082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

282 Chapter 8 Matrix algebra
(3) Finally, the inverse ofAisgiven by
A −1 = adj(A)
|A|
Example8.30 Findthe inverse of
⎛ ⎞
1−20
A = ⎝ 3 15⎠
−1 23
⎛ ⎞
13−1
Solution A T = ⎝−21 2⎠
05 3
ReplacingeachelementofA T byits cofactor, wefind
⎛ ⎞
−7 6 −10
adj(A) = ⎝−14 3 −5⎠
70 7
ThedeterminantofAisgiven by
|A|=1
∣ 1 5
∣ ∣ ∣∣∣ 2 3∣ − (−2) 3 5
∣∣∣ −1 3∣ +0 3 1
−1 2∣
Therefore,
= (1)(−7) + (2)(14)
= 21
A −1 = adj(A)
|A|
⎛ ⎞
= 1 −7 6 −10
⎝−14 3 −5⎠
21
70 7
Note thatthis solution should be checked by formingAA −1 togiveI.
It is clear that should |A| = 0 then no inverse will exist since then the quantity 1/|A| is
undefined.Recallthatsuchamatrix issaidtobesingular.
Forany squarematrixA, the following statements are equivalent:
|A|=0
Aissingular
Ahas no inverse