Question

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|

Answer 1

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,

{{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.