![Inverse using CH Theorem
Step 1: For a Given Non-singular Square Matrix ‘A’ of order ‘n’, Find its Characteristic equation
|A-λIn| = 0.
Expand the Determinant such that it is reduced in the below format:
λn + a1λn-1 + a2λn-2 + . . . + anIn = 0, where a1, a2, . . . , an are real Constants.
Step 2: As per Cayley-Hamilton’s theorem, the above Characteristic equation of ‘A’ is satisfied
by itself, Hence:
An + a1An-1 + a2An-2 + . . . + anIn = 0
Step 3: Multiplying by A-1 on Both Sides of the above Equation reduces it to:
An-1 + a1An-2 + a2An-3 + . . . an-1In + anA-1 = 0
Step 4: Find A-1 by simplifying and reordering the terms of the above equation, then A-1 is:
A-1 =
−1
an
[ An-1 + a1An-2 + a1An-3 + . . . An-1In ]](https://image.slidesharecdn.com/inverseusingch-230516034444-bd0ec0bf/85/inverse-using-CH-pptx-1-320.jpg)
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inverse using CH.pptx
- 1. Inverse using CH Theorem Step 1: For a Given Non-singular Square Matrix ‘A’ of order ‘n’, Find its Characteristic equation |A-λIn| = 0. Expand the Determinant such that it is reduced in the below format: λn + a1λn-1 + a2λn-2 + . . . + anIn = 0, where a1, a2, . . . , an are real Constants. Step 2: As per Cayley-Hamilton’s theorem, the above Characteristic equation of ‘A’ is satisfied by itself, Hence: An + a1An-1 + a2An-2 + . . . + anIn = 0 Step 3: Multiplying by A-1 on Both Sides of the above Equation reduces it to: An-1 + a1An-2 + a2An-3 + . . . an-1In + anA-1 = 0 Step 4: Find A-1 by simplifying and reordering the terms of the above equation, then A-1 is: A-1 = −1 an [ An-1 + a1An-2 + a1An-3 + . . . An-1In ]