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प्रश्न
Find the inverse of the following matrix:
उत्तर
\[ C = \begin{bmatrix}a & b \\ c & \frac{1 + bc}{a}\end{bmatrix}\]
\[\left| C \right| = 1 + bc - bc = 1 \neq 0\]
C is a singular matrix; therefore, it is invertible .
\[\text{ Let }C_{ij}\text{ be a cofactor of }c_{ij}\text{ in C. }\]
Now,
\[ C_{11} = \frac{1 + bc}{a} \]
\[ C_{12} = - c\]
\[ C_{21} = - b\]
\[ C_{22} = a\]
\[adjC = \begin{bmatrix}\frac{1 + bc}{a} & - c \\ - b & a\end{bmatrix}^T = \begin{bmatrix}\frac{1 + bc}{a} & - b \\ - c & a\end{bmatrix}\]
\[ \therefore C^{- 1} = \frac{1}{\left| C \right|}adjC = \begin{bmatrix}\frac{1 + bc}{a} & - b \\ - c & a\end{bmatrix}\]
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