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Question: Evaluate the value of the matrix: \[\Delta =\left| \begin{matrix} \cos \alpha \cos \beta & \cos...

Evaluate the value of the matrix: Δ=∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣\Delta =\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

Hint: Use the fact that the value of a 3×33\times 3 determinant of the form ∣abc def ghi ∣\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right| is a(ei−fh)−b(di−gf)+c(dh−eg)a\left( ei-fh \right)-b\left( di-gf \right)+c(dh-eg). Substitute the value of the variables by comparing it with the given matrix. Simplify the value of the matrix using the trigonometric identity cos⁡2x+sin⁡2x=1{{\cos }^{2}}x+{{\sin }^{2}}x=1.

Step-by-step answer:
We have to calculate the value of the matrix Δ=∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣\Delta =\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 observe that this matrix has 3 rows and 3columns. Thus, it is a 3×33\times 3 matrix.
We will now calculate the value of the given matrix using the fact that the value of the matrix of the form ∣abc def ghi ∣\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right| is a(ei−fh)−b(di−gf)+c(dh−eg)a\left( ei-fh \right)-b\left( di-gf \right)+c(dh-eg).
We will now substitute the values a=cos⁡αcos⁡β,b=cos⁡αsin⁡β,c=−sin⁡αa=\cos \alpha \cos \beta ,b=\cos \alpha \sin \beta ,c=-\sin \alpha , d=−sin⁡β,e=cos⁡β,f=0d=-\sin \beta ,e=\cos \beta ,f=0 and g=sin⁡αcos⁡β,h=sin⁡αsin⁡β,i=cos⁡αg=\sin \alpha \cos \beta ,h=\sin \alpha \sin \beta ,i=\cos \alpha in each row of the matrix.
Thus, we have Δ=cos⁡αcos⁡β(cos⁡αcos⁡β−0(cos⁡αcos⁡β))−cos⁡αsin⁡β(−cos⁡αsin⁡β−0(sin⁡αcos⁡2β))−sin⁡α(−sin⁡αsin⁡2β−sin⁡αcos⁡2β)\Delta =\cos \alpha \cos \beta \left( \cos \alpha \cos \beta -0\left( \cos \alpha \cos \beta \right) \right)-\cos \alpha \sin \beta \left( -\cos \alpha \sin \beta -0\left( \sin \alpha {{\cos }^{2}}\beta \right) \right)-\sin \alpha \left( -\sin \alpha {{\sin }^{2}}\beta -\sin \alpha {{\cos }^{2}}\beta \right).
Simplifying the above expression, we have Δ=cos⁡2αcos⁡2β+cos⁡2αsin⁡2β+sin⁡2αsin⁡2β+sin⁡2αcos⁡2β\Delta ={{\cos }^{2}}\alpha {{\cos }^{2}}\beta +{{\cos }^{2}}\alpha {{\sin }^{2}}\beta +{{\sin }^{2}}\alpha {{\sin }^{2}}\beta +{{\sin }^{2}}\alpha {{\cos }^{2}}\beta .
Rearranging the terms of the above equation, we have Δ=cos⁡2α(cos⁡2β+sin⁡2β)+sin⁡2α(cos⁡2β+sin⁡2β)\Delta ={{\cos }^{2}}\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)+{{\sin }^{2}}\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right).
We know the trigonometric identity cos⁡2x+sin⁡2x=1{{\cos }^{2}}x+{{\sin }^{2}}x=1.
Thus, we have Δ=cos⁡2α(1)+sin⁡2α(1)=cos⁡2α+sin⁡2α\Delta ={{\cos }^{2}}\alpha \left( 1 \right)+{{\sin }^{2}}\alpha \left( 1 \right)={{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha .
So, we have Δ=cos⁡2α+sin⁡2α=1\Delta ={{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1.
Hence, the value of the matrix Δ=∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α −sin⁡βcos⁡β0 sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α ∣\Delta =\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 can expand the matrix along any row or column. However, in each case, the value of the matrix remains the same. A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. The dimension of a matrix is written as the product of the number of rows and columns. Two matrices can be added or subtracted element by element if they have the same dimension.