# Calculate the determinant of $3\times 3$ matrix with $\sin x$ and powers of $\cos x$

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?

• it is ok: you got the right sign in the middle – Avitus Dec 11 '13 at 13:14
• Everything looks fine to me. – Salech Rubenstein Dec 11 '13 at 13:15
• That is correct. There are probably some simplifications are trig properties you can apply. – apnorton Dec 11 '13 at 13:18
• You could also expand using the last column because it contains one zero. This way you have to calculate one less $2 \times 2$ determinants but yeah :) – Slugger Dec 11 '13 at 13:18

Subtract the first row from the second and third row and develop relative the first column we find $$\det A=\begin{vmatrix} \sin x & \cos^2x & 1 \\ \sin x & \cos x & 0 \\ \sin x & 1 & 1 \end{vmatrix}=\begin{vmatrix} \sin x & \cos^2x & 1 \\ 0& \cos x(1-\cos x) & -1 \\ 0& 1-\cos^2x & 0 \end{vmatrix}=\sin x(1-\cos^2x).$$
$$\left[A\right]=\begin{vmatrix} \sin x & \cos^2x & 1\\ \sin x & \cos x & 0 \\ \sin x & 1 & 1 \end{vmatrix}\begin{vmatrix} \sin x & \cos^2x \\ \sin x & \cos x \\ \sin x & 1 \end{vmatrix}=\sin x\cos x+\sin x-\sin x\cos x -\sin x\cos^2x=\\=\sin x\left(1-\cos^2 x\right)=\\=\sin x\cdot \sin^2x=\sin^3 x$$