MATH 304 Linear Algebra Lecture 7 Inverse matrix continued Identity matrix Definition The identity matrix or unit matrix is a diagonal matrix with all diagonal entries equal to 1 The n n identity matrix is denoted In or simply I 1 0 0 1 0 I1 1 I2 I3 0 1 0 0 1 0 0 1 1 0 0 0 1 0 In general I 0 0 1 Theorem Let A be an arbitrary m n matrix Then Im A AIn A Inverse matrix Definition Let A be an n n matrix The inverse of A is an n n matrix denoted A 1 such that AA 1 A 1A I If A 1 exists then the matrix A is called invertible Otherwise A is called singular Let A and B be n n matrices If A is invertible then we can divide B by A left division A 1B right division BA 1 Remark There is no notation for the matrix division and the notion is not used often Basic properties of inverse matrices If B A 1 then A B 1 In other words if A is invertible so is A 1 and A A 1 1 The inverse matrix if it exists is unique Moreover if AB CA I for some n n matrices B and C then B C A 1 Indeed B IB CA B C AB CI C If n n matrices A and B are invertible so is AB and AB 1 B 1A 1 B 1 A 1 AB B 1 A 1 A B B 1 IB B 1 B I AB B 1 A 1 A BB 1 A 1 AIA 1 AA 1 I 1 1 Similarly A1A2 Ak 1 A 1 k A2 A1 Inverting diagonal matrices Theorem A diagonal matrix D diag d1 dn is invertible if and only if all diagonal entries are nonzero di 6 0 for 1 i n If D is invertible then D 1 diag d1 1 dn 1 1 d1 0 0 0 d 0 2 0 0 dn d1 1 0 0 0 d2 1 0 0 0 1 dn Inverting 2 2 matrices Definition The determinant of a 2 2 matrix a b A is det A ad bc c d a b Theorem A matrix A is invertible if c d and only if det A 6 0 If det A 6 0 then 1 1 a b d b c d a ad bc c Problem Solve a system 4x 3y 5 3x 2y 1 This system is equivalent to a matrix equation Ax b 4 3 x 5 where A x b 3 2 y 1 We have det A 1 6 0 Hence A is invertible Ax b A 1 Ax A 1 b A 1 A x A 1 b x A 1 b Conversely x A 1 b Ax A A 1 b AA 1 b b 1 1 4 3 5 x 2 3 5 13 y 3 2 1 4 1 19 1 3 System of n linear equations in n variables a11x1 a12x2 a1n xn b1 a21x1 a22x2 a2n xn b2 Ax b an1 x1 an2 x2 ann xn bn where a11 a12 a a 21 22 A an1 an2 a1n a2n ann b1 x1 b x 2 2 x b bn xn Theorem If the matrix A is invertible then the system has a unique solution which is x A 1b General results on inverse matrices Theorem 1 Given a square matrix A the following are equivalent i A is invertible ii x 0 is the only solution of the matrix equation Ax 0 iii the row echelon form of A has no zero rows iv the reduced row echelon form of A is the identity matrix Theorem 2 Suppose that a sequence of elementary row operations converts a matrix A into the identity matrix Then the same sequence of operations converts the identity matrix into the inverse matrix A 1 Theorem 3 For any n n matrices A and B BA I AB I Row echelon form of a square matrix invertible case noninvertible case 3 2 0 Example A 1 0 1 2 3 0 To check whether A is invertible we convert it to row echelon form Interchange the 1st row with the 2nd row 1 0 1 3 2 0 2 3 0 Add 3 times the 1st row to the 2nd row 1 0 1 0 2 3 2 3 0 Add 2 times the 1st row to the 3rd row 1 0 1 0 2 3 0 3 2 Multiply the 2nd row by 0 5 1 0 1 0 1 1 5 0 3 2 Add 1 0 0 3 times the 2nd row to the 3rd row 0 1 1 1 5 0 2 5 Multiply the 3rd row by 0 4 1 0 1 0 1 1 5 0 0 1 We already know that the matrix A is invertible Let s proceed towards reduced row echelon form Add 1 0 0 1 5 times the 3rd row to the 2nd row 0 1 1 0 0 1 Add 1 0 0 1 times the 3rd row to the 1st row 0 0 1 0 0 1 To obtain A 1 we need to apply the following sequence of elementary row operations to the identity matrix interchange the 1st row with the 2nd row add 3 times the 1st row to the 2nd row add 2 times the 1st row to the 3rd row multiply the 2nd row by 0 5 add 3 times the 2nd row to the 3rd row multiply the 3rd row by 0 4 add 1 5 times the 3rd row to the 2nd row add 1 times the 3rd row to the 1st row A convenient way to compute the inverse matrix A 1 is to merge the matrices A and I into one 3 6 matrix A I and apply elementary row operations to this new matrix 1 0 0 3 2 0 A 1 0 1 I 0 1 0 2 3 0 0 0 1 3 2 0 1 0 0 A I 1 0 1 0 1 0 2 3 0 0 0 1 3 2 0 1 0 0 1 0 1 0 1 0 2 3 0 0 0 1 Interchange the 1st 1 0 1 0 1 3 2 0 1 0 2 3 0 0 0 row with the 2nd row 0 0 1 Add 3 times the 1st row to the 2nd row 1 0 1 0 1 0 0 2 3 1 3 0 2 3 0 0 0 1 Add 2 times the 1st row to the 3rd row 1 0 1 0 1 0 0 2 3 1 3 0 0 3 2 0 2 1 Multiply the 2nd row by 0 5 1 0 1 0 1 0 0 1 1 5 0 5 1 5 0 0 3 2 0 2 1 Add 1 0 0 3 times the 2nd row 0 1 0 1 1 1 5 0 5 …
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