<p> Section 4 Inverse Matrix </p><p>1. Definition:</p><p>Definition of inverse matrix: An n  n matrix A is called nonsingular or invertible if there exists an n  n matrix B such that </p><p>, AB  BA  I n</p><p> n  n where I n is a identity matrix. The matrix B is called an inverse of A. If there exists no such matrix B, then A is called singular or noninvertible. is called a odd permutation. </p><p>Theorem: If A is an invertible matrix, then its inverse is unique. </p><p>[proof:] Suppose B and C are inverses of A. Then,</p><p>BA  CA  I n  B  BI n  B(AC)  (BA)C  I nC  C .</p><p>Note: Since the inverse of a nonsingular matrix A is unique, we denoted the inverse of A as A1 .</p><p>Note: If A is not a square matrix, then  there might be more than one matrix L such that LA  I (or AL  I) .  there might be some matrix U such that </p><p>1 UA  I but AU  I</p><p>Example:</p><p>Let </p><p> 1 1    A  1 0  .  3 1</p><p>Then,  there are infinite number of matrices L such that LA  I , for example 1 3 1 4 15 4 L    or L    . 2 5 1 7 25 6 1 3 1  As L    , 2 5 1</p><p> 3 8 2    LA  I but AL  1  3 1  0 .  1 4 2 </p><p>2. Computation of A1 :</p><p>1. Using Gauss-Jordan reduction:</p><p>The procedure for computing the inverse of a n  n matrix A:</p><p>1. Form the n  2n augmented matrix </p><p>2 a11 a12 ⋯ a1n ⋮ 1 0 ⋯ 0   a21 a22 ⋯ a2n ⋮ 0 1 ⋯ 0 A ⋮ I     n  ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋮ ⋱ ⋮   an1 an2 ⋯ ann ⋮ 0 0 ⋯ 1 and transform the augmented matrix to the matrix C ⋮ D in reduced row echelon form via elementary row operations. </p><p>2. If 1 (a) C  I n , then A  D . 1 (b) C  I n , then A is singular and A does not exist. </p><p>Example:</p><p> 1 1  2   To find the inverse of A   2  3  5 , we can employ the procedure 1 3 5  introduced above. </p><p>1.</p><p> 1 1  2 ⋮ 1 0 0  ⋮   2  3  5 0 1 0 . 1 3 5 ⋮ 0 0 1</p><p>1 1  2 ⋮ 1 0 0 (3)(3)(1)   ⋮ (2)(2)2*(1)  0 1 1  2 1 0 0 2 3 ⋮ 1 0 1</p><p>1 1  2 ⋮ 1 0 0   (2)1*(2) ⋮  0 1 1 2 1 0 0 2 3 ⋮ 1 0 1</p><p>3 1 0 1 ⋮ 3 1 0 (1)(1)(2)   ⋮ (3)(3)2*(2)  0 1 1 2 1 0 0 0 1 ⋮  3 2 1</p><p>1 0 0 ⋮ 0 1 1  (1)(1)(3)   ⋮ (2)(2)(3) 0 1 0 5  3 1 0 0 1 ⋮  3 2 1 </p><p>2. The inverse of A is </p><p> 0 1 1    5  3 1 .  3 2 1</p><p>Example:</p><p>1 1 1   Find the inverse of A  0 2 3 if it exists. 5 5 1</p><p>[solution:]</p><p>1. Form the augmented matrix  1 1  2 ⋮ 1 0 0 A | I   2  3  5 ⋮ 0 1 0  3    . 1 3 5 ⋮ 0 0 1 And the transformed matrix in reduced row echelon form is 1 0 0 ⋮ 13/8 1/ 2 1/8  ⋮  0 1 0 15/ 8 1/ 2 3/ 8  0 0 1 ⋮ 5/ 4 0 1/ 4</p><p>2. The inverse of A is </p><p>4  13/8 1/ 2 1/8   15/8 1/ 2 3/8  .  5/ 4 0 1/ 4</p><p>Example:</p><p>1 2  3   Find the inverse of A  1  2 1  if it exists. 5  2  3</p><p>[solution:]</p><p>1. Form the augmented matrix 1 2  3 ⋮ 1 0 0 A | I  1  2 1 ⋮ 0 1 0  3    . 5  2  3 ⋮ 0 0 1 And the transformed matrix in reduced row echelon form is 1 0 1 ⋮ 1/ 2 1/ 2 0  ⋮  0 1 1 1/ 4 1/ 4 0 0 0 0 ⋮  2  3 1</p><p>2. A is singular!!</p><p>2. Using the adjoint adj(A) of a matrix:</p><p>As det(A)  0 , then</p><p>1 adj(A) A  . det(A) Note: adj(A)A  det(A)I n is always true.</p><p>5 Note: As det(A)  0  A is nonsingular.</p><p>3. Properties of The Inverse Matrix:</p><p>The inverse matrix of an n  n nonsingular matrix A has the following important properties:</p><p>1 1. A1   A .</p><p>1 t 1. At   A1  2. If A is symmetric, So is its inverse. 3. AB1  B1 A1 4. If C is an invertible matrix, then  AC  BC  A  B.  CA  CB  A  B . 1 5. As A  I  exists, then </p><p>I  A  A2 ⋯  An1  An  I A  I 1  A  I 1 An  I .</p><p>[proof of 2]</p><p> t t A1  At  AA1   I t  I similarly, </p><p> t t At A1   A1A  I t  I .</p><p>[proof of 3:]</p><p>By property 2, </p><p>6 t 1 A1   At   A1 .</p><p>[proof of 4:]</p><p>1 1 1 1 1 B A AB  B A AB  B IB  I . Similarly, 1 1 1 1 1 ABB A  ABB A  AIA  I .</p><p>[proof of 5:]</p><p>Multiplied by the inverse of C, then ACC 1  AI  A  BCC 1  BI  B . Similarly, C 1CA  IA  A  C 1CB  IB  B .</p><p>[proof of 6:]</p><p>I  A  A2 ⋯ An1 A  I   A  A2 ⋯ An  I  A  A2 ⋯ An1   An  I . Multiplied by A  I 1 on both sides, we have </p><p>1 A  A2  ⋯ An1  An  I A  I 1 . I  A  A2 ⋯ An1  A  I 1 An  I  can be obtained by using similar procedure. </p><p>Example:</p><p>Prove that I  AB1  I  AI  BA1 B . [proof:]</p><p>7 I  AI  BA1 BI  AB  I  AB  AI  BA1 B  AI  BA1 BAB  I  AB  AI  BA1  I  BA1 BAB  I  AB  AI  BA1 I  BAB  I  AB  AIB  I  AB  AB  I</p><p>Similar procedure can be used to obtain I  ABI  AI  BA1 B I .</p><p>4. Left and Right Inverses:</p><p>Definition of left inverse: For a matrix A, LA  I but AL  I , with more than one such L. Then, the matrices L are called left inverse of A. </p><p>Definition of right inverse: For a matrix A, AR  I but RA  I , with more than one such R. Then, the matrices R are called left inverse of A. </p><p>Theorem:</p><p> r  c A matrix Arc has left inverses only if r  c .</p><p>[proof:]</p><p>We prove that a contradictory result can be obtained as r  c and</p><p>8 Arc having a left inverse. For r  c , let </p><p>Arc  X rr Yr(cr) </p><p>Then, suppose</p><p> M rr  Lcr    N(cr)r  is the left inverse of Arc . Then,</p><p> M rr  Lcr Arc   X rr Yr(cr)  N(cr)r  . MX MY  Irr 0      Icc     NX NY   0 I(cr)(cr)  Thus, MX  I, MY  0, NX  0, NY  I.</p><p>Since MX  I and both M and X are square matrices, then M  X 1 . Therefore, MY  X 1Y  0 multiplied by X  XX 1Y  Y  X 0  0 . However, NY  N0  0  I . </p><p>It is contradictory. Therefore, as r  c , Arc has no left inverse.</p><p>9 Theorem:</p><p> r  c A matrix Arc has left inverses only if r  c .</p><p>10</p>