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  • Please Solve RD Sharma Class 12 Chapter 4 Algebra of Matrices Exercise Very Short Asnwer Question 26 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 4 Algebra of Matrices Exercise Very Short Asnwer Question 26 Maths Textbook Solution.

Answers (1)

Answer: \lambda=\left ( -1 \right )^{n}

Hint: Here to solve this we use the basic concept of skew symmetric matrix

Given: \left ( A^{n} \right )^{T}=\lambda A'' \: \lambda=?

Solution: A is skew symmetric

\begin{aligned} &A^{T}=-A \\ &\qquad \begin{aligned} \left(A^{n}\right)^{T}=\left(A^{n-1} \times A^{1}\right)^{T} &=A^{T} \times\left(A^{n-1}\right)^{T} \quad\left[(A B)^{T}=B^{T} A^{T}\right] \\ =(-A) \times\left(A^{n-2} \times A\right)^{T} &=(-A) A^{T}\left(A^{n-2}\right)^{T} \\ =(-A)(-A)\left(A^{n-3} \times A\right)^{T} &=(-A)(-A)\left(A^{1} A^{1}\right)^{T} \end{aligned} \\ &\qquad(-1)^{n-2} \times A^{n-2}\left(A^{T} \times A^{T}\right) =(-1)^{n-2} \times A^{n-2}(A \times A) \quad= \\ &(-1)^{n}(-1)^{-2} A^{n} \quad=(-1)^{n} \frac{1}{(-1)^{2}} \times A^{n}=(-1)^{n} A^{n} \end{aligned}

So, \lambda=\left ( -1 \right )^{n}

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