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.