If $A+B+C=0$, then prove that the value of the determinant is $0$.

Question

I'll state the question from my textbook below:

If $A+B+C=0$, the prove that $\begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.

This is how I tried solving the problem:

$LHS = \begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix}$

$= 1(1- \cos^2 A) - \cos C (\cos C - \cos A \cos B) + \cos B (\cos A \cos C - \cos B)$

$= 1 + 2 \cos A \cos B \cos C - (\cos^2 A +\cos^2 B + \cos^2 C)$

I don't know how to proceed further. I tried using the fact that $A+B+C=0$ but it didn't lead to anything I could solve. I don't know where is it supposed to be used.

Also, I read a solution to this problem somewhere in which the term $(\cos^2 A +\cos^2 B + \cos^2 C)$ was replaced by $1 + 2 \cos A \cos B \cos C$ as $A+B+C=0$. Are these two terms equal for the given condition? Also, is there a way to prove the statement without using this fact?

Any help would be appreciated.

Answer 1

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$\cos^2A+\cos^2B+\cos^2C =1+\cos^2A-\sin^2B+\cos^2C$$

$$=1+\cos(A+B)\cos(A-B)+\cos^2C$$

$$=1-\cos C\cos(A-B)+\cos^2C\text{ as }\cos(A+B)=\cos(\pi-C)=?$$

$$=1-\cos C[\cos(A-B)-\cos C]$$

$\cos(A-B)-\cos C=\cos(A-B)+\cos(A+B)=?$

Answer 2

It's worth noting that we are dealing with a Gram determinant of three unit vectors in $3$-space. This determinant is $0$ if and only if the vectors are linearly dependent, that is, lying in the same plane. The condition $A+B+C=0$ ensures this.

It is also worth noting that the determinant always equals

$$4 \sin s \sin(s-A) \sin (s-B) \sin (s-C)$$ where $s= \frac{A+B+C}{2}$, so we get a necessary and sufficient condition for the determinant to be $0$, as above.

Answer 3

When at least one of $A,B$ or $C$ is not an integer multiple of $\pi$, the vector $(\sin A,\sin B,\sin C)$ is nonzero. However, as $A+B+C=0$, the sum-of-angle formula implies that $$ \pmatrix{1&\cos C&\cos B \\ \cos C&1&\cos A\\ \cos B&\cos A& 1} \pmatrix{\sin A\\ \sin B\\ \sin C} =0 $$ and hence the matrix is singular.

In case all of $A,B,C$ are integer multiples of $\pi$, you may complete the proof by a continuity argument, or by noticing that the matrix must have at least two identical rows, because up to a permutation, $(\cos A,\cos B,\cos C)$ is either $(1,1,1)$ or $(1,-1,-1)$.

Answer 4

If one of the angles, say $A=0$, then two rows become identical and the determinant becomes zero.

Else we multiply the rows in order by $\sin A, \sin B, \sin C$ and we get

\begin{vmatrix}\sin A & \sin A\cos C & \sin A\cos B \\ \sin B\cos C & \sin B & \sin B\cos A \\ \sin C\cos B & \sin C\cos A & \sin C \end{vmatrix}

Now with $R_1 \rightarrow R_1+R_2+R_3$ we see for instance that $$a_{11} = \sin A +\sin B \cos C+\sin C \cos B = \sin A+\sin (B+C) = \sin A - \sin A =0$$

i.e. the first row becomes null and hence the determinant equals zero.