Prove that : A2−6A−17I2=0, where A= [23−34] and hence if det(A−1)=m1.Find m

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Prove that : A^2 - 6A - 17I_2 = 0, where A = {bmatrix} 2 & -3

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Prove that : $A^{2} - 6 A - 17 I_{2} = 0$ , where $A = [23 - 3 4]$ and hence if det $(A^{- 1}) = \frac{1}{m}$ .Find $m$

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Benjamin Discussed

Prove that :

A^{2} - 6 A - 17 I_{2} = 0

, where

A = [23 - 3 4]

and hence if det

(A^{- 1}) = \frac{1}{m}

.Find

m

14 mins ago

Discuss this question LIVE

14 mins ago

A = [23 - 3 4] A^{2} = [23 - 3 4] \times [23 - 3 4] = [4 - 9 6 + 12 - 6 - 12 - 9 + 16] = [- 5 18 - 18 7] 6 A = [1218 - 18 24] A^{2} - 6 A + 17 I_{2} = [- 5 - 12 + 17 18 - 18 - 18 + 18 7 - 24 + 17]

= [0000] = 0

[Proved]

A^{- 1} = \frac{1}{17} [43 - 3 2]^{T} = \frac{1}{17} [4 - 3 32] d e t (A^{- 1}) = \frac{8 + 9}{17 \times 17} = \frac{1}{17}

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Question Text	Prove that : $A^{2} - 6 A - 17 I_{2} = 0$ , where $A = [23 - 3 4]$ and hence if det $(A^{- 1}) = \frac{1}{m}$ .Find $m$
Answer Type	Text solution:1
Upvotes	150