Inverses and Determinants

Transcription

1 Engineering Mathematics 1 Fall 017 Inverses and Determinants I begin finding the inverse of a matrix; namely 1 4 The inverse, if it exists, will be of the form where AA 1 I; which works out to ( 1 4 A 1 x u y v x u y v that is ( x + y u + v x + 4y u + 4v ( 1 0 ( 1 0 We have to solve two systems of linear equations, both having system matrix A: and x + y 1 x + 4y 0 u + v 0 u + 4v 1 For the first one we have to row reduce (if we use the method of row reduction, which we do 1 1, 4 0 for the second, ( We can save a lot of time by row reducing both simultaneously; in other words row reducing ( If we get the first columns to row reduce to the identity matrix, then the third column will be the solution of while the fourth column will be the solution ( x A y u of v ( u A v In other words, the third and fourth columns wil constitute A 1 Here we go The operations applied are III ( (1, II 1 (, III (1 ( : ( 1 0 ( 0 1, ; 1 1 ( x y

2 And we are done with One sees that, in fact, A Now for a more complicated case Exercise Determine the inverse of the following matrix ( 1 0 Solution We augment the matrix by the identity matrix, and row reduce to reduced row echelon form The operations can proceed as follows (there is always more than one way of getting there; the end product should be the same / 1 1/4 1/ / 1 /4 1/ / 1 1/4 1/ / 1/4 1/4 1/ / 1/ 1 0 /4 /8 5/8 /4 0 /4 1/8 1/8 1/4 0 0 / 1/4 1/4 1/ / 1/ 1/6 1 0 / 1 /4 1/ / 1 1/4 1/ / 1/ / 1/ 1 0 /4 /8 5/8 /4 0 /4 1/8 1/8 1/4 0 0 / 1/4 1/4 1/ /8 11/4 7/8 1/ /8 1/4 /8 1/4 0 /4 1/4 1/4 1/4 0 0 / 1/ 1/6 The inverse is /8 11/4 7/8 1/4 1/8 1/4 /8 1/4 1/4 1/4 1/4 1/4 / 1/ 1/6 1 Determinants If A is a square n n matrix, one assigns to it a number; the determinant of A, or det A One of the problems we have here and in all courses at our level, is that there are many ways of computing determinants and one should know more than one But at this level it isn t easy to show that all lead to the same result I start with a constructive definition det A is a number satisfying: 1 deti 1 (The determinant of the identity matrix is 1 If à is obtained from A by exchanging two (different rows (row operation I ij with i j, then det(ã det A

3 If à is obtained from A by multiplying a row by c (row operation I c(i, then det(ã c det A or equivalently, and better, det 1 det(ã (assuming c 0 c 4 If à is obtained from A by adding to one row another row times a constant(row operation I (i+c(j with i j, then det(ã det A 5 If A has two equal rows, or a row of zeros, then det 0 With these properties it is easy to compute determinants But we might want to notice a few shortcuts I Suppose the matrix has all diagonal entries equal to 1 and all entries below the diagonal equal to 0 Here is an example of such a matrix of order 5 5 The asterisks represent entries that play no major role in this; they can be replaced by any numbers one wishes to replace them by It should be clear that by using only row operations of the third kind we can reduce such a matrix to the identity matrix; such row operations do not change the value of the determinant; since the final matrix, the identity matrix, has determinant 1, we see that det 1 Conclusion: If a matrix has all entries below the main diagonal equal to 0, and all diagonal entries to 1, then its determinant is 1 II Suppose the matrix is upper triangular; this means that all entries below the diagonal are 0 Here is an example of a 6 6 such matrix a 11 0 a 0 0 a a a a 66 Then det a 11 a a a 44 a 55 a 66 The determinant of an upper triangular matrix is the product of its diagonal entries Here is why Suppose first no diagonal entry is 0 Then multiplying the first row by 1/a 11 places a 1 in the first diagonal position and changes the determinant by a factor of 1/a 11 Next, multiplying the second row by 1/a places a 1 in the second diagonal position and changes the determinant by a factor of 1/a If the matrix is n n, after n of these operations the determinant has changed by a factor of (a 11 a nn 1, and the matrix is a matrix that, by point I, has determinant 1 That means that the original determinant must be a 11 a nn If one of a ii 0 I ll leave it as an exercise to show that the matrix row reduces to a matrix with at least one row equal to 0, so the determinant is 0; again the product of the diagonal entries Armed with these facts we can now confidently go out into the world and compute determinants We will do it by row reduction, but we don t need to go all the way to the identity matrix We shoot for upper triangular Let s begin with an easy example Compute det I will start by exchanging rows 1 and Since this changes the determinant by a factor of 1, we have det det

4 4 Next I subtract from row, row 1 times 4 This does not change the determinant so det det Next I add to row, row times /8 This does not change the determinant so det det /8 We have reached upper triangularity, so finally 0 4 det ( Here is a slightly harder one Are the computations right? Compute det Solution By row reduction to an upper triangular matrix: det det det Determinants and Linear Systems det ( 4 5 ( Consider a linear system of the form Ax b, where A is an n n matrix We saw it has a unique solution for every choice of the vector b if and only if we can row reduce A to the identity matrix Some row operations change the determinant of a matrix, others leave it as it is But the only change is a change in sign or multiplication by a non zero number So if the determinant is not zero, it stays not zero If at the end you have a non-zero determinant, it was non zero to begin with The conclusion is that A row reduces to the identity matrix if and only if det A 0 If you consider the row reduction method for computing inverses, it iseasy to see why the following three theorems are true: Theorem 1 The square matrix A is invertible if and only if det A 0 Theorem The linear system Ax b has a unique solution x for each b if and only if det A 0 In particular, if det A 0 and b is vector with all entries 0, we write b 0, then the only solution of Ax 0 is x 0 Theorem The linear system Ax 0 has a non-zero solution; ie, a solution not all of whose entries are 0, if and only if det 0 Another important property of determinants This property is important and probably not very obvious The only decent proofs depend on a more sophisticated approach to determinants, as done in a good undergraduate linear algebra course Theorem 4 Let A, B be square matrices of the same order Then det(ab det(a det(b

5 5 4 Laplace s expansion Here is a more precise way to define the determinant of a matrix It is also constructive I want to emphasize that the result is the same as before; this is not a new function, merely an alternative definition We begin with matrices and define a b det ad bc c d For example ( 5 7 det The definition is recursive It tells you how to define/compute an n n determinant assuming you know how to compute (n 1 (n 1 determinants For this we need some notation/terminology Suppose (a ij 1 i,j n is an n n matrix If 1 i, j n, the (i, j-th minor of A is the determinant of the (n 1 (n 1 matrix obtained by crossing out the i-th row and j-th column of A The (i, j-th minor of A will be denoted by M ij (A So assume we are at the beginning of our knowledge of determinants We know what a determinant is, so we know what the minors of a matrix are For example if , 1 0 there are 9 minors They are (check! M 11 (A det 14, M 0 1 (A det 1, M 1 1 (A det 7, M 1 (A det 4, M 0 det 1, M 1 (A det, M 1 det, M 7 1 det 1, M 0 1 (A det The determinant can now be recursively defined as follows If (a ij 1 i,j n then det(a a 11 M 11 (A a 1 M 1 (A ± +( 1 n+1 a 1n M 1n (A a 1j ( 1 j+1 M 1j (A So if we know how to compute determinants, we can use this formula to compute determinants But now, knowing how to compute determinants, we can compute 4 4 determinants And so forth For example, for the matrix A given above, we get det a 11 M 11 a 1 M 1 + a 1 M ( 7 9 As an exercise, check you get the same result when row reducing Laplace s expansion can be useful if the matrix has a lot of zero entries For small matrices it is probably just as good as row reduction; by small I mean or Since a determinant does only change in sign if one exchanges rows, it is not surprising that one does not have to use the first row to expand, as in the definition That is, one also has det(a ( a 1 M 1 (A a M (A ± +( 1 n+1 a n M n (A a j ( 1 j+ M j (A In general, any row can be used; one adds and subtracts row entries by corresponding minors, beginning with +a i1 M i1 if i is odd (rows 1,, 5, etc, with a i1 M i1 if i is even That is: It holds for i 1,, n that det(a a ij ( 1 i+j M ij (A j1 j1 j1

6 6 There is one more matrix object that we may haveto consider, the transpose of a matrix If (a ij is of order m n, the transpose of A, denoted by A T, is the n m matrix whose rows are the columns of A (and whose columns are the rows of A; that is, A T (a ji For example, if then A T ( One can prove,we ll accept it: If A is a square matrix, then, det(a T det A This means that most of what I said about rows and determinants is also true if you replace the word row by column In particular, one can expand by a column One has also It holds for j 1,, n that det(a a ij ( 1 i+j M ij (A i1 To sort of summarize, and repeat: Determinants in nutshell; maybe a coconut shell If A is an n n matrix one way of defining the determinant is by expansion by the first row This is a recursive definition, meaning that to compute a determinant of a given order you have to compute first determinants of lower order The definition is then, if (a ij is an n n matrix, det a 11 M 11 a 1 M ( 1 n+1 a 1n M 1n, where M ij M ij (A is the determinant of the (n 1 (n 1 matrix obtained by crossing out the i-th row and the j-th column of A Once one has the definition one can prove (we ll just accept it that any row can be substituted for the first one, as long as you switch signs accordingly at the front; that is, one also has in general det a 1 M 1 + a M + ( 1 n+ a n M n a 1 M 1 a M + + ( 1 n+ a n M n, det ( 1 i+j a ij M ij holds for all i, 1 i n One also sees that one can replace rows by columns; holds for all j, 1 j n det j1 ( 1 i+j a ij M ij i1 Computing a determinant by one of these expansions works probably best if the matrix is of order at most or has a lot of zero entries One then expands by the row or column with the largest number of zero entries Otherwise row reduction works best Probably Certainly from a computational point of view Computing det A by row reduction you shoot for an upper triangular matrix Starting with A you perform row operations Let us say that the first row operation changed A to A 1 You write det det A 1 if it was a row operation of the third kind, det 1 c det A 1 if the row operation consisted in multiplying a row of A by c, and det det A 1 if the row operation was an exchange of two rows Then you do the same to A 1 Once the matrix is in upper triangular form, the determinant is the product of the diagonal elements Here is an illustration Can you follow it? Figure out the row operations involved?

7 7 Compute Solution det det ( det A, where det det Since the determinant of a matrix and its transpose are the same, one can also do column operations; one can shoot for lower triangular The determinant of a lower triangular matrix is also the product of its diagonal elements 5 Another Method for Computing Inverses Suppose A is an n n square matrix The adjunct matrix of A is the matrix A whose (i, j th entry is ( 1 i+j M j,i (A (notice the reversal of subindices For example, for the matrix A whose minors we calculated above, namely , 1 0 we get (check that this is right A The main property of the adjunct matrix is that , AA A (det ai For example, you can verify that if A is the matrix displayed above, then AA ( 9I This shows that the matrix has determinant equal to 9 One then has the following theorem, which provides a relatively decent way to compute inverses of small matrices, a very inefficient way for larger matrices: Theorem 5 Assume A is a square matrix and assume det A 0 (so A is invertible Then A 1 1 det A A For example, suppose we wnt to compute A 1, where ( Well, there is such a thing as a 1 1 matrix; it is simply a number in parentheses: (a The determinant is just the number, det(a a So the minors of this matrix are M 1 1(A 7, M 1 (A 5, M 1 (A, M (A 1

8 8 Remembering 1 that we have to multiply them by ( 1 i+j, which means thathalf of them, alternatingly, get their signs changed, we get A We also have det (1( 7 ((5 so that A 1 1 ( One calls ( 1 i+j M ij (A the (i, j-th cofactor of A The adjunct can the be defined as the transpose of the cofactor matrix