Question #162152

Suppose that a 4 × 4 matrix A has eigenvalues λ1 = 1, λ2 = − 4, λ3 = 6, and λ4 = − 6. Use the following method to find det (A).

If

p(λ) = det (λI − A) = λn + c1λn − 1 + ⋯ + cn

So, on setting λ = 0, we obtain that

det (− A) = cn or det (A) = (− 1)ncn

det (A) = ?


1
Expert's answer
2021-02-24T07:56:47-0500

Solution: λ1=1, λ2=4, λ3=6, λ4=6det(λIA)=λn+c1λn1+c2λn2+....+cnHere n=4, det(λIA)=λ4+c1λ3+c2λ2+c3λ+c4Forλ=1,det(IA)=(1)4+c1(1)3+c2(1)2+c3(1)+c4                   =1+c1+c2+c3+c4.........................................(I)Forλ=4,det(4IA)=(4)4+c1(4)3+c2(4)2+c3(4)+c4                   =25664c1+16c24c3+c4.........................................(II)Forλ=6,det(6IA)=(6)4+c1(6)3+c2(6)2+c3(6)+c4                   =1296+216c1+36c2+6c3+c4.........................................(III)Forλ=6,det(6IA)=(6)4+c1(6)3+c2(6)2+c3(6)+c4                         =1296216c1+36c26c3+c4.........................................(IV)Rewrite the equation (I),(II),(III) and (IV),c4+c1+c2+c3=1..........................................................(I)c464c1+16c24c3=256..............................................(II)c4+216c1+36c2+6c3=1296...........................................(III)c4216c1+36c26c3=1296...........................................(IV)Solving (I),(II),(III) and (IV),we get ,c4=144,c1=3,c2=40 and c3=108det(λIA) becomes,det(λIA)=λ4+3λ340λ2108λ+144For λ=0det(A)=c4=144 or det(A)=(1)4.144=144det(A)=144Note:Their are several methods to solve system of linear equations.Solution: ~\lambda_1=1,~\lambda_2=-4,~\lambda_3=6,~\lambda_4=-6 \\det(\lambda I-A)=\lambda^n +c_1\lambda^{n-1}+c_2 \lambda^{n-2}+....+c_n \\Here ~ n=4, \\ \therefore ~det(\lambda I-A)=\lambda^4 +c_1\lambda^3+c_2 \lambda^2+c_3 \lambda+c_4 \\ For \lambda=1, \\det( I-A)=(1)^4 +c_1(1)^3+c_2 (1)^2+c_3 (1)+c_4 \\~~~~~~~~~~~~~~~~~~~=1 +c_1+c_2 +c_3 +c_4.........................................(I) \\ For \lambda=-4, \\det( -4I-A)=(-4)^4 +c_1(-4)^3+c_2 (-4)^2+c_3 (-4)+c_4 \\~~~~~~~~~~~~~~~~~~~=256 -64c_1+16c_2 -4c_3 +c_4.........................................(II) \\ For \lambda=6, \\det( 6I-A)=(6)^4 +c_1(6)^3+c_2 (6)^2+c_3 (6)+c_4 \\~~~~~~~~~~~~~~~~~~~=1296 +216c_1+36c_2 +6c_3 +c_4.........................................(III) \\ \\ For \lambda=-6, \\det( -6I-A)=(-6)^4 +c_1(-6)^3+c_2 (-6)^2+c_3 (-6)+c_4 \\~~~~~~~~~~~~~~~~~~~~~~~~~=1296 -216c_1+36c_2 -6c_3 +c_4.........................................(IV) \\Re-write~ the ~equation ~(I),(II),(III)~and~(IV), \\c_4+c_1+c_2 +c_3 =-1..........................................................(I)^* \\c_4-64c_1+16c_2-4c_3=-256..............................................(II)^* \\c_4+216c_1+36c_2+6c_3=-1296...........................................(III)^* \\c_4-216c_1+36c_2-6c_3=-1296...........................................(IV)^* \\Solving~ (I)^*,(II)^*,(III)^*~and~(IV)^*, we~get~, \\c_4=144,c_1=3,c_2=-40~and~c_3=-108 \\ \therefore det(\lambda I-A) ~becomes, \\det(\lambda I-A)=\lambda^4 +3\lambda^3-40 \lambda^2-108 \lambda+144 \\For~ \lambda=0 \\ det(-A)=c_4=144 ~or~det(A)=(-1)^4 .144=144 \\ \therefore det(A)=144 \\Note: Their ~are~ several~ methods ~to ~solve~ system ~of ~linear ~equations.


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