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If `{:A=[(-4,-1),(3,1)]:}`, then the determint of the matrix `(A^2016-2A^2015-A^2014)`,isA. 2014B. 2016C. `-175`D. `-25` |
Answer» Correct Answer - B Let `B=A^2016-2A^2015-A^2014`. Then, `B=A^2014(A^2-2A-1)` `rArr absB=absA^2014abs(A^2-2A-1)` Now,`A=[(-4,-1),(3,1)]` `:.{:A^2-2A-I=[(-4,-1),(3,1)][(-4,-1),(3,1)]-2[(-4,-1),(3,1)]-[(1,0),(0,1)]:}` `{:= [(13,3),(-9,-2)]+[(8,2),(-6,-2)]+[(-1,0),(0,-1)]=[(20,5),(-15,-5)]:}` `rArrabs(A^2-2A-I)={:[(20,5),(-15,-5)]=-25:}` `and absA={:abs((-4,-1),(3,1))=-1:}` Hence,`absB=(-1)^2014=-25` |
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