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Question: Evaluate the value of \(\Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alph...

Evaluate the value of \Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\\ { - \sin \alpha }&0&{\sin \beta } \\\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right| .

Explanation

Solution

Here, we have to find the value of \Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\\ { - \sin \alpha }&0&{\sin \beta } \\\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right| .
To find the value of any determinant of order 3, say A = \left| {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right| , we do A = a \cdot \left| {\begin{array}{*{20}{c}} e&f; \\\ h&i; \end{array}} \right| - b \cdot \left| {\begin{array}{*{20}{c}} d&f; \\\ g&i; \end{array}} \right| + c \cdot \left| {\begin{array}{*{20}{c}} d&e; \\\ g&h; \end{array}} \right| .
So, using the above method we need to evaluate \Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\\ { - \sin \alpha }&0&{\sin \beta } \\\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right| .

Complete step-by-step answer:
Here, we are asked to find the value of \Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\\ { - \sin \alpha }&0&{\sin \beta } \\\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right| .
Let A = \left| {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right| be any determinant of order 3.
Now, to find the value of determinant of order 3, we do as A = a \cdot \left| {\begin{array}{*{20}{c}} e&f; \\\ h&i; \end{array}} \right| - b \cdot \left| {\begin{array}{*{20}{c}} d&f; \\\ g&i; \end{array}} \right| + c \cdot \left| {\begin{array}{*{20}{c}} d&e; \\\ g&h; \end{array}} \right| .
Similarly, we will find the value of the determinant \Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\\ { - \sin \alpha }&0&{\sin \beta } \\\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right| as \Delta = 0 \cdot \left| {\begin{array}{*{20}{c}} 0&{\sin \beta } \\\ { - \sin \beta }&0 \end{array}} \right| - \sin \alpha \cdot \left| {\begin{array}{*{20}{c}} { - \sin \alpha }&{\sin \beta } \\\ {\cos \alpha }&0 \end{array}} \right| + \left( { - \cos \alpha } \right) \cdot \left| {\begin{array}{*{20}{c}} { - \sin \alpha }&0 \\\ {\cos \alpha }&{ - \sin \beta } \end{array}} \right|
Δ=0[0(sinβ)(sinβ)]sinα[0(cosα)(sinβ)]cosα[(sinα)(sinβ)0] Δ=0sinα(cosαsinβ)cosα[sinαsinβ] Δ=sinαsinβcosαsinαsinβcosα Δ=0  \Rightarrow \Delta = 0\left[ {0 - \left( {\sin \beta } \right)\left( { - \sin \beta } \right)} \right] - \sin \alpha \left[ {0 - \left( {\cos \alpha } \right)\left( {\sin \beta } \right)} \right] - \cos \alpha \left[ {\left( { - \sin \alpha } \right)\left( { - \sin \beta } \right) - 0} \right] \\\ \Rightarrow \Delta = 0 - \sin \alpha \left( { - \cos \alpha \sin \beta } \right) - \cos \alpha \left[ {\sin \alpha \sin \beta } \right] \\\ \Rightarrow \Delta = \sin \alpha \sin \beta \cos \alpha - \sin \alpha \sin \beta \cos \alpha \\\ \Rightarrow \Delta = 0 \\\

Thus, the value of Δ=0\Delta = 0 .

Note: Here, the direct approach to evaluate the value of \Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\\ { - \sin \alpha }&0&{\sin \beta } \\\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right| , can be as:
If the given matrix or determinant is skew-symmetric, then the value of its determinant is always 0 if the order of matrix or determinant is an odd number.
Here, the given determinant is \Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\\ { - \sin \alpha }&0&{\sin \beta } \\\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right| .
The given determinant is skew-symmetric and order is 3, which is an odd number.
So, the value of \Delta = \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha }&{ - \cos \alpha } \\\ { - \sin \alpha }&0&{\sin \beta } \\\ {\cos \alpha }&{ - \sin \beta }&0 \end{array}} \right| directly becomes 0.
Primary diagonal of matrix or determinant:
The elements in the form of aij,i=j{a_{ij}},i = j , form a diagonal of any square matrix or determinant. That diagonal is called the primary diagonal of the given matrix or determinant.
Skew-symmetric:
The matrix given in the form of \left[ {\begin{array}{*{20}{c}} 0&a;&{ - b} \\\ { - a}&0&c; \\\ b&{ - c}&0 \end{array}} \right] is called a skew-symmetric matrix.