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  • Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 9

Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 9

Answers (1)

 

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

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

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

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

              \begin{aligned} &A=I A \\ &A=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -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} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 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 & -1 & \frac{4}{3} \\ 2 & -3 & 4 \\ 0 & -1 & 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 & -1 & \frac{4}{3} \\ 0 & -1 & \frac{4}{3} \\ 0 & -1 & 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-R_{2}

              \Rightarrow\left[\begin{array}{ccc} 1 & -1 & \frac{4}{3} \\ 0 & 1 & \frac{-4}{3} \\ 0 & -1 & 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

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

            

Applying R_{3} \rightarrow-3 R_{3}

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

Applying R_{2} \rightarrow R_{2}+\frac{4}{3} R_{3}

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

        

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

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