CUET · MATHS · PYQ PAPER 2023
The inverse of the matrix \(A=\left[\begin{array}{cc}a & b \\ c & \frac{1+b c}{a}\end{array}\right]\) is:
- A \(\left[\begin{array}{cc}-a & c \\ b & \frac{-1-b c}{a}\end{array}\right]\)
- B \(\left[\begin{array}{cc}\frac{1+b c}{a} & -b \\ -c & a\end{array}\right]\)
- C \(\left[\begin{array}{cc}c & \frac{1+b c}{a} \\ a & b\end{array}\right]\)
- D \(\left[\begin{array}{cc}\frac{1+b c}{a} & -c \\ -b & a\end{array}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[\begin{array}{cc}\frac{1+b c}{a} & -b \\ -c & a\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(\det(A) = a\left(\frac{1+bc}{a}\right) - bc = 1+bc-bc = 1\) \(A^{-1} = \frac{1}{\det(A)}\text{adj}(A) = \frac{1}{1}\left[\begin{array}{cc}\frac{1+bc}{a} & -b \\ -c & a\end{array}\right] = \left[\begin{array}{cc}\frac{1+bc}{a} & -b \\ -c & a\end{array}\right]\)
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