1.

For the matrix `A= [[3,2],[1,1]]`, find the numbers a and b such that `A^2+aA+bI=0.`

Answer» `A = [[3,2],[1,1]]`
`:. A^2 = [[3,2],[1,1]] [[3,2],[1,1]] = [[9+2,6+2],[3+1,2+1]] = [[11,8],[4,3]]`
`aA = a [[3,2],[1,1]] = [[3a,2a],[a,a]]`
`bI = b[[1,0],[0,1]] = [[b,0],[0,b]]`
`:. A^2-aA+bI = [[11,8],[4,3]]+[[3a,2a],[a,a]]+ [[b,0],[0,b]] `
It is given that `A^2-aA+bI = 0`
`:. [[11,8],[4,3]]+[[3a,2a],[a,a]]+ [[b,0],[0,b]] = [[0,0],[0,0]]`
`=>11+3a+b = 0 => 3a+b = -11->(1)`
`=>3+a+b = 0 => a = -b - 3`
Putting this value of `a` in (1),
`3(-b-3)+b = -11`
`=> - 3b -9 +b = -11`
`=>-2b = -2 => b = 1`
`:. a = -1-3 = -4`
`:. a = -4 and b = 1`


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