Question

The inverse of matrix \(\left[ \begin{array}{ccc}1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1\end{array} \right]\) is ...

• A. \(\left[ \begin{array}{ccc}1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1\end{array} \right]\)
• B. \(\left[ \begin{array}{ccc}1 & p & p^2 \\ q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2\end{array} \right]\)
• C. \(\left[ \begin{array}{ccc}1 & -p & p^2 \\ -q & 1+pq & -(p+p^2q) \\ q^2 & -(q+pq^2) & 1+pq+p^2q^2\end{array} \right]\)
• D. \(\left[ \begin{array}{ccc}1 & -p & p^2 \\ -q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2\end{array} \right]\)

Hint:

Find the determinant first, then the cofactors; here the determinant is 1.

Answer 1

The Correct Option is C

Step 1: Check Entries of M times Option (C):
Call option (C) $N$. Row 1 of $M$ times column 1 of $N$: $(1+pq)(1)+p(-q)+0=1$.

Step 2: More Entries:
Row 1 of $M$ times column 2 of $N$: $(1+pq)(-p)+p(1+pq)+0=0$. Row 1 of $M$ times column 3: $(1+pq)p^2+p\big(-(p+p^2q)\big)+0=p^2+p^3q-p^2-p^3q=0$.
Row 3 of $M$ is $(0,q,1)$. Times column 3 of $N$: $q\big(-(p+p^2q)\big)+1\cdot(1+pq+p^2q^2)=-pq-p^2q^2+1+pq+p^2q^2=1$.

Step 3: Compare With Other Options:
Option (B) fails the first entry check in column 2: row 1 of $M$ times $(p,1+pq,q+pq^2)$ gives $(1+pq)p+p(1+pq)=2p(1+pq)$, not 0. So (C) is the inverse.

Final Answer:
Option (C). \[ \boxed{\text{(C)}} \]