Question
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| .
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=0−sinα(−cosαsinβ)−cosα[sinαsinβ] ⇒Δ=sinαsinβcosα−sinαsinβcosα ⇒Δ=0
Thus, the value of Δ=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 , 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.