Matrices & Determinants
Adjoint
MMTS_Full_Test_19
Grade 12
Question:
If $A$ is a square matrix of order $n$, then $(\underbrace{\text{adj adj}\cdots\text{adj}A}_{(n-1)\text{ times}})\cdot(\underbrace{\text{adj adj}\cdots\text{adj}A}_{n\text{ times}})$ is equal to
$|A^{n-1}|^{(n-1)^2}$
$|A^{-1}|^{(n-1)(n-1)}$
$\dfrac{1}{n}|A|^{(n-1)^{(n-1)}}\cdot I_{n\times n}$
$|A|^{(n-1)^{(n-1)}}\cdot I_{n\times n}$
Step-by-Step Solution
Key Concept: $\text{adj}^k(A)$ has determinant $|A|^{(n-1)^k}$
$|\text{adj}^{n-1}A|=|A|^{(n-1)^{n-1}}$. Product $=|A|^{(n-1)^{n-1}}I$.
Correct Answer: 4