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  • Need solution for RD Sharma maths class 12 chapter Adjoint and inverse of matrix exercise 6.2 question 15

 Need solution for RD Sharma maths class 12 chapter Adjoint and inverse of matrix exercise 6.2 question 15

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Answer:  A^{-1}=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ 3 & 5 & 9 \end{array}\right]

Hint :Here we use basic inverse elementary operation

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

Solution: A = IA

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

Applying \mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}+3 \mathrm{R}_{{1}} and R_{3} \Rightarrow R_{3}-2 R_{1} we get

\left[\begin{array}{ccc} 1 & 3 & -2 \\ 0 & 9 & -7 \\ 0 & -5 & 4 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{array}\right] A 

Applying \mathrm{R}_{2} \Rightarrow 1 / 9 \mathrm{R}_{2}

\left[\begin{array}{ccc} 1 & 3 & -2 \\ 0 & 1 & -7 / 9 \\ 0 & -5 & 4 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 / 3 & 1 / 9 & 0 \\ -2 & 0 & 1 \end{array}\right] A

Applying \mathrm{R}_{{3}} \Rightarrow 9 \mathrm{R}_{3}

\left[\begin{array}{ccc} 1 & 0 & 1 / 3 \\ 0 & 1 & -7 / 9 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & -1 / 3 & 0 \\ 1 / 3 & 1 / 9 & 0 \\ -3 & 5 & 9 \end{array}\right] A

Applying \mathrm{R}_{1} \Rightarrow \mathrm{R}_{1}-1 / 3 \mathrm{R}_{3} and \mathrm{R}_{2} \Rightarrow \mathrm{R}_{2}+7 / 9 \mathrm{R}_{3}

\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{array}\right] A

I=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{array}\right] A

So A^{-1}=\left[\begin{array}{ccc} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{array}\right]

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