2.2 The Inverse of a Matrix

Theorem 5 proof Take any b in Rn. A solution exists because A^-1b is substituted for x,

then Ax=A(A^-1b) =(AA^-1)b =Ib =b so A-1b is a solution.

an nxn matrix A is said to be invertible if...

there is an nxn matrix C such that CA = I and AC = I where I=In the nxn identity matrix

An invertible matrix A is row equivalent to an identity matrix and we can find A^-1 by

watching the row reduction of A to I

How to know if A does NOT have an inverse

1. IF [A I] is NOT row equivalent to [I A^-1] 2.

An elementary matrix is

A matrix that is obtained by performing a single elementary row operation on an identity matrix Ex: E1 = [], E2 = [], E3 = [],....

Theorem 7

An nxn matrix A is invertible if and only if A is row equivalent to In, and in this case, any sequence of elementary row operations that reduces A to In also transforms In into A^-1

CA = I and AC = I in this case, C is an inverse of A. C is uniquely determined by A because if B were another inverse of A, then B = ...????

BI = B(AC) = (BA)C = IC = C This unique inverse is denoted by A^-1, so that A^-1A = I and AA^-1 = I

Algorithm for finding A^-1

Row reduce the augmented matrix [A I]. If A is row equivalent to I, then [A I] is row equivalent to [I A^-1]. Otherwise, A does not have an inverse

so A-1b is a solution. Prove the solution is unique

Show that if u is any solution, then u, infact, must be A^-1b. Indeed, if Au=b, we can multiply both sides by A^-1 and obtain A^-1Au=A^-1b Iu=A^-1b and u=A^-1b

True/False Elementary matrices are invertible

True. Since row operation are invertible, if E is produced by a row operation on I, then there is another row operation of th same type that changes E back into I. Hence there is an elementary matrix F such that FE=I. since E and F correspond to reverse operations, EF=I too

Theorem 4

a 2x2 matrix A is invertible if and only if the determinant A does not equal 0

A matrix that is invertible is called

a nonsingular matrix

A matrix that is NOT invertible is sometimes called

a singular matrix

Theorem 6 proof

a) find a matrix C (aka A^-1) such that A^-1C=I and CA^-1=I These equations are satisfied with A in place of C. Hence, A^-1 is invertible, and A is its invrese. b) Compute (AB)(B^-1A^-1)=A(BB^-1)A^-1=AIA^-1 =AA^-1 = I Also, (B^-1A^-1)(AB)=I c) Use Theorem 3(d), from left to right, (A^-1)^TA^T=(AA^-1)^T=I^T=I Similarly, A^T(A^-1)^T = I^T = I . Hence, A^T is invertible, and its inverse is (A^-1)T

Theorem 6

a. If A is an invertible matrix, then A^-1 is invertible and (A^-1)^-1=A b. If A and B are nxn matrices, then so is AB, and the inverse of AB is the product of the inverses of A and B in the reverse order. That is, (AB)^-1=B^-1A^-1 c. If A is an invertible matrix, then so is A^T, and the inverse A^T is the transpose of A^-1. That is, (A^T)^-1=(A^-1)^T

Theorem 5

if A is an invertible nxn matrix, then for each b in Rn, the equation Ax=b has the unique solution x=A^-1b *Review proof

Theorem 7 proof

just read it...

If an elementary row operation is performed on an mxn matrix A, the resulting matrix can be written as EA, where the mxm matrix E is created by....

performing the same row operation on Im

The inverse of the elementary matrix E is...

the elementary matrix of the same type that transforms E back into I

Definition: The product of nxn invertible matrices is invertible, and the inverse is

the product of their inverses in reverse order