If ${\text{A = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{2}{3}}&1&{\dfrac{5}{3}} \\
  {\dfrac{1}{3}}&{\dfrac{2}{3}}&{\dfrac{4}{3}} \\
  {\dfrac{7}{3}}&2&{\dfrac{2}{3}}
\end{array}} \right]$ and ${\text{B = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{2}{5}}&{\dfrac{3}{5}}&1 \\
  {\dfrac{1}{5}}&{\dfrac{2}{5}}&{\dfrac{4}{5}} \\
  {\dfrac{7}{5}}&{\dfrac{6}{5}}&{\dfrac{2}{5}}
\end{array}} \right]$ then compute 3A – 5B.

Hint: To find the value of 3A – 5B, we individually compute the values of 3A and 5B and find their difference. Arithmetic operations can be performed on matrices, such as addition, subtraction, multiplication and division. Just like we perform them on any term.

Complete step-by-step answer:
Given data,
${\text{A = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{2}{3}}&1&{\dfrac{5}{3}} \\
  {\dfrac{1}{3}}&{\dfrac{2}{3}}&{\dfrac{4}{3}} \\
  {\dfrac{7}{3}}&2&{\dfrac{2}{3}}
\end{array}} \right]$
${\text{B = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{2}{5}}&{\dfrac{3}{5}}&1 \\
  {\dfrac{1}{5}}&{\dfrac{2}{5}}&{\dfrac{4}{5}} \\
  {\dfrac{7}{5}}&{\dfrac{6}{5}}&{\dfrac{2}{5}}
\end{array}} \right]$

We know, if we multiply a scalar quantity to a matrix, every term inside the matrix gets multiplied by the scalar quantity, giving us a new matrix.
If we subtract or add one matrix to another, each of the corresponding term in one matrix to another matrix, undergo the respective operation giving us a new matrix.

Let us first compute 3A, every element in A gets multiplied by 3
$
   \Rightarrow {\text{3A = 3}} \times \left[ {\begin{array}{*{20}{c}}
  {\dfrac{2}{3}}&1&{\dfrac{5}{3}} \\
  {\dfrac{1}{3}}&{\dfrac{2}{3}}&{\dfrac{4}{3}} \\
  {\dfrac{7}{3}}&2&{\dfrac{2}{3}}
\end{array}} \right] \\
   \Rightarrow 3{\text{A = }}\left[ {\begin{array}{*{20}{c}}
  2&3&5 \\
  1&2&4 \\
  7&6&2
\end{array}} \right] \\
$


Similarly, let us compute 5B, every element in B gets multiplied by 5
$
   \Rightarrow {\text{5B = 5}} \times \left[ {\begin{array}{*{20}{c}}
  {\dfrac{2}{5}}&{\dfrac{3}{5}}&1 \\
  {\dfrac{1}{5}}&{\dfrac{2}{5}}&{\dfrac{4}{5}} \\
  {\dfrac{7}{5}}&{\dfrac{6}{5}}&{\dfrac{2}{5}}
\end{array}} \right] \\
   \Rightarrow 5{\text{B = }}\left[ {\begin{array}{*{20}{c}}
  2&3&5 \\
  1&2&4 \\
  7&6&2
\end{array}} \right] \\
$


Now 3A – 5B is given as,
$
   \Rightarrow {\text{3A - 5B = }}\left[ {\begin{array}{*{20}{c}}
  2&3&5 \\
  1&2&4 \\
  7&6&2
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  2&3&5 \\
  1&2&4 \\
  7&6&2
\end{array}} \right] \\
   \Rightarrow {\text{3A - 5B = }}\left[ {\begin{array}{*{20}{c}}
  {2 - 2}&{3 - 3}&{5 - 5} \\
  {1 - 1}&{2 - 2}&{4 - 4} \\
  {7 - 7}&{6 - 6}&{2 - 2}
\end{array}} \right] \\
   \Rightarrow {\text{3A - 5B = }}\left[ {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&0 \\
  0&0&0
\end{array}} \right] \\
$

Therefore 3A – 5B = $\left[ {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&0 \\
  0&0&0
\end{array}} \right]$

Note – In order to solve this type of problems the key is to know how to perform arithmetic operations on matrices. A matrix can be multiplied or divided with a scalar quantity by just performing the operation using the scalar on each term of the matrix but multiplication or division between two matrices follows a different procedure.
In contrast, a matrix can be added or subtracted with another matrix by performing the operation on each of the corresponding terms of both the matrices but we cannot add/subtract a scalar quantity to a matrix by just adding/subtracting it to every term of the matrix.