Question

For a $3 \times 3$ invertible matrix $A$ satisfying the characteristic equation $A^3+pA^2+qA-rI_3=0$, which of the following is/are true [Here $tr(A) = \text{trace of matrix } A$, $det(A) = \text{determinant value of matrix } A$]

• A. p = -tr(A)
• B. q = \frac{(tr(A))^2 - tr(A^2)}{2}
• C. r = det(A)
• D. A^{-1} = \frac{-2(A - tr(A) I_3)}{(tr(A))^2 - tr(A^2)}

Answer 1

Answer: (A), (B), (C)

p = -tr(A), q = \frac{(tr(A))^2 - tr(A^2)}{2}, r = det(A)

Answer 2

Using the Cayley–Hamilton theorem for a $3\times3$ matrix $A$, its characteristic polynomial is

$$\chi_A(\lambda)=\lambda^3 - \bigl(\mathrm{tr}\,A\bigr)\lambda^2+q\,\lambda - \det A.$$

By Cayley–Hamilton, $A^3 - (\mathrm{tr}\,A)A^2 + q\,A - (\det A)\,I_3=0$.

1. Comparing with $A^3 + pA^2 + qA - rI_3=0$ gives
   • $p = -\,\mathrm{tr}\,A$.
2. The coefficient $q$ equals the sum of principal $2\times2$ minors of $A$, which satisfies $$q = \frac{(\mathrm{tr}\,A)^2 - \mathrm{tr}(A^2)}{2}.$$
3. The constant term $r$ equals $\det A$.
4. One can also solve for $A^{-1}$ from the CH relation: $$0 = A^3 - (\mathrm{tr}\,A)A^2 + qA - \det(A)\,I_3 \;\Longrightarrow\; A^{-1} = \frac{A^2 - (\mathrm{tr}\,A)\,A + q\,I_3}{\det A},$$ which is not the form given in option D. Hence D is false.