square matrices because it is obvious that that (11) is ot possible because dimesio reasos Example 3 For matrices we ca show 1 1 1 3 A 0 1 0 5 6 B 0 1 1 0 AB 0 1 but reverse is ot of the same dimesio 1 3 3 BA 0 5 6 0 5 6 The matrices that have iverse are called sigular matrices We also call them ivertible What is the importace of iverse? We ca ow solve square systems of equatios easily Example 4 x x 4 1 x 3x 1 or 1 x1 4 1 3 x
-1 writte i the format AX C ca easily be solved by X = A C Now take a look we have the same A matrix from example #1 thus we ca use the iverse give there ad thus we have x1 3 4 x 1 1 x1 16 x 6 ad therefore we have immediate solutio, x1 16, x 6 As a matter of fact whe the system of equatios is cosistet ad has uique solutio the matrix of the system is sigular ad vice versa We shall show this fact later Theorem 1 If matrices AB, are ivertible the so are the matrices T 1 1 A A AB 1 1 1 3 T A ad AB ad the followig holds, T B A Proof: Usig properties of traspose matrices from the previous lecture we have 1 T 1 T 1 1 T T T A A AA I I T T T A A A A I I ad thus we have the first property i (1) You ca verify (11) for the secod property just by usig associativity for matrix product as follows 1 1 1 1 1 1 1 1 AB B A A BB A I B A AB B A A B I Thus we are left with the oly problem how to fid a iverse of the matrix of such a system! There are may ways to do that, the simple way is to use elemetary row matrix operatios to trasform the matrix ito uit matrix while doig the same trasformatios with uit matrix The uit matrix will trasform ito iverse matrix (see the ext example)! (1)
We shall justify this fact i the Theorem 3 after the followig examples Example 5 (example #1 revisited) We shall fid the iverse of matrix 1 A 1 3 To do that with elemetary row operatios while at the same time workig with uit matrix we do the followig: (a) write both matrices (A ad uit matrix) i the same elarged matrix, 1 1 0 1 3 0 1 (b) ow trasform A ito uit matrix with elemetary row operatios while doig exactly the same row operatios with the uit matrix i the other half of elarged matrix, 1 1 0 1 3 0 1 1 1 1 0 0 1 1 1 1 1 1 0 3 0 1 1 1 1 (c) the iverse is writte i the other half of the elarged matrix, 1 3 A 1 1 4
Example 6 (example # revisited) Fid the iverse of 0 1 A 3 0 1 0 (a) 0 1 1 0 0 3 0 0 1 0 1 0 0 0 1 (b) 1 0 1/ 1/ 0 0 1 / 3 0 0 1 0 1 0 0 0 1 3 1 0 1/ 1/ 0 0 1 0 0 1/ 3 / 1 0 3 0 1 1 1 0 1 1 3 1 1 0 1/ 1/ 0 0 0 1 1 1 0 1 0 0 1/ 3 / 1 0 1 3 5
1 0 1/ 1/ 0 0 1 0 1 1 1 0 1 0 0 1/ 3 / 1 0 3 1 0 1/ 1/ 0 0 1 0 1 1 1 0 1 0 0 1 3 0 3 1 0 0 1 0 1/ 0 1 0 4 1 3 0 0 1 3 0 3 1 3 (c) 1 0 3 0 1 A 4 1 We call matrices elemetary if they are made by performig oe elemetary row operatio o idetity matrix This is how they might look i 3 3 case 1 0 0 1 0 0 1 0 0 E1 0 0 1 0 3 0 3 0 1 0 E E 0 1 0 0 0 1 0 3 1 Here the first matrix have secod ad third row exchaged, the secod has secod row multiplied with scalar 3, the third has triple secod row added to the third Now very importat property to otice is that if we multiply ay matrix with elemetary matrix (from the left) we shall get the same chage Here is oe example: 1 0 0 a b c a b c 0 1 0 d e f d e f 0 3 1 g h k 3d h 3e h 3 f k 6
It should also be obvious that elemetary matrices are ivertible sice the elemetary row operatios are reversible Example 7 E 1 0 0 1 0 0 0 0 1 0 0 1 E 0 1 0 0 1 0 1 1 1 E 1 0 0 1 0 0 1 0 3 0 0 0 E 3 0 0 1 0 0 1 1 E 1 0 0 1 0 0 0 1 0 0 1 0 E 0 3 1 0 3 1 1 3 3 We shall call two matrices row-equivalet if oe ca be reached from the other by elemetary row operatios The previous meas the followig is true Theorem If square matrices A ad B are row-equivalet the there is a matrix E such that 1 E B A EA B ad Proof: From the previous discussio E is simply a product of elemetary row matrices eeded to reach B while the iverse of E would be the product of iverses of these i reverse order This meas the followig theorem justifies our matrix iversio algorithm used i examples #5 ad #6 Theorem 3 If E is a matrix obtaied by performig elemetary row operatios o I that are the same ad i the same order are the oes performed o matrix A eeded to reach I the E is the iverse of A 7
Proof: By the coditio we have we have the iverse for E ad the followig follows which shows what we eeded to prove EA I Sice E is ivertible (as a product of ivertible matrices) the 1 1 1 1 E E I E EA E E A IA A Importat ote: If two liear systems of equatios have augmeted matrices that are row equivalet the the system have the same solutios This is obvious from G-J Elimiatio procedure that is doe by elemetary row operatios The ext theorem is a summary of may results we already had Theorem 4 A square matrix A is ivertible is equivalet to ay of the followig: T T 1 A is ivertible [The traspose of the iverse of A is the iverse of A This is easy to show] A is row-equivalet to idetity matrix 3 A has pivot positios 4 System Ax 0 has oly trivial solutio 5 System Ax b is cosistet for ay b [It is easy to see that #4 implies #5 give what we kow about G-J Elimiatio procedure If #5 holds the we have #3 because if we would have less tha pivot positios we would have at least oe complete 0 row which would force oe variable to be 0 meaig #5 would fail] 6 Liear trasformatio give by A matrix is 1-1 [This is equivalet to #4] 7 The colums of A are liearly idepedet [Agai, equivalet to #4] 8 The colums of A spa [This is equivalet to #5] 9 The liear trasformatio give by A is oto [Equivalet to #8] Importat ote: Because of #6 ad #9 i the previous theorem we see that oto ad 1-1 otio is the same otio for liear trasformatio with stadard square matrices Theorem 5 Let A be a matrix If there is a square matrix B so that BA I or AB I the A is ivertible ad B is the iverse [Meaig oly oe side multiplicatio iverse is eough to show there is both] 8
Proof Assume that AB I Sice b b b 1k k A results with k-th colum of AB the liear trasformatio k give by A would provide all the elemetary vectors eeded to spa all of We have #8 i the previous theorem Thus A is ivertible Now B A 1 A B A 1 AB A 1 [I case that BA I we ca simply show that B is ivertible ad that the iverse is A] //// QED We ca add these two properties to the list i Theorem 4 as two more equivalets to ivertibility of a matrix ***** Alterative method to fid iverse i case of matrices We ca fid iverse matrix of square matrix by way of formula as follows: Give matrix a b A c d the iverse is simply give by A 1 1 d b ad bc c a For example give (example #1 agai) 1 A 1 3 9
3 1 1 d b 1 3 A ad bc c a 1 3 1 1 1 1 1 ule of thumb to remember: 1 Switch the mai diagoal Chage the sigs o the other diagoal 3 Divide with ad bc (this is also called a determiat of the matrix) ***** Solvig a system of equatios usig iverse matrix Example 7 Solve the followig system usig iverse matrix Usig matrices we have 3x7y 10 x5y 3 7 x 10 5 y We eed to fid iverse matrix of the system 1 3 7 1 5 7 5 7 5 35 7 3 3 Fially, we are ready to apply this o the system, x 5 7 10 36 y 3 14 The solutio is therefore x 36, y 14 or alteratively36, 14 Check the system to make sure this is a correct solutio Example 8 Solve the followig system usig iverse matrix 10
Usig matrices we have x y z 0 3x y 8 x 3y 4z 6 1 1 1 x 0 3 1 0 y 8 3 4 z 6 We eed to fid iverse matrix of the system, here give 4 1 1 1 1 1 1 9 9 9 4 1 3 1 0 3 3 3 3 4 7 5 4 9 9 9 Now we are have to apply this o the system, 4 1 1 14 9 9 9 0 9 x 4 1 10 y 8 3 3 3 3 z 6 7 5 4 16 9 9 9 9 14 10 16 14 10 16 The solutio is therefore x, y, z or alteratively,, 9 3 9 9 3 9 system to make sure this is a correct solutio Check the Homework: Check olie 11