Question

The value of the determinant is
$\left|\begin{array}{ccc}cos(A-P)& cos(A-Q)& cos(A-R)\\ cos(B-P)& cos(B-Q)& cos(B-R)\\ cos(C-P)& cos(C-Q)& cos(C-R)\end{array}\right|$

• A. zero for all values of P, Q, R, A, B, and C.
• B. one for all values of P, Q, R, A, B, and C.
• C. zero only if P= Q= R= A= B= C.
• D. one only if P= Q= R= A= B= C.

Answer 1

The correct option is A zero for all values of P, Q, R, A, B, and C.

Let $\Delta =\left|\begin{array}{ccc}cos(A-P)& cos(A-Q)& cos(A-R)\\ cos(B-P)& cos(B-Q)& cos(B-R)\\ cos(C-P)& cos(C-Q)& cos(C-R)\end{array}\right|$
$\Rightarrow \Delta =\left|\begin{array}{ccc}cosAcosP+sinAsinP& cos(A-Q)& cos(A-R)\\ cosBcosP+sinBsinP& cos(B-Q)& cos(B-R)\\ cosCcosP+sinCsinP& cos(C-Q)& cos(C-R)\end{array}\right|$
$\Rightarrow \Delta =\left|\begin{array}{ccc}cosAcosP& cos(A-Q)& cos(A-R)\\ cosBcosP& cos(B-Q)& cos(B-R)\\ cosCcosP& cos(C-Q)& cos(C-R)\end{array}\right|+\left|\begin{array}{ccc}sinAsinP& cos(A-Q)& cos(A-R)\\ sinBsinP& cos(B-Q)& cos(B-R)\\ sinCsinP& cos(C-Q)& cos(C-R)\end{array}\right|$
$\Rightarrow \Delta =cosP\left|\begin{array}{ccc}cosA& cos(A-Q)& cos(A-R)\\ cosB& cos(B-Q)& cos(B-R)\\ cosC& cos(C-Q)& cos(C-R)\end{array}\right|+sinP\left|\begin{array}{ccc}sinA& cos(A-Q)& cos(A-R)\\ sinB& cos(B-Q)& cos(B-R)\\ sinC& cos(C-Q)& cos(C-R)\end{array}\right|$
Applying $C_2\to C_2-C_{1}cosQ,C_3\to C_3-C_{1}cosR$ in first determinant and $C_2\to C_2-C_1$ sin Q and in second determinant
$\Rightarrow \Delta =cosP\left|\begin{array}{ccc}cosA& sinAsinQ& sinAsinR\\ cosB& sinBsinQ& sinBsinR\\ cosC& sinCsinQ& sinCsinR\end{array}\right|+sinP\left|\begin{array}{ccc}sinA& cosAcosQ& cosAcosR\\ sinB& cosBcosQ& cosBcosR\\ sinC& cosCcosQ& cosCcosR\end{array}\right|\Delta =cosPsinQsinR\left|\begin{array}{ccc}cosA& sinA& sinA\\ cosB& sinB& sinB\\ cosC& sinC& sinC\end{array}\right|+sinPcosQcosR\left|\begin{array}{ccc}sinA& cosA& cosA\\ sinB& cosB& cosB\\ sinC& cosC& cosC\end{array}\right|$
$\Delta$ = 0 + 0 = 0