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If P = \( \begin{bmatrix} 3 &4 \\[0.3em] 2 & -1 \\[0.3em] 0 & 5 \end{bmatrix}\) and P = \( \begin{bmatrix} 7 &-5 \\[0.3em] -4 & 0 \\[0.3em] 2 & 6 \end{bmatrix},\) verify that (P + Q)’ = (P’ + Q’).

P = [(3,4)(2,-1)(0,5)]

P = [(7,-5)(-4,0)(2,6)]

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Given P = \( \begin{bmatrix} 3&4 \\[0.3em] 2 & -1 \\[0.3em] 0 & 5 \end{bmatrix}\) and Q = \( \begin{bmatrix} 7&-5 \\[0.3em] -4& 0 \\[0.3em] 2 & 6 \end{bmatrix}\)

To Prove: (P + Q)’ = P’ + Q’

Proof: Let us consider R = P + Q,

LHS = R \(\Rightarrow\) (P + Q)’

LHS = \( \begin{bmatrix} 10& -2 & 2 \\[0.3em] -1& -1 & 11\\[0.3em] \end{bmatrix}\)

To find RHS, we will first find the transpose of matrix P and Q

RHS = P’ + Q’

LHS = RHS

Hence proved.

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