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Question
If A = `[[4 2],[-1 1]]`
, prove that (A − 2I) (A − 3I) = O
Solution
Given: (A−2 I)(A−3 I)
⇒ (A−2I)(A−3I)= `([[4 2],[-1 1]]-2[[1 0],[-1 1]])` `([[4 2],[-1 1]]-3[[1 0],[0 1]])`
⇒ (A−2I)(A−3I)=`([[4 2],[-1 1]]-[[2 0],[0 2]])` `([[4 2],[-1 1]]-[[3 0],[0 3]])`
⇒ (A−2I)(A−3I)= `[[4-2 2-0],[-1-0 1-2]] [[4-3 2-0],[-1-0 1-3]]`
⇒ (A−2I)(A−3I)=`[[2 2],[-1 -1]],[[1 2],[-1 -2]]`
⇒ (A−2I)(A−3I)= `[[2-2 4-4],[-1+1 -2+2]]`
⇒ (A−2I) A−3I)= `[[0 0],[0 0]]`
⇒ (A−2I)(A−3I) =0
Hence proved.
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