Question

If $\alpha ,\beta ,\gamma$ are three real numbers then the matrix A given below is
$\left[\begin{array}{ccc}1& cos(\alpha -\beta )& cos(\alpha -\gamma \\ cos(\beta -\gamma )& 1& cos(\beta -\gamma )\\ cos(\gamma -\alpha )& cos(\gamma -\beta )& 1\end{array}\right]$

• A. singular
• B. symmetric
• C. invertible
• D. none

Answer 1

Answer: (A), (B)

singular
symmetric

Since $a_{ij}=a_{ji}\forall i,j.\therefore$ A is symmetric

Put $\beta -\gamma =A,\gamma -\alpha =B,\alpha -\beta =C$

So that $A+B+C=0$

or $B+C=-A$

$\therefore$ $cos(B+C)=cos(-A)=cosA$

$sin(B+C)=-sinA$ ............. (1)

$|A|$ = $(1-co{s}^{2}A)-cosC(cosC-cosAcosB)+cosB[cosCcosA-cosB]$

=$si{n}^{2}A-cosC\left[cos\right(A+B)-cosAcosB]+cosB[cosCcosA-cos(C+A\left)\right]$ by (1)

=$si{n}^{2}A-cosC[-sinAsinB]+cosB\left(sinCsinA\right)$

= $si{n}^{2}A+sinAsin(B+C)$

= $si{n}^{2}A+sinA(-sinA)=0$ by (1)

Since $|A|=0$, the matrix $A$ is singular.