Question #216171

Two of the eigenvalues of a 3*3 matrix A are 2 and 3 and the determinant is 48. Find the third eigenvalue and the trace (A)


1
Expert's answer
2021-07-12T15:01:40-0400

Determinant is the product of eigenvalues


detA=λ1λ2λ3=48\det A=\lambda_1\lambda_2\lambda_3=48

Given λ1=2,λ2=3.\lambda_1=2, \lambda_2=3. Then


λ3=detAλ1λ2=482(3)=8\lambda_3=\dfrac{\det A}{\lambda_1\lambda_2}=\dfrac{48}{2(3)}=8

tr(A)=λ1+λ2+λ3=2+3+8=13tr(A)=\lambda_1+\lambda_2+\lambda_3=2+3+8=13


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