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  • Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 29 maths textbook solution

Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 29 maths textbook solution

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VSQ : 29

Answer : \left[\begin{array}{ll} 2 & 5 \\ 5 & 4 \end{array}\right]=\left[\begin{array}{ll} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]

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

and column operation C_{2} \rightarrow C_{2}+2 C_{1} \\

Hint : you must know about the concept of elementary transform

Solution : The Matrix is in the form X=AB

Elementary opration can be apply only on X and B

\begin{aligned} &\text { So },\left[\begin{array}{ll} 2 & 1 \\ 2 & 0 \end{array}\right]=\left[\begin{array}{ll} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 1 \end{array}\right] \\ &C_{2} \rightarrow C_{2}+2 C_{1} \\ &{\left[\begin{array}{ll} 2 & 1+4 \\ 2 & 0+4 \end{array}\right]=\left[\begin{array}{ll} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 0+2 \\ -1 & 1-2 \end{array}\right]} \\ &{\left[\begin{array}{ll} 2 & 5 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]} \end{aligned}

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