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Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 13

Answers (1)

 

Answer: \left[\begin{array}{ccc} -2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ 4 & \frac{-1}{2} & -2 \end{array}\right]

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

Given: \left[\begin{array}{ccc} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{array}\right]

Solution: Let A=\left[\begin{array}{ccc} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{array}\right]

             \begin{aligned} &A=I A \\ &A=\left[\begin{array}{ccc} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \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} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \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} & 2 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A

Applying

\begin{aligned} &R_{2} \rightarrow R_{2}-4 R_{1} \\ &R_{3} \rightarrow R_{3}-3 R_{1} \\ &\qquad \quad \Rightarrow\left[\begin{array}{ccc} 1 & -\frac{1}{2} & 2 \\ 0 & 2 & -6 \\ 0 & -\frac{1}{2} & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -2 & 1 & 0 \\ -\frac{3}{2} & 0 & 1 \end{array}\right] A \end{aligned} 

             

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

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

Applying

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

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

              

Applying

            \begin{aligned} &R_{1} \rightarrow R_{1}-\frac{1}{2} R_{3} \\ &R_{2} \rightarrow R_{2}+3 R_{3} \\ &\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ 4 & \frac{-1}{2} & -2 \end{array}\right] A \end{aligned}

            So, A^{-1}=\left[\begin{array}{ccc} -2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ 4 & \frac{-1}{2} & -2 \end{array}\right]

 

 

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