Determinant involvoing sides and angles of a triangle

Question

If $a, b, c$ are the sides of $\triangle ABC$ and $A, B,C$ are respectively the angles opposite to them, then find the value of

$$ \begin{vmatrix} a^2 & b \sin A & c \cos A \\ b\sin A & 1 & \cos (B-C) \\ c\sin A & \cos (B-C) & 1 \\ \end{vmatrix} $$

I tried using the property $$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} $$ and $c^2 = a^2 +b^2 -2ab \cos C$

But the expressions won't factorize or simplify nicely. What other property should be used to simplify the expressions in order to obtain the determinant ?

Answer 1

Here's a solution featuring matrix manipulations.

(I suspect that the problem's source has a typo in that the $\cos A$ in the upper-right corner should be $\sin A$. I'm going to proceed with steps that would make more sense in that context.)

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First, using the Law of Sines to write

$$a = d \sin A \qquad b = d \sin B \qquad c = d \sin C$$ where $d$ is the circumdiameter of $\triangle ABC$, we have

$$\left| \begin{matrix} d^2 \sin^2 A & d \sin A \sin B & d \cos A \sin C \\ d \sin A \sin B & 1 & \cos(B-C) \\ d \sin A \sin C & \cos(B-C) & 1 \end{matrix} \right| \tag{1}$$

Factoring-out $d \sin A$ from the first row and first column ...

$$d^2 \sin^2 A\;\left| \begin{matrix} 1 & \sin B & \cot A \sin C \\ \sin B & 1 & \cos(B-C) \\ \sin C & \cos(B-C) & 1 \end{matrix} \right| \tag{2}$$

Now, subtract $(\sin B \cdot \text{column}\;1)$ from $\text{column 2}$, noting that $\cos(B-C) = \cos B \cos C + \sin B \sin C$:

$$d^2 \sin^2 A\;\left| \begin{matrix} 1 & 0 & \cot A \sin C \\ \sin B & 1-\sin^2 B & \cos(B-C) \\ \sin C & \cos B \cos C & 1 \end{matrix} \right| \tag{3}$$

Since $1-\sin^2 B = \cos^2 B$, we can factor-out $\cos B$ from column $2$:

$$d^2 \sin^2 A \cos B\;\left| \begin{matrix} 1 & 0 & \cot A \sin C \\ \sin B & \cos B & \cos(B-C) \\ \sin C & \cos C & 1 \end{matrix} \right| \tag{4}$$

We can do similarly by subtracting $(\sin C \cdot \text{column}\;1)$ from $\text{column}\;3$:

$$d^2 \sin^2 A \cos B \cos C\;\left| \begin{matrix} 1 & 0 & \tan C(\cot A -1 ) \\ \sin B & \cos B & \cos B\\ \sin C & \cos C & \cos C \end{matrix} \right| \tag{5}$$

Now, if the original problem's top-right corner had been $c \sin A$, that corner would now be $0$, making the second and third columns match; in that case, the determinant would be $0$.

Taking the problem as presented, however, we must continue ...

Subtracting $\text{column}\;2$ from $\text{column}\;3$: $$d^2 \sin^2 A \cos B \cos C\;\left| \begin{matrix} 1 & 0 & \tan C(\cot A -1 ) \\ \sin B & \cos B & 0\\ \sin C & \cos C & 0 \end{matrix} \right| \tag{6}$$

Finally, expand the determinant by the last column to get $$d^2 \sin^2 A \cos B \cos C \cdot \tan C (\cot A - 1 ) (\sin B \cos C - \cos B \sin C) $$ $$= d^2 \sin A \cos B \sin C (\cos A - \sin A ) \sin(B-C)$$

$$a c \cos B (\cos A - \sin A ) \sin(B-C) \tag{$\star$}$$