Evaluate the given expression: ${\left( { - 4} \right)^{ - 2}}$
A. $\dfrac{1}{{ - 16}}$
B. $\dfrac{1}{{16}}$
C.-16
D.16

Hint: It is given in the question that we have to Evaluate: ${\left( { - 4} \right)^{ - 2}}$ .
Then, apply the property ${a^{ - m}} = \dfrac{1}{{{a^m}}}$ in the given question.
Thus, find the value of ${\left( { - 4} \right)^{ - 2}}$ by using the above property.

Complete step-by-step answer:
It is given in the question that we have to Evaluate: ${\left( { - 4} \right)^{ - 2}}$ .
 $ = {\left( { - 4} \right)^{ - 2}}$
By using property ${a^{ - m}} = \dfrac{1}{{{a^m}}}$ in the above question, we get
 $ = \dfrac{1}{{{{\left( { - 4} \right)}^2}}}$
 $ = \dfrac{1}{{\left( { - 4} \right) \times \left( { - 4} \right)}}$
 $ = \dfrac{1}{{16}}$
 ${\left( { - 4} \right)^{ - 2}} = \dfrac{1}{{16}}$

Note: Some properties of exponents:
 $
  {a^m} \times {a^n} = {a^{m + n}} \\
  {a^n} \times {b^n} = {\left( {a \times b} \right)^n} \\
  \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}} \\
  \dfrac{{{a^n}}}{{{b^n}}} = {\left( {\dfrac{a}{b}} \right)^n} \\
  {\left( {{a^m}} \right)^n} = {a^{m \times n}} \\
  \sqrt[n]{{{a^m}}} = {a^{\dfrac{m}{n}}} \\
  \sqrt[n]{a} = {a^{\dfrac{1}{n}}} \\
  {a^{ - n}} = \dfrac{1}{{{a^n}}} \\
  {a^0} = 1 \\
 $