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Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6 point 2 question 5

Answers (1)

    

Answer: \left[\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right]

Hint: Here, we use the concept of matrix inverse using elementary row operation

Given: A=\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]

Solution: Let A=\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]

                   A = IA

                    A=\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right], I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]

                    \Rightarrow\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A

Applying R_{1} \rightarrow \frac{1}{3} R_{1}

              \Rightarrow\left[\begin{array}{ll} 1 & \frac{10}{3} \\ 2 & 7 \end{array}\right]=\left[\begin{array}{ll} \frac{1}{3} & 0 \\ 0 & 1 \end{array}\right] A

Applying  R_{2} \rightarrow R_{2}-2 R_{1}

            \Rightarrow\left[\begin{array}{ll} 1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{array}\right]=\left[\begin{array}{cc} \frac{1}{3} & 0 \\ \frac{-2}{3} & 1 \end{array}\right] A

             

Applying R_2\rightarrow 3R_2

              \Rightarrow\left[\begin{array}{ll} 1 & \frac{10}{3} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} \frac{1}{3} & 0 \\ -2 & 3 \end{array}\right] A

Applying R_{1} \rightarrow R_{1}-\frac{10}{3} R_{2}

              \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right]

 

Hence, \left[\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right]

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