Matrices & Determinants
Matrix polynomial recurrence
nta_pyq_2023_jan
Grade None
Question:
Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix A such that $A^2 = 3A + \alpha I$. If $A^4 = 21A + \beta I$, then
\alpha = 1
\alpha = 4
\beta = 8
\beta = -8
Step-by-Step Solution
Key Concept: Use the recurrence $A^2 = 3A + \alpha I$ repeatedly to express $A^4$ in terms of $A$ and $I$
$A^2 = 3A + \alpha I$. $A^3 = 3A^2 + \alpha A = 3(3A+\alpha I)+\alpha A = (9+\alpha)A + 3\alpha I$. $A^4 = (9+\alpha)A^2 + 3\alpha A = (9+\alpha)(3A+\alpha I)+3\alpha A = A(27+6\alpha)+\alpha(9+\alpha)I$. Comparing with $A^4 = 21A + \beta I$: $27+6\alpha = 21 \Rightarrow \alpha = -1$ and $\beta = \alpha(9+\alpha) = -1(8) = -8$. Answer: (4)
Correct Answer: $\beta = -8$