Question

Q5. Consider the square matrix A - 6 4 3 (a) Show that the characteristic polynomial of A# (X) = x-91-2. (6 pts) (b) Compute the matrix B-A 9A 21. (5 pts) (c) Show that A2 9A-21, for the given matrix A. (5 pts) (d) Is it possible to use the equation A? (Justify your answer) (5 pts) 9A 21, to incl the inverse of the given matrix A

Answer 1

Solutions) The given matrix is A = -5 6 -4 Solution (a) The characteristic equation of A is det (A - x [2) → det o 6:33 - 1% 6 4 1 = 0 6 -5. . 0 » det = 0 + [ -4 3 0

6-1 → det - 5 +0 - 4+0 3-1 6-1 - 5 ㅋ =0 - 4 3-) 20 = 0 6-1)(3-1)-2 Ź 18 – 31 - 61+12-20 = 0 学 12_91-2 = 0 There fore characteristic equation of A is 1291-2 = 0 Answer solution (b) Given that 6 5 B = A? 9 A - 2 Iz, cohere A = Then 6 A2 6 -51 4 ; L-4 36 62+20 - 30 -15 => A²= 1-24-12 20+) 56 -45 → A2= 1-36 29

Therefore 56 0 -45 6 b 2 BE -36 29 در -4 O 2 |-54 45 56 - 45 + + -2 O → B = 29 -36 36 -27 56-54 - 2 -45 +45 7 B = -36 + 36 29-27-2 B [:: ? Answer O Therefore A2-9 A - 212 [ = (1) Solution (C) From (1), we get A2-9 A-2 I2 = [: [::] +2I2 → A2-9 A = > A²_9A = 2 Iz Answer)

solution (d) Given that A? - 9 A = 2 I2 by A → A2. AT-9 A- A' = 2I, A Mltiplying to both sides A (A_A"- 9A. AY= 2(12.A) since → A. I2 – 9 IZ = 2 A A-A=I2 and Iz à = A -9 I2 = 2 A since A•Iz =A Since 6 -5 A= { (A-912) = 21 + A' = ţ (A-912) a n-6 -51-10 À = $ 33+[:-)] » A' = 6-9 -5+0 5] 4 3 9 4 -4+0 -3 - 介 -4 - 3 om at MIN ㅋ 올

3. 2 이어 2 2 - 3 (Answer