Question

# Show that $\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\\end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)$ using the properties of determinants and row and column operations of determinants.

Hint: We will expand the determinant keeping the basic rule of determinants in mind. Generally, if we have a determinant to be solved as $\left| \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} \right|$, it is evaluated or expanded as following $a\left\{ \left( ei-fh \right) \right\}-b\left\{ \left( di-fg \right) \right\}+c\left\{ \left( dh-eg \right) \right\}$.
We have given a determinant as $\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|$ and we have to show that this determinant, on evaluating, is equal to $\left( a-b \right)\left( b-c \right)\left( c-a \right)$. To show this we will expand the determinant keeping the basic rule of determinant in mind. For example, if we have a determinant to be solved as $\left| \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} \right|$, it is evaluated or expanded as following $a\left\{ \left( ei-fh \right) \right\}-b\left\{ \left( di-fg \right) \right\}+c\left\{ \left( dh-eg \right) \right\}$. We have also to keep in mind the sign rule of determinant in mind which is $\left| \begin{matrix} + & - & + \\ - & + & - \\ + & - & + \\ \end{matrix} \right|$.
Now, given determinant $\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|$, performing the following two row operations ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$ and ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$ on the given determinant, we get $\Delta =\left| \begin{matrix} 0 & a-b & {{a}^{2}}-{{b}^{2}} \\ 0 & b-c & {{b}^{2}}-{{c}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|$.
Now we know that ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$, therefore using it we get $\Delta =\left| \begin{matrix} 0 & a-b & \left( a-b \right)\left( a+b \right) \\ 0 & b-c & \left( b-c \right)\left( b+c \right) \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|$ now we will expand the determinant along column 1, C1 to get $\Delta =1\left[ \left( a-b \right)\left( b-c \right)\left( b+c \right)-\left( b-c \right)\left( a-b \right)\left( a+b \right) \right]$, solving further $\Delta =\left( b-c \right)\left[ \left( a-b \right)\left( b+c \right)-\left( a-b \right)\left( a+b \right) \right]$, opening the brackets we get $\Delta =\left( b-c \right)\left[ ab+ac-{{b}^{2}}-bc-{{a}^{2}}-ab+ab+{{b}^{2}} \right]$, cancelling the similar terms we get $\Delta =\left( b-c \right)\left[ ac+ab-bc-{{a}^{2}} \right]$, taking $\left( c-a \right)$ as common factor out of the middle bracket, we get $\Delta =\left( b-c \right)\left[ a\left( c-a \right)-b\left( c-a \right) \right]$, $\Delta =\left( b-c \right)\left[ \left( c-a \right)\left( a-b \right) \right]$, Therefore , finally we get $\Delta =\left( a-b \right)\left( b-c \right)\left( c-a \right)$.