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if A=[[1,2],[2,3]] and A^2-kA-I2=0 then ...

if `A=[[1,2],[2,3]]` and `A^2-kA-I_2=0` then the value of `k` is

A

4

B

-4

C

8

D

-8

Text Solution

Verified by Experts

The correct Answer is:
A

`A=[{:(1,2),(2,3):}]`
`A^(2)=[{:(1,2),(2,3):}][{:(1,2),(2,3):}]=[{:(1+4,2+6),(2+6,4+9):}]=[{:(5,8),(8,13):}]`
`A^(2)-kA - I_(2)=[{:(5,8),(8,13):}]-[{:(k,2k),(2k,3k):}]-[{:(1,0),(0,1):}]`
`=[{:(5-k-1,8-2k-0),(8-2k-0,12-3k-1):}]`
`=[{:(4-k,8-2k),(8-2k,12-3k):}]`
Given, `A^(2) - kA - I_(2) = 0`
`therefore 4 - k = 0 rArr k = 4`
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