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  • Please solve rd sharma class 12 Chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 6 maths textbook solution

Please solve rd  sharma class 12 Chapter 6   Adjoint and Inverse of Matrix excercise 6.2 question 6 maths textbook solution

Answers (1)

Ex-6.2_6:

Answer: \left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right]

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

Given: \left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]

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

              A = IA

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

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

Applying R_{1} \leftrightarrow R_{2}

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

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

              \Rightarrow\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right] A

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

              \Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right] A

Applying R_{3} \rightarrow R_{3}+5 R_{2}

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

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

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

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

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

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

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

So,\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right]

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