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  • Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 17

Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 17

Answers (1)

Answer: \left[\begin{array}{ccc} 3 & -2 & 1 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right]

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

Given: A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right]

Solution: A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right]

              \begin{aligned} &A=I A \\ &A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right], I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}

               \Rightarrow\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A

Applying R_{2} \rightarrow R_{2}+R_{3}

              \Rightarrow\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ -2 & -4 & -5 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] A

Applying R_{3} \rightarrow 2 R_{1}+R_{3}

              \Rightarrow\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] A

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

              \Rightarrow\left[\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -5 & 0 & -3 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] A

Applyinng R_{2} \rightarrow R_{2}-2 R_{3}

             \Rightarrow\left[\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -5 & 0 & -3 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right] A

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

                \Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -2 & 1 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right] A

                So,A^{-1}=\left[\begin{array}{ccc} 3 & -2 & 1 \\ -4 & 1 & -2 \\ 2 & 0 & 1 \end{array}\right]

 

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