Matrices & Determinants
Non-singular matrix with A² = Aᵀ, A ≠ I
MJAT_TS2_P1
Grade 12
Question:
Let $A$ be a $3\times 3$ non-singular matrix with real entries such that $A^2 = A^T$, where $A^T$ is the transpose of $A$. If $A\neq I$, then which of the following statements is/are TRUE?
A) $\det(A) = 1$
B) $\mathrm{tr}(A) = 0$
C) $A^2 + A + I = O$
D) $\det(A^2 + A + I) = 0$
Step-by-Step Solution
Key Concept: From $A^2=A^T$: $|A^2|=|A^T|\Rightarrow|A|^2=|A|\Rightarrow|A|(|A|-1)=0$. Since $A$ invertible, $|A|=1$ (A). From $A^4=A$ and $A^3=I$ (shown by: $(A^2)^T=A\Rightarrow(A^T)^T=A\Rightarrow A^4=A$, and $A^4=A\Rightarrow A^3=I$). From $A^3=I$ and $A\neq I$: $(A-I)(A^2+A+I)=0\Rightarrow A^2+A+I=O$ only if $A-I$ is not invertible... but we need to check.
$|A|=1$ (A ✓). $A^3=I\Rightarrow$ eigenvalues are cube roots of unity. $A\neq I\Rightarrow$ no eigenvalue is $1$ (otherwise $(A-I)$ singular but $A^2+A+I$ invertible, contradiction). So eigenvalues are $\omega,\omega^2,\omega^3$ or similar sum-zero set: $\mathrm{tr}(A)=0$ (B ✓). $A^2+A+I=O$ (C and D ✓). Answer: A, B, D.
Correct Answer: ABD