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Question
\[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\]show that A2 − 5A + 7I = O use this to find A4.
Solution
Given : A=`[[ 3,1],[-1,2]]`
`⇒A2= A A`
`⇒A^2=``[[ 3,1],[-1,2]]` `[[ 3,1],[-1,2]]`
`⇒A^2`=`[[9-1,3+2],[-3-2,-1+4]]`
`⇒A^2= ``[[8,5],[-5,3]]`
`A^2−5A+7I`
`⇒A^2−5A+7I=``[[8,5],[-5,3]]-5[[3,1],[-1,2]]+7[[1,0],[0,1]]`
`⇒A^2−5A+7I=` `[[8,5],[-5,3]]-[[15,5],[-5,10]]+[[7,0],[0,7]]`
`⇒A^2−5A+7I=` `[[8-15+7,5-5+0],[-5+5+0,3-10+7]]`
`⇒A^2−5A+7I=` `[[0,0],[0,0]]=0`
Hence proved.
Given: `A^2−5A+7I=0`
`⇒A^2=5A−7I` ...(1)
`⇒A^3=A(5A−7I)` (Multilpying by A on both sides)
`⇒A^3=5A^2−7AI`
`⇒A^3=5(5A−7I)−7A` [From eq. (1)]
`⇒A^3=25A−35I−7A`
`⇒A^3=18A−35I`
`⇒A^4=(18A−35I)`A (Multilpying by A on both sides)
`⇒A^4=18A^2−35A`
`⇒A^4=18(5A−7I)−35A` [From eq. (1)]
`⇒A^4=90A−126I−35A`
`⇒A^4=55A−126I`
`⇒A^4=55``[[3,1],[-1,2]]-126` `[[1,0],[0,1]]`
`⇒A^4= ``[[165,55],[-55,110]]-[[126,0],[0,126]]`
`⇒A^4=``[[165-126,55-0],[-55-0,110-126]]`
`⇒A4= ``[[39,55],[-55,-16]]`
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