Question

Evaluate the following determinant $\left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right|$ .

Answer 1

Hint: In this problem, we need to evaluate the given determinant $\left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right|$ . We first need to know how to evaluate a determinant. We evaluate a determinant by adding the product of an element and its co-factor along a single row or column. So, for the given determinant, we evaluate along the third column as, $\begin{align}
  & \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right| \\
 & =-\sin \alpha \times {{\left( -1 \right)}^{1+3}}\left| \begin{matrix}
   -\sin \beta & \cos \beta \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta \\
\end{matrix} \right|+0+\cos \alpha \times {{\left( -1 \right)}^{3+3}}\left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta \\
   -\sin \beta & \cos \beta \\
\end{matrix} \right| \\
\end{align}$
Again, breaking down these determinants and then simplifying the final answer, we arrive at the answer.

Complete step by step answer:
In this problem, we need to evaluate,
$\left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right|$
The cofactor of an element ${{a}_{ij}}$ is ${{\left( -1 \right)}^{i+j}}{{M}_{ij}}$ where, ${{M}_{ij}}$ is the minor of the element ${{a}_{ij}}$ found out by removing the corresponding row and column of the element and taking the remaining determinant. We evaluate a determinant by adding the product of an element and its co-factor along a single row or column. So, for the given determinant, we evaluate along the third column as,
$\begin{align}
  & \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right| \\
 & =-\sin \alpha \times {{\left( -1 \right)}^{1+3}}\left| \begin{matrix}
   -\sin \beta & \cos \beta \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta \\
\end{matrix} \right|+0+\cos \alpha \times {{\left( -1 \right)}^{3+3}}\left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta \\
   -\sin \beta & \cos \beta \\
\end{matrix} \right| \\
\end{align}$
We now simplify the above expression and get,
$\begin{align}
  & \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right| \\
 & =-\sin \alpha \left| \begin{matrix}
   -\sin \beta & \cos \beta \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta \\
\end{matrix} \right|+\cos \alpha \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta \\
   -\sin \beta & \cos \beta \\
\end{matrix} \right| \\
\end{align}$
We again repeat the same procedure by adding the product of an element and its co-factor along a single row or column to evaluate the remaining $2\times 2$ determinant. We then get,
$\begin{align}
  & \Rightarrow \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right| \\
 & =-\sin \alpha \left[ -\sin \beta \left( \sin \alpha \sin \beta \right)-\cos \beta \left( \sin \alpha \cos \beta \right) \right] \\
 & +\cos \alpha \left[ \cos \alpha \cos \beta \cos \beta -\cos \alpha \sin \beta \left( -\sin \beta \right) \right] \\
\end{align}$
Multiplying the terms, we get,
$\begin{align}
  & \Rightarrow \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right| \\
 & =-\sin \alpha \left[ -{{\sin }^{2}}\beta \sin \alpha -\sin \alpha {{\cos }^{2}}\beta \right]+\cos \alpha \left[ \cos \alpha {{\cos }^{2}}\beta +\cos \alpha {{\sin }^{2}}\beta \right] \\
\end{align}$
Now, taking $-\sin \alpha $ common from the first term and $\cos \alpha $ common from the second term, we get,
$\begin{align}
  & \Rightarrow \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right| \\
 & =-\sin \alpha \left[ -\sin \alpha \left( {{\sin }^{2}}\beta +{{\cos }^{2}}\beta \right) \right]+\cos \alpha \left[ \cos \alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right) \right] \\
\end{align}$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ . So, the above expression becomes,
$\begin{align}
  & \Rightarrow \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right| \\
 & =-\sin \alpha \left[ -\sin \alpha \right]+\cos \alpha \left[ \cos \alpha \right]={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\
\end{align}$
Again, using the same trigonometric identity, we get,
$\Rightarrow \left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right|=1$
Thus, we can conclude that the value of the determinant $\left| \begin{matrix}
   \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
   -\sin \beta & \cos \beta & 0 \\
   \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \\
\end{matrix} \right|$ is $1$ .

Note: We must be thorough with the concepts of minors and cofactors. We also must keep in mind to multiply an additional $-1$ for the odd position terms. Evaluating a determinant is a long process, so we must remain patient and should flawlessly solve the problem.