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  • Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 18

Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 18

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Answer: \left[\begin{array}{ccc} 0 & 1 & -3 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]

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

Given: \left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]

Solution : Let A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]

              \begin{aligned} &A=I A \\ &A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \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 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{aligned}

Applying \mathrm{R}_{1} \leftrightarrow \mathrm{R}_{3}

\left[\begin{array}{ccc} 1 & 1 & -2 \\ 3 & 2 & -4 \\ 2 & -3 & 5 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right] A

Applying \mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}-3 \mathrm{R}_{1} and \mathrm{R}_{3} \Rightarrow \mathrm{R}_{3}-2 \mathrm{R}_{1}

\left[\begin{array}{ccc} 1 & 1 & -2 \\ 0 & -1 & 2 \\ 0 & -5 & 9 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & -3 \\ 1 & 0 & -2 \end{array}\right] A

Applying \mathrm{R}_{2} \Rightarrow-\mathrm{R}_{2} and \mathrm{R}_{1}=\mathrm{R}_{1+} \mathrm{R}_{2}

\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & -5 & 9 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -3 \\ 0 & -1 & 3 \\ 1 & 0 & -2 \end{array}\right] A

Applying \mathrm{R}_{3} \Rightarrow \mathrm{R}_{3}+5 \mathrm{R}_{2}

\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -3 \\ 0 & -1 & 3 \\ 1 & -5 & 13 \end{array}\right] A

Applying \mathrm{R}_{3} \Rightarrow-\mathrm{R}_{3}

\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -3 \\ 0 & -1 & 3 \\ -1 & 5 & -13 \end{array}\right] A

Applying \mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}+2 \mathrm{R}_{3}

\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -3 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right] A

So, A^{-1}=\left[\begin{array}{ccc} 0 & 1 & -3 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]

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