Question

prove without expanding that deteminant with row 1 as 0 a -b ; row 2 as -a 0 -c ; row 3 as b c 0 = 0.

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}0& a& -b\\ -a& 0& -c\\ b& c& 0\end{array}\right] \\ Hencewecansaythatitisaskewsymmemtricmatrixoforder3\times 3as{a}_{ij}=-a_{ji} \\ Nowdeterminantofaskewmatrixiszero \\ Proof:- \\ Asweknowdet\left(A\right)=det\left(A^T\right)anddet\left(A^T\right)=det(-A)(forskewsymmetricmatrix) \\ Nowdet(-A)=(-1{)}^{n}det(A) \\ Hencedet(A)=(-1{)}^{n}det\left(A\right) \\ heren=3whichisodd \\ \Rightarrow det\left(A\right)=-det\left(A\right) \\ \Rightarrow 2det\left(A\right)=0 \\ \Rightarrow det\left(A\right)=0 \\ Hencethedeterminantofthegivenmatrixis0asitisaskewsymmetricmatrixofordern\times andnisodd \end{gathered}$$