GATE 2023 MA – Question 60
A=(a_ij) is real 3×3 with A(1,2,1)ᵀ=2(1,2,1)ᵀ, A(0,1,1)ᵀ=2(0,1,1)ᵀ and A(−1,1,0)ᵀ=4(−1,1,0)ᵀ. If m is its minimal-polynomial degree, find a₁₁+a₂₁+a₃₁+m.
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Correct answer: 4 to 4
Explanation
The three supplied eigenvectors are independent, so A is diagonalizable with minimal polynomial (λ−2)(λ−4), degree m=2. Express e₁=(v₁−v₂−v₃)/2. Then Ae₁=v₁−v₂−2v₃=(3,−1,0)ᵀ. The first-column sum is 2, and adding m gives **4**.