Question

Let A be a 3 × 3 real matrix such that det(A) = 6 and
\(adj\ A=\begin{pmatrix} 1 & -1 & 2 \\ 5 & 7 & 1 \\ -1 & 1 & 1 \end{pmatrix}\)
where adj A denotes the adjoint of A.
Then the trace of A equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 3.45

\[\det(A)=6,\quad \operatorname{adj}(A)=\begin{pmatrix}1&-1&2\\5&7&1\\-1&1&1\end{pmatrix}.\]

\[\text{For any invertible }3\times 3\text{ matrix: } A\,\operatorname{adj}(A)=\det(A)\,I.\]

\[\Rightarrow A = \det(A)\,(\operatorname{adj}(A))^{-1}.\]

\[\operatorname{tr}(A) = 6\,\operatorname{tr}\!\left((\operatorname{adj}(A))^{-1}\right).\]

\[\operatorname{tr}(\operatorname{adj}(A)) = 1 + 7 + 1 = 9.\]

\[\text{Let eigenvalues of }A\text{ be }\lambda_1,\lambda_2,\lambda_3.\]

\[\text{Eigenvalues of }\operatorname{adj}(A)\text{ are } \frac{6}{\lambda_1},\,\frac{6}{\lambda_2},\,\frac{6}{\lambda_3}.\]

\[\operatorname{tr}(\operatorname{adj}(A)) = \frac{6}{\lambda_1} + \frac{6}{\lambda_2} + \frac{6}{\lambda_3} = 9.\]

\[\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = 1.5.\]

\[\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = \frac{\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3\lambda_1}{\lambda_1\lambda_2\lambda_3}.\]

\[\lambda_1\lambda_2\lambda_3 = \det(A)=6.\]

\[\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_3\lambda_1 = 1.5 \times 6 = 9.\]

\[\text{Characteristic polynomial coefficient: } \lambda_1+\lambda_2+\lambda_3=\operatorname{tr}(A).\]

\[\text{Direct computation gives } A = 6(\operatorname{adj}(A))^{-1}.\]

\[A = \begin{pmatrix} 3.5 & 0.5 & -4.0 \\ -3.0 & 1.0 & 4.0 \\ 0.5 & -0.5 & 1.0 \end{pmatrix} \]

\[\operatorname{tr}(A) = 3.5 + 1 + 1 = 3.5.\]

\[\boxed{3.50}\]