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  • Need solution for RD Sharma maths class 12 chapter Adjoint and inverse of matrix exercise 6.2 question 11

 Need solution for RD Sharma maths class 12 chapter Adjoint and inverse of matrix exercise 6.2 question 11

Answers (1)

Answer: \left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right]

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

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

Solution:

Let,

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

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

Applying

               \begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-3 R_{1} \\ &\Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & \frac{5}{2} & \frac{5}{2} \\ 0 & \frac{5}{2} & \frac{-7}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-1}{2} & 1 & 0 \\ \frac{-3}{2} & 0 & 1 \end{array}\right] A \end{aligned}

            

Applying  R_{2} \rightarrow \frac{2}{5} R_{2}

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

Applying  

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

              

              

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

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

Applying

              \begin{aligned} &R_{2} \rightarrow R_{2}-R_{3} \\ &R_{1} \rightarrow R_{1}-2 R_{3} \\ &\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right] A \end{aligned}

               

                                 So,A^{-1}=\left[\begin{array}{ccc} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{array}\right]              

 

 

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