Applied Matrix Algebra Lecture Notes Section 2.2. Gerald Höhn Department of Mathematics, Kansas State University

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1 Applied Matrix Algebra Lecture Notes Section 22 Gerald Höhn Department of Mathematics, Kansas State University September, 216

2 Chapter 2 Matrices 22 Inverses Let (S) a 11 x 1 + a 12 x 2 + +a 1n x n = b 1 a m1 x 1 + a m2 x 2 + +a mn x n = b m be a linear system Then (S) is equivalent to the matrix equation Ax = b where a 11 a 1n x 1 b 1 A =, x =, b = a m1 a mn x n b m How to solve it with matrix algebra? We like x = A 1 b Does this makes sense? For real numbers a there exists a unique x such that ax = xa = 1 Namely, x = 1 a = a 1, the multiplicative inverse of a Definition 221 (Inverse) A square matrix A of size n is called invertible (or nonsingular) if there exists a square matrix X of size n such that AX = I n and XA = I n The matrix X is called an inverse for A If no such matrix exists, A is called singular (or noninvertible) Examples: O 1 1 = [] or more generally O = O n n is singular since OX = O I n for any X 1

3 has the inverse A = is singular since AX = I 2 would imply AAX = AI 2 and so OX = A or O = A which is false Theorem 222 If the inverse X of a square matrix A exists it is uniquely determined Proof: Suppose X and Y are both inverses for A Then AX = I n and YA = I n We get Y = YI n = Y(AX) = (YA)X = I n X = X Notation: The unique inverse of an invertible matrix A is denoted by A 1 If A is singular we say A 1 does not exist Theorem 223 If for the square matrix A there is a unique matrix X with AX = I n (a right inverse) then XA = I n and A is invertible with inverse A 1 = X Proof: Let Y = XA+X I n Then AY = AXA+AX AI n = I n A+I n A = I n Since X is by assumption the unique matrix with that property we have X = Y = XA+X I n and so XA = I n Remark 224 We will see later that the same result holds without the assumption that X is unique How to compute A 1 if it exists? Let e i = 1 n 1 for i = 1,, n, where the only entry 1 is in row i Then I n = [e 1 e 2 e n ], where [e 1 e 2 e n ] denotes the juxtaposition of the column vectors e i Let x i be the solution of the equation Ax i = e i for i = 1,, n Then A[x 1 x 2 x n ] = [e 1 e 2 e n ] = I n, ie X = [x 1 x 2 x n ] is a right inverse of A Assume now ranka = n To solve the equation Ax i = e i for all i, we consider the augmented matrix M = [A I n ] and bring it into the reduced row echelon form M Since 2

4 ranka = n one has M = 1 1 x 1 x 2 x n = [I n X] We also see that x i is the unique solution of the equation Ax i = e i Thus X is the unique solution of AX = I n and hence X = A 1 This shows: Theorem 225 If A is an n n matrix with ranka = n then A is invertible and the reduced row echelon form of [A I n ] is [I n A 1 ] This gives a practical way to compute the inverse /2 1/2 Example: Let A = Then [A I 2 ] = /2 5/ /2 1/ Thus A = 5 2 The used elementary row operation were: 1 2 r 1 r 1, 5r 1 + r 2 r 2, 2r 2 r 2 and 1 2 r 2 +r 1 r 1 What if ranka < n? Theorem 226 An n n matrix A is invertible if and only if ranka = n Proof : If ranka < n, the last row of the row echelon form of A of A is identical zero (< n pivot elements) The row operation reducing A to A reduces [A I n ] to [A B] with a matrix B Since ranki n = n, also rankb = n Hence [A B] = p with a nonzero entry p in a certain column j of the last row of B It follows that the linear system Ax j = e j is inconsistent because ranka < rank[a e j ] Thus AX = I n has no solution and therefore A is not invertible Together with the previous theorem we are done Theorem 227 Let A and B square matrices of the same size and s be a nonzero scalar Then: (P)1 If A is invertible, then A 1 is invertible and (A 1 ) 1 = A (P2) If A and B are invertible, then AB is invertible and (AB) 1 = B 1 A 1 3

5 (P3) If A is invertible, then sa is invertible and (sa) 1 = s 1 A 1 (P4) A is invertible if and only if A t is invertible and (A t ) 1 = (A 1 ) t Proof: (P1) If A is invertible, then AA 1 = I n and A 1 A = I n, ie A is an inverse for A 1 (P2) (AB)(B 1 A 1 ) = A(BB 1 )A 1 = AI n A 1 = AA 1 = I n and similarly (B 1 A 1 )(AB) = I n, ie B 1 A 1 is the inverse for AB (P3) (sa)(s 1 A 1 ) = (s s 1 )(AA 1 ) = I n = (s 1 A 1 )(sa), ie s 1 A 1 is the inverse of sa (P4) If A is invertible then AA 1 = A 1 A = I n Transposing gives (A 1 ) t A t = A t (A 1 ) t = I t n = I n, ie (A 1 ) t is the inverse of A t Similarly, using (A t ) t = A it follows that with A t also A is invertible The following theorem can be checked easily by a direct calculation a b Theorem 228 Let A = be a 2 2 matrix Then A is invertible if and only c d if the determinant det(a) := ad bc is non-zero In this case A 1 1 d b = det(a) c a Example: A = [ Elementary Row Operations ], det(a) = = 1, and hence A 1 = [ Definition 229 An elementary matrix E is the result of a single elementary row operation applied to I n Examples: There are three types of elementary row operations We list examples for each type for n = 3 r 2 r 3 E 1 = 1 One easily checks: 5r 2 r 2 E 2 = r 3 +2r 2 r 3 E 3 = ] 4

6 Theorem 221 Performing a single row operation on an m n matrix A is the same as multiplying A from the left with the corresponding elementary matrix E: A EA Example: We apply the elementary row operation r 3 +2r 2 r 3 to the matrix A = 3 4 B = 3 4 On the other hand 5 6 E 3 A = = B Corollary 2211 The reduced row echelon form A of a matrix A is obtained by multiplying A from the left with a product F = E k E 2 E 1 of elementary matrices E 1, E 2,, E k Theorem 2212 Every elementary matrix E is invertible and the inverse E 1 is again an elementary matrix Proof: Let F be the elementary matrix corresponding to the inverse elementary row operation corresponding to E Then FE = I n and also EF = I n Example: The elementary row operation r 3 + 2r 2 r 3 has the inverse elementary operation r 3 2r 2 r 3 corresponding two the two matrices E 3 = 1 and F = Theorem 2213 Every invertible matrix A is the product of elementary matrices Proof: Since A is invertible A has rank n and therefore A = I n By the Corollary, there exist elementary matrices E 1, E 2,, E k such that I n = E k E 2 E 1 A Hence A = E 1 1 E 1 2 E 1 k and the result follows from the previous theorem 2 1 Example: A = From the first example of this section, we find / /2 I 2 = A and so A = [ 2 1 ][ ][ 1 1/2 ][ 1 1/2 1 ] 5