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  • Need solution for RD Sharma maths class 12 chapter Adjoint and inverse of matrix exercise 6.2 question 8

 Need solution for RD Sharma maths class 12 chapter Adjoint and inverse of matrix exercise 6 point 2 question 8

Answers (1)

Answer: \left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{array}\right]

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

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

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

              A = IA

              \begin{aligned} &A=\left[\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \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 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \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}{lll} 1 & \frac{3}{2} & \frac{1}{2} \\ 2 & 4 & 1 \\ 3 & 7 & 2 \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}-2 R_{1} \\ R_{3} & \rightarrow R_{3}-3 R_{1} \\ \Rightarrow\left[\begin{array}{lll} 1 & \frac{3}{2} & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & \frac{5}{2} & \frac{1}{2} \end{array}\right] &=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -1 & 1 & 0 \\ \frac{-3}{2} & 0 & 1 \end{array}\right] A \end{aligned}

 

Applying

            

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

Applying    R_{3} \rightarrow 2 R_{3}

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

Applying   R_{1} \rightarrow R_{1}-\frac{1}{2} 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 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{array}\right] A

            So,A^{-1}=\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2 \end{array}\right]

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