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Question: Evaluate the following determinant \(\left| \begin{matrix} \cos \alpha \cos \beta & \cos \alpha...

Evaluate the following determinant ∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣\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| .

Explanation

Solution

In this problem, we need to evaluate the given determinant ∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣\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, ∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣ =−sin⁡α×(−1)1+3∣−sin⁡βcos⁡β sin⁡αcos⁡βsin⁡αsin⁡β ∣+0+cos⁡α×(−1)3+3∣cos⁡αcos⁡βcos⁡αsin⁡β −sin⁡βcos⁡β ∣ \begin{aligned} & \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{aligned}
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,
∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣\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 aij{{a}_{ij}} is (−1)i+jMij{{\left( -1 \right)}^{i+j}}{{M}_{ij}} where, Mij{{M}_{ij}} is the minor of the element aij{{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,
∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣ =−sin⁡α×(−1)1+3∣−sin⁡βcos⁡β sin⁡αcos⁡βsin⁡αsin⁡β ∣+0+cos⁡α×(−1)3+3∣cos⁡αcos⁡βcos⁡αsin⁡β −sin⁡βcos⁡β ∣ \begin{aligned} & \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{aligned}
We now simplify the above expression and get,
∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣ =−sin⁡α∣−sin⁡βcos⁡β sin⁡αcos⁡βsin⁡αsin⁡β ∣+cos⁡α∣cos⁡αcos⁡βcos⁡αsin⁡β −sin⁡βcos⁡β ∣ \begin{aligned} & \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{aligned}
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 2×22\times 2 determinant. We then get,
⇒∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣ =−sin⁡α[−sin⁡β(sin⁡αsin⁡β)−cos⁡β(sin⁡αcos⁡β)] +cos⁡α[cos⁡αcos⁡βcos⁡β−cos⁡αsin⁡β(−sin⁡β)] \begin{aligned} & \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{aligned}
Multiplying the terms, we get,
⇒∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣ =−sin⁡α[−sin⁡2βsin⁡α−sin⁡αcos⁡2β]+cos⁡α[cos⁡αcos⁡2β+cos⁡αsin⁡2β] \begin{aligned} & \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{aligned}
Now, taking −sin⁡α-\sin \alpha common from the first term and cos⁡α\cos \alpha common from the second term, we get,
⇒∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣ =−sin⁡α[−sin⁡α(sin⁡2β+cos⁡2β)]+cos⁡α[cos⁡α(cos⁡2β+sin⁡2β)] \begin{aligned} & \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{aligned}
We know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 . So, the above expression becomes,
⇒∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣ =−sin⁡α[−sin⁡α]+cos⁡α[cos⁡α]=sin⁡2α+cos⁡2α \begin{aligned} & \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{aligned}
Again, using the same trigonometric identity, we get,
⇒∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣=1\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 ∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣\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 11 .

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