Matrices & Determinants
Matrices
nta_pyq_2025_jan
Grade 12
Question:
If $A$ and $B$ are non-singular matrices of the same order, then the inverse of $A\bigl(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\bigr)^{-1}B$ is equal to:
$AB^{-1}+A^{-1}B$
$\operatorname{adj}(B^{-1})+\operatorname{adj}(A^{-1})$
$\dfrac{|A|}{|B|}BA^{-1}+AB^{-1}$
$\dfrac{1}{|AB|}\bigl(\operatorname{adj}(B)+\operatorname{adj}(A)\bigr)$
Step-by-Step Solution
Key Concept: Use $\operatorname{adj}(M^{-1})=\dfrac{1}{|M|}M$ (from $\operatorname{adj}(M)=|M|M^{-1}$). Then $M\cdot\operatorname{adj}(M)=|M|I$ collapses cross-products.
$\bigl[A\,(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1}))^{-1}B\bigr]^{-1} = B^{-1}\bigl(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\bigr)A^{-1}.$
Use $\operatorname{adj}(A^{-1})\,A^{-1}=|A^{-1}|I=\dfrac{1}{|A|}I$, so
$B^{-1}\operatorname{adj}(A^{-1})A^{-1}=\dfrac{1}{|A|}B^{-1}=\dfrac{1}{|A|}\cdot\dfrac{\operatorname{adj}(B)}{|B|}=\dfrac{\operatorname{adj}(B)}{|AB|}.$
Similarly $B^{-1}\operatorname{adj}(B^{-1})A^{-1}=\dfrac{1}{|B|}A^{-1}=\dfrac{\operatorname{adj}(A)}{|AB|}.$
Sum: $\dfrac{1}{|AB|}\bigl(\operatorname{adj}(B)+\operatorname{adj}(A)\bigr).$
Correct Answer: 4