How to calculate the determinant of this matrix
$A=\begin{bmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{bmatrix}$
$$\left[A\right]=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{vmatrix}=\\=\sin x\begin{vmatrix}
\cos x& 0\\
1 & 1
\end{vmatrix}-\cos^2x\begin{vmatrix}
\sin x & 0 \\
\sin x & 1
\end{vmatrix}+\begin{vmatrix}
\sin x & \cos x\\
\sin x & 1\\\end{vmatrix}=\\=\sin x\cos x-\cos^2x\sin x+\sin x-\sin x\cos x =\\=\sin x\left(\cos x-\cos^2x+1-\cos x\right)=\sin x \left(1-\cos ^2x\right)=\\=\sin x\cdot \sin^2x=\sin^3x$$
The path is something like this? I'm using the wrong rule?
Best Answer
The calculation will be easier if you showed zeros:
Subtract the first row from the second and third row and develop relative the first column we find $$\det A=\begin{vmatrix} \sin x & \cos^2x & 1 \\ \sin x & \cos x & 0 \\ \sin x & 1 & 1 \end{vmatrix}=\begin{vmatrix} \sin x & \cos^2x & 1 \\ 0& \cos x(1-\cos x) & -1 \\ 0& 1-\cos^2x & 0 \end{vmatrix}=\sin x(1-\cos^2x).$$