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: Determinant:
For \(M\) with rows \((1+pq,\,p,\,0)\), \((q,\,1+pq,\,p)\), \((0,\,q,\,1)\), expand along the first row:
\[ \det M=(1+pq)\left[(1+pq)(1)-pq\right]-p\left[q(1)-0\right]=(1+pq)-pq=1 \]
Since \(\det M=1\), \(M^{-1}=\text{adj}\,M\).

Step 2: Cofactors:
\(C_{11}=(1+pq)-pq=1\), \(C_{12}=-(q\cdot1-0)=-q\), \(C_{13}=q^2\).
\(C_{21}=-(p\cdot1-0)=-p\), \(C_{22}=1+pq\), \(C_{23}=-\big(q(1+pq)-0\big)=-(q+pq^2)\).
\(C_{31}=p^2\), \(C_{32}=-\big((1+pq)p-0\big)=-(p+p^2q)\), \(C_{33}=(1+pq)^2-pq=1+pq+p^2q^2\).

Step 3: Adjoint:
The inverse is the transpose of the cofactor matrix. Its rows are:
\((1,\,-p,\,p^2)\), \((-q,\,1+pq,\,-(p+p^2q))\), \((q^2,\,-(q+pq^2),\,1+pq+p^2q^2)\).

Step 4: Match the Option:
This is exactly option (C). Option (B) has first row \((1,p,p^2)\) with no minus signs, so it differs. Option (D) has the right signs in the first row and column but \(+(p+p^2q)\) and \(+(q+pq^2)\) in the lower right, where we found minus signs. Option (A) is \(M\) itself, which is not equal to its own inverse because \(M^2\neq I\).

Final Answer:
The inverse is option (C). \[ \boxed{\text{(C)}} \]