MAT223H1 Final: MAT223H1 - Final Exam - Part 1 of 2
Document Summary
If there is a matrix b st. ba=i an inverse of a multiply b on the left matrix then: = . A is invertible if there is a matrix b st. ab=i and ba=i. We call b the inverse of a, devoted by 1. If a, b are 2x2 matrices st. ab=0. Then a and b are both not invertible. Proof: if a is invertible then there is a c st. ca=i (0) = () If a is a 2x3 matrix then b has to be 3x2 matrix. If ab=0, then b=0 (where 0 is the zero matrix) 1() = 1(: let be any vector, then the system = has a unique solution. Reduce a to its rre form: []~[] (identity matrix) Cannot be reduced to 2 (contains a zero row) 1 does not exist (a is not invertible) It has no non trivial solutions (no parameters / free variables)
