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  • Please solve rd sharma class 12 Chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 4 maths textbook solution

Please solve rd  sharma class 12 Chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 4  maths textbook solution

Answers (1)

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

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

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

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

A = IA

A=\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right], I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]

\Rightarrow\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A

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

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

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

              \Rightarrow\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ -1 & 2 \end{array}\right] A

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

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

 

Hence, A^{-1}=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]

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