Show that $\det\left[\begin{smallmatrix}1&\cos a&\cos b\\ \cos a&1&\cos(a+b) \\ \cos b&\cos(a+b)&1 \end{smallmatrix}\right]=0$

Question

I am unable to show - without expanding, by using determinant properties - that $$\det\begin{bmatrix} 1 &\cos a &\cos b\\ \cos a &1 &\cos(a+b) \\ \cos b &\cos(a+b) &1 \end{bmatrix}=0$$

I am using trigonometric identities to solve this but I don't understand what would be the next step.

Answer 1

This determinant is clearly zero.

The matrix is the Gram matrix of the three unit vectors $(1,0)$, $(\cos a,\sin a)$ and $(\cos b,-\sin b)$ in the plane. Your matrix equals $AA^T$ where $$A=\pmatrix{1&0\\\cos a&\sin a\\\cos b&-\sin b}$$ and so it is singular.

Answer 2

Here we go: directly $$\begin{align}\begin{vmatrix} 1 &\cos a &\cos b\\ \cos a &1 &\cos(a+b) \\ \cos b &\cos(a+b) &1 \end{vmatrix}&=\\&= [1+2\cos a\cos b\cos(a+b)]-[\cos^2b+\cos^2a+\cos^2(a+b)]\\&= 1+\cos(a+b)\cdot [2\cos a\cos b-\cos (a+b)]-\cos^2a-\cos^2b\\&= 1+\cos(a+b)\cos(a-b)-\cos^2a-\cos^2b\\&= 1+\frac12[\cos(2a)+\cos(2b)]-\cos^2a-\cos^2b\\&= 1+\frac12[2\cos^2a-1+2\cos^2b-1]-\cos^2a-\cos^2b\\&=0\end{align}$$

Another method: triangulation $$\begin{vmatrix} 1 &\cos a &\cos b\\ \cos a &1 &\cos(a+b) \\ \cos b &\cos(a+b) &1 \end{vmatrix}\stackrel{(-\cos a)R_1+R_2\to R_2; \\(-\cos b)R_1+R_3\to R_3}{=}\\ \begin{vmatrix} 1 &\cos a &\cos b\\ 0 &1-\cos^2a &\cos(a+b)-\cos a\cos b \\ 0 &\cos(a+b)-\cos a\cos b &1-\cos^2b \end{vmatrix}=\\ \begin{vmatrix} 1 &\cos a &\cos b\\ 0 &\sin^2a &-\sin a\sin b \\ 0 &-\sin a\sin b &\sin^2b \end{vmatrix}\stackrel{\frac{\sin b}{\sin a}\cdot R_2+R_3\to R_3}{=}\\ \begin{vmatrix} 1 &\cos a &\cos b\\ 0 &\sin^2a &-\sin a\sin b \\ 0 &0 &0 \end{vmatrix}=0.$$

Answer 3

To flesh out my comment from earlier,$$\Delta:=\left|\begin{array}{ccc} 1 & \cos a & \cos b\\ \cos a & 1 & \cos\left(a+b\right)\\ \cos b & \cos\left(a+b\right) & 1 \end{array}\right|=\left|\begin{array}{ccc} 1 & \cos a & \cos b\\ \cos a & 1 & \cos\left(a+b\right)\\ 0 & -\sin a\sin b & \sin^{2}b \end{array}\right|=\left|\begin{array}{ccc} 1 & \cos a & 0\\ \cos a & 1 & -\sin a\sin b\\ 0 & -\sin a\sin b & \sin^{2}b \end{array}\right|.$$We may as well also simplify the second row/column:

$$\Delta=\left|\begin{array}{ccc} 1 & 0 & 0\\ \cos a & \sin^{2}a & -\sin a\sin b\\ 0 & -\sin a\sin b & \sin^{2}b \end{array}\right|=\left|\begin{array}{ccc} 1 & 0 & 0\\ 0 & \sin^{2}a & -\sin a\sin b\\ 0 & -\sin a\sin b & \sin^{2}b \end{array}\right|.$$Well, now we just have an easy $2\times 2$ determinant.

Answer 4

Multiply the given matrix by

$$\begin{bmatrix}\sin (a+b)\\-\sin b \\ -\sin a \end{bmatrix}$$

When you carry out the matrix multiplication, the identities for the sine of a sum and sine of a difference render all components of the product vector $=0$.

This of course fails when $a$ and $b$ are multiples of $\pi$, die to the proposed eigenvector being zero; but then the matrix trivially has all rows proportional to each other. The breakdown of this eigenvector argument correlates with the null space becoming two-dimensional so that the zero-eigenvalue eigenvector is nonunique.