Answer:

Given, \[matrix\text{ }A\text{ }=\text{ }matrix\text{ }B\]

Then, at that point, their comparing components are equivalent.

Thus, we have

\[{{a}_{11}}~=\text{ }{{b}_{11}};\]\[a\text{ }+\text{ }4\text{ }=\text{ }2a\text{ }+\text{ }2\text{ }\Rightarrow \text{ }a\text{ }=\text{ }2\]

\[{{a}_{12}}~=\text{ }{{b}_{12}};~\]\[~3b\text{ }=\text{ }{{b}^{2}}~+\text{ }2\text{ }\Rightarrow \text{ }{{b}^{2}}-\text{ }3b\text{ }+\text{ }2~\]\[=\text{ }0\text{ }\Rightarrow \text{ }b\text{ }=\text{ }1,\text{ }2\]

\[{{a}_{22}}~=\text{ }{{b}_{22}};~\]\[~-6\text{ }=\text{ }{{b}^{2}}-\text{ }5b\text{ }\Rightarrow \text{ }{{b}^{2}}-\text{ }5b\text{ }+\text{ }6~\]\[=\text{ }0\text{ }\Rightarrow \text{ }b\text{ }=\text{ }2,\text{ }3\]

Thus, \[a\text{ }=\text{ }2\text{ }and\text{ }b\text{ }=\text{ }2~\] (normal worth)