   Question Answers

# 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 $a\left\{ \left( ei-fh \right) \right\}-b\left\{ \left( di-fg \right) \right\}+c\left\{ \left( dh-eg \right) \right\}$.

Complete step-by-step solution -
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 $\Delta =\left( a-b \right)\left( b-c \right)\left( c-a \right)$.
Thus LHS=RHS, Hence proved.

Note: Usually students expand the given determinant as it is and end up making mistakes. It takes up a lot of time and chances of missing terms or changing the sign of terms is also higher. Therefore it is recommended to do some row operation on row and column of the determinant to make it easier to solve our question.

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