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Need solution for RD Sharma maths Class 12 Chapter 5 Determinants Exercise VSQ Question 56 maths textbook solution.

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Answer: -1

Hint: Here we use basic concept of determinant of matrix

Given: A \text { is } 3 \times 3 \text { matrix invertable then } \mathrm{n}=3

\rightarrow \operatorname{Det}\left(A^{-1}\right)=(\operatorname{Det} A)^{k}=?

Solution :

\rightarrow Because of A is invertible

\mathrm{AA}^{-1}=I_{3 \times 3}

\rightarrowTake determinant on both sides

\begin{aligned} &\left|\mathrm{AA}^{-1}\right|=\left|I_{3 \times 3}\right| \\ &\left.\left|\mathrm{AA}^{-1}\right|=1 \quad \text { (Because of }|I| \neq 1\right) \\ &|A| \times\left|A^{-1}\right|=1 \\ &\left|A^{-1}\right|=\frac{1}{|A|}=(|A|)^{-1} \\ &\rightarrow \text { So, here }(|A|)^{k}=|A|^{-1} \\ &k=-1 \end{aligned}

 

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