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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 30 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 30 maths textbook solution

Answers (1)

VSQ : 30

Answer : \left[\begin{array}{ll} 2 & 3 \\ 3 & 7 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 17 & -7 \end{array}\right]

Given : \left[\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 9 & -4 \end{array}\right]

and elementary operation R_{2}\rightarrow R_{2}+R_{1}

Hint : In multiplication row operation is equivalent to left multiplication 

Solution :

 \left[\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 9 & -4 \end{array}\right]

 

By applying operation R_{2}\rightarrow R_{2}+R_{1}

\left[\begin{array}{ll} 2 & 3 \\ 3 & 7 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 17 & -7 \end{array}\right]

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