LA2.2.10

$A+B=\begin{bmatrix} 2 & 6 & 5 \\ 2 & 5 & 4 \\ 5 & 16 & 13\end{bmatrix}\,$

Applying $R_2-2R_1,R_3-\frac{5}{2}R_1\,$ we get

$\begin{bmatrix} 2 & 6 & 5 \\ 0 & -7 & -6 \\ 0 & -1 & \frac{1}{2}\end{bmatrix}\,$

Hence $r(A+B)=3\,$

Now $AB=\begin{bmatrix} 23 & 23 & 23 \\ 12 & 12 & 12 \\ 58 & 58 & 58\end{bmatrix}\,$

Applying $\frac{1}{23}R_1,R_2-12R_1,R_3-58R_1\,$ we get

$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\,$

Therefore the rank of $AB\,$ is equal to 1.