Answer to Question #162152 in Linear Algebra for Ricca

Question

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 + c₁λ^{n − 1} + ⋯ + c_n

So, on setting λ = 0, we obtain that

det (− A) = c_n or det (A) = (− 1)ⁿc_n

det (A) = ?

Answer 1

$Solution: ~\lambda_1=1,~\lambda_2=-4,~\lambda_3=6,~\lambda_4=-6 \\det(\lambda I-A)=\lambda^n +c_1\lambda^{n-1}+c_2 \lambda^{n-2}+....+c_n \\Here ~ n=4, \\ \therefore ~det(\lambda I-A)=\lambda^4 +c_1\lambda^3+c_2 \lambda^2+c_3 \lambda+c_4 \\ For \lambda=1, \\det( I-A)=(1)^4 +c_1(1)^3+c_2 (1)^2+c_3 (1)+c_4 \\~~~~~~~~~~~~~~~~~~~=1 +c_1+c_2 +c_3 +c_4.........................................(I) \\ For \lambda=-4, \\det( -4I-A)=(-4)^4 +c_1(-4)^3+c_2 (-4)^2+c_3 (-4)+c_4 \\~~~~~~~~~~~~~~~~~~~=256 -64c_1+16c_2 -4c_3 +c_4.........................................(II) \\ For \lambda=6, \\det( 6I-A)=(6)^4 +c_1(6)^3+c_2 (6)^2+c_3 (6)+c_4 \\~~~~~~~~~~~~~~~~~~~=1296 +216c_1+36c_2 +6c_3 +c_4.........................................(III) \\ \\ For \lambda=-6, \\det( -6I-A)=(-6)^4 +c_1(-6)^3+c_2 (-6)^2+c_3 (-6)+c_4 \\~~~~~~~~~~~~~~~~~~~~~~~~~=1296 -216c_1+36c_2 -6c_3 +c_4.........................................(IV) \\Re-write~ the ~equation ~(I),(II),(III)~and~(IV), \\c_4+c_1+c_2 +c_3 =-1..........................................................(I)^* \\c_4-64c_1+16c_2-4c_3=-256..............................................(II)^* \\c_4+216c_1+36c_2+6c_3=-1296...........................................(III)^* \\c_4-216c_1+36c_2-6c_3=-1296...........................................(IV)^* \\Solving~ (I)^*,(II)^*,(III)^*~and~(IV)^*, we~get~, \\c_4=144,c_1=3,c_2=-40~and~c_3=-108 \\ \therefore det(\lambda I-A) ~becomes, \\det(\lambda I-A)=\lambda^4 +3\lambda^3-40 \lambda^2-108 \lambda+144 \\For~ \lambda=0 \\ det(-A)=c_4=144 ~or~det(A)=(-1)^4 .144=144 \\ \therefore det(A)=144 \\Note: Their ~are~ several~ methods ~to ~solve~ system ~of ~linear ~equations.$