Answer: 5
Let the characteristic polynomial of the 3×3 matrix M be P(λ)=λ3−4λ2+aλ+30.
Since λ=2 is an eigenvalue of M, it must be a root of P(λ). Substituting λ=2 into P(λ)=0 gives:
23−4(2)2+a(2)+30=08−16+2a+30=0⟹2a+22=0⟹a=−11Let the three eigenvalues of M be λ1,λ2, and λ3, with λ1=2. Using the properties of the roots of a polynomial:
- Sum of eigenvalues: λ1+λ2+λ3=4⟹2+λ2+λ3=4⟹λ2+λ3=2
- Product of eigenvalues: λ1λ2λ3=−30⟹2λ2λ3=−30⟹λ2λ3=−15
Solving the quadratic equation x2−2x−15=0 yields the other two eigenvalues:
(x−5)(x+3)=0⟹x=5 or x=−3Thus, the eigenvalues of M are 2,5, and −3. Their absolute values are ∣2∣=2, ∣5∣=5, and ∣−3∣=3. The largest absolute value is 5.