LA2.2.10

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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.

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