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  • Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 10

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

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Answer: \left[\begin{array}{ccc} \frac{-4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{-1}{2} & \frac{-1}{6} \\ \frac{5}{6} & \frac{-1}{2} & \frac{1}{6} \end{array}\right]

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

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

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

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

Applying

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

Applying R_{2} \rightarrow-R_{2}

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

Applying 

             \begin{aligned} &R_{1} \rightarrow R_{1}-2 R_{2} \\ &R_{3} \rightarrow R_{3}+3 R_{2} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 6 \end{array}\right]=\left[\begin{array}{ccc} -3 & 2 & 0 \\ 2 & -1 & 0 \\ 5 & -3 & 1 \end{array}\right] A \end{aligned}

Applying  R_{3} \rightarrow \frac{1}{6} R_{3}

              \Rightarrow\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & 2 & 0 \\ 2 & -1 & 0 \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{array}\right] A

Applying

              \begin{aligned} &R_{1} \rightarrow R_{1}+2 R_{3} \\ &R_{2} \rightarrow R_{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{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{-1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{array}\right] A \end{aligned}

              So,A^{-1}=\left[\begin{array}{ccc} -\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{-1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{array}\right]

              

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