The value of the determinant \[\left| {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }}
\end{array}} \right|\]

Hint: We are asked to find the determinant of the given matrix. Recall the formula to find the determinant of a matrix. We observe that trigonometric terms are there in the given matrix, and use some trigonometric identities for simplification.

Complete step-by-step answer:
Given the matrix \[\left( {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }}
\end{array}} \right)\]
Let \[A = \left( {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }}
\end{array}} \right)\]
Determinant of matrix is written as,
 \[\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right| = \left( {{a_{11}}{a_{22}} - {a_{12}}{a_{21}}} \right)\]
Using this formula for matrix A, we get
 \[\left| {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }}
\end{array}} \right| = \cos {80^ \circ }\sin {10^ \circ } - ( - \cos {10^ \circ })\sin {80^ \circ }\]
 \[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }}
\end{array}} \right| = \cos {80^ \circ }\sin {10^ \circ } + \cos {10^ \circ }\sin {80^ \circ }\] (i)
We have the trigonometric identity for \[\sin (A + B)\] as
 \[\sin (A + B) = \sin A\cos B + \cos A\sin B\]
Using this trigonometric identity in equation (i), we get
 \[\left| {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }}
\end{array}} \right| = \sin ({80^ \circ } + {10^ \circ }) \\
   \Rightarrow \left| {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }}
\end{array}} \right| = \sin {90^ \circ } \]

 \[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }}\\
\end{array}} \right| = 1\]
Therefore, the value of determinant of the given matrix \[\left( {\begin{array}{*{20}{c}}
  {\cos {{80}^ \circ }}&{ - \cos {{10}^ \circ }} \\
  {\sin {{80}^ \circ }}&{\sin {{10}^ \circ }} \\
\end{array}} \right)\] is \[1\] .
So, the correct answer is “1”.

Note: A matrix can be defined as a rectangular array of ‘m’ number of rows and ‘n’ number of columns. A square matrix has equal numbers of rows and columns. The matrix here given is a square matrix or we can say \[2 \times 2\] square matrix. There are few properties which matrices obey are commutative property of addition, associative property of addition, associative property of multiplication, distributive property.
Determinant of a matrix is also an important concept that you should remember as in many problems you might be asked to find the determinant of a given question. The determinant of a matrix can tell us about the properties of a matrix and help us to find the inverse of a matrix.
Here, a \[2 \times 2\] square matrix was given, if we are a given a \[3 \times 3\] square matrix, then determinant will be,
 \[\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{21}}}
\end{array}} \right| = {a_{11}}({a_{22}}{a_{33}} - {a_{23}}{a_{32}}) - {a_{12}}({a_{21}}{a_{21}} - {a_{23}}{a_{31}}) + {a_{13}}({a_{21}}{a_{32}} - {a_{22}}{a_{31}})\] .
Similarly, we can find the determinant for a \[4 \times 4\] square matrix. While solving questions involving matrices, always remember these basic points.