Question

# Show that $$\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 } & b^{ 2 } & c^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)$$.

Solution

## $$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ {a}^{2} & {b}^{2} & {c}^{2} \end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right)$$Solving determinant, we have$$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ {a}^{2} & {b}^{2} & {c}^{2} \end{vmatrix}$$Applying $${C}_{2} \rightarrow {C}_{2} - {C}_{1}$$ and $${C}_{3} \rightarrow {C}_{3} - {C}_{1}$$$$= \begin{vmatrix} 1 & 0 & 0 \\ a & b - a & c - a \\ {a}^{2} & {b}^{2} - {a}^{2} & {c}^{2} - {a}^{2} \end{vmatrix}$$Taking $$\left( c - a \right)$$ and $$\left( b - a \right)$$ as common from $${C}_{3}$$ and $${C}_{2}$$ respectively.$$= \left( b - a \right) \left( c - a \right) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ {a}^{2} & b + a & c + a \end{vmatrix}$$Expanding the determinant along $${R}_{1}$$, we have$$= \left( b - a \right) \left( c - a \right) \left( 1 \left( c + a - b - a \right) - 0 + 0 \right)$$$$= \left( b - a \right) \left( c - a \right) \left( c - b \right)$$$$= \left( a - b \right) \left( b - c \right) \left( c - a \right)$$$$=$$ R.H.S.Hence proved.AApplied Mathematics

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