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Need solution for RD Sharma Maths Class 12 Chapter Algebra of Matrices Excercise 4.4 Question 6 Subquestion (i).

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Answer: \left (AB \right )^{T}=B^{T}A^{T}

Given: A=\begin{bmatrix} 2 & 1 &3 \\ 4 & 1 & 0 \end{bmatrix}, B=\begin{bmatrix} 1 &-1 \\ 0 & 2\\ 5 & 0 \end{bmatrix}

To prove:\left ( AB \right )^{T}=B^{T}A^{T}

Hint: The A^{T} of matrix A can be obtained by reflecting the elements along it’s main diagonal.

Solution:

               A^{T}=\begin{bmatrix} 2 &4 \\ 1 &1 \\ 3 & 0 \end{bmatrix}, B^{T}=\begin{bmatrix} 1 & 0 &5 \\ -1 & 2& 0 \end{bmatrix}

               \left ( AB \right )^{T}=B^{T}A^{T}

               \left ( \begin{bmatrix} 2 & 1 &3 \\ 4 &1 & 0 \end{bmatrix}\begin{bmatrix} 1 &-1 \\ 0 &2 \\ 5 & 0 \end{bmatrix} \right )^{T}=\begin{bmatrix} 1& 0 &5 \\ -1 &2 & 0 \end{bmatrix}\begin{bmatrix} 2 &4 \\ 1 &1 \\ 3 & 0 \end{bmatrix}

               \left ( \begin{bmatrix} 2+0+15& -2+2+0\\ 4+0+0 & -4+2+0 \end{bmatrix} \right )^{T}=\begin{bmatrix} 2+0+15 &4+0+0 \\ -2+2+0 & -4+2+0 \end{bmatrix}

               \begin{bmatrix} 17 &0 \\ 4 & -2 \end{bmatrix}^{T}=\begin{bmatrix} 17 &4 \\ 0& -2 \end{bmatrix}

                \begin{bmatrix} 17 &0 \\ 4 & -2 \end{bmatrix}^{T}=\begin{bmatrix} 17 &4 \\ 0& -2 \end{bmatrix}

∴LHS=RHS

Hence, \left ( AB \right )^{T}=B^{T}A^{T}, is proved.

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