Matrices & Determinants
Matrix satisfying A² - 16A - 17I = 0 with integer entries
MJAT_TS6_P1
Grade 12
Question:
Let $A=\begin{pmatrix}c&b\\a&d\end{pmatrix}$ where $a,b,c,d$ are positive integers in ascending order, exactly two are prime and pairwise coprime. $A$ satisfies $A^2-16A-17I=\mathbf{0}$. If $B=\begin{pmatrix}b&c\\a&d\end{pmatrix}$, then $\det(B)$ equals:
Step-by-Step Solution
Key Concept: From $A^2-16A-17I=0$: eigenvalues satisfy $\lambda^2-16\lambda-17=0\Rightarrow(\lambda-17)(\lambda+1)=0$. So eigenvalues are $17$ and $-1$. $\text{tr}(A)=a+d=16$, $\det(A)=-17$. With $a<b<c<d$ positive integers, two prime, coprime.
$\det(B)=\mathbf{43}$.
Correct Answer: 43