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  • Please solve rd sharma class 12 Chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 14 maths textbook solution

Please solve rd  sharma class 12 Chapter 6   Adjoint and Inverse of Matrix excercise 6.2 question 14 maths textbook solution

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Answer: \left[\begin{array}{ccc} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right]

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

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

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

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

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

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

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

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

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

              \Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array}\right] A

Applying R_{3} \rightarrow R_{3}-4 R_{2}

              \Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & \frac{1}{9} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ \frac{8}{9} & \frac{-4}{3} & 1 \end{array}\right] A

Applying R_{3} \rightarrow 9 R_{3}

              \Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ 8 & -12 & 9 \end{array}\right] A

Applying

                \begin{aligned} &R_{1} \rightarrow R_{1}+\frac{1}{3} R_{3} \\ &R_{2} \rightarrow R_{2}-\frac{2}{9} R_{3} \end{aligned}

                \Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right] A

                So,A^{-1}=\left[\begin{array}{ccc} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right]

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