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If A = \( \begin{bmatrix} 3& 2& -1 \\[0.3em] -5& 0 & -6 \\[0.3em] \end{bmatrix}\) and B = \( \begin{bmatrix} -4& -5& -2 \\[0.3em] 3&1 & 8 \\[0.3em] \end{bmatrix}\) , verify that (A + B)’ = (A’ + B’).

A = [(3,2,-1)(-5,0,-6)]

B = [(-4,-5,-2)(3,1,8)]

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Given A = \( \begin{bmatrix} 3& 2& -1 \\[0.3em] -5& 0 & -6 \\[0.3em] \end{bmatrix}\) and B = \( \begin{bmatrix} -4& -5& -2 \\[0.3em] 3&1 & 8 \\[0.3em] \end{bmatrix}\) 

To Prove: (A + B)’ = A’ + B’

Proof: Let us consider C = A + B

Now LHS = C’

To find RHS, we will find transpose of matrix A and B

LHS = RHS

Hence proved.

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