Matrices & Determinants
Symmetric matrix from transpose equation
nta_pyq_2023_jan
Grade 12
Question:
If P is a $3 \times 3$ real matrix such that $P^T = aP + (a-1)I$, where $a > 1$, then
P is a singular matrix
$|\text{Adj}P| > 1$
$|\text{Adj}P| = \frac{1}{2}$
$|\text{Adj}P| = 1$
Step-by-Step Solution
Key Concept: Transpose both sides to get another equation; combine to deduce $P = P^T$, then substitute back to find P
$P^T = aP + (a-1)I$. Transposing: $P = aP^T + (a-1)I = a(aP+(a-1)I)+(a-1)I = a^2P + (a-1)(a+1)I$. $(1-a^2)P = (a^2-1)I \Rightarrow P = -I$ (for $a \neq \pm 1$, $a>1$). $|P| = |-I| = -1$ (for $3\times3$), so $|\text{Adj}P| = |P|^{n-1} = (-1)^2 = 1$. Answer: (4)
Correct Answer: $|\text{Adj}P| = 1$