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Without expanding the determinant, prove that: \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|=\left|\begin{array}{lll} 1 & b c & b+c \\ 1 & c a & c+a \\ 1 & a b & a+b \end{array}\right|
CBSE | Class 12 | Determinants | Exercise 6B | Maths | RS Aggarwal
Without expanding the determinant, prove that: \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|=\left|\begin{array}{lll} 1 & b c & b+c \\ 1 & c a & c+a \\ 1 & a b & a+b \end{array}\right|

Solution:

\text { Operating } R 1 \rightarrow R 1^{-} R_{3}, R_{2 \rightarrow R_{2}-R 3}
\begin{array}{l} \left|\begin{array}{ccc} 0 & \mathrm{a}-\mathrm{c} & \mathrm{a}^{2}-\mathrm{c}^{2} \\ 0 & \mathrm{~b}-\mathrm{c} & \mathrm{b}^{2}-\mathrm{c}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2} \end{array}\right|=\left|\begin{array}{ccc} 0 & \mathrm{~b} \mathrm{c}-\mathrm{ab} & \mathrm{b}+\mathrm{c}-\mathrm{a}-\mathrm{b} \\ 0 & \mathrm{ac}-\mathrm{ab} & \mathrm{c}+\mathrm{a}-\mathrm{a}-\mathrm{b} \\ 1 & \mathrm{ab} & \mathrm{a}+\mathrm{b} \end{array}\right| \\ \rightarrow\left|\begin{array}{ccc} 0 & \mathrm{a}-\mathrm{c} & (\mathrm{a}-\mathrm{c})(\mathrm{a}+\mathrm{c}) \\ 0 & \mathrm{~b}-\mathrm{c} & (\mathrm{b}-\mathrm{c})(\mathrm{b}+\mathrm{c}) \\ 1 & \mathrm{c} & \mathrm{c}^{2} \end{array}\right|=\left|\begin{array}{ccc} 0 & -\mathrm{b}(\mathrm{a}-\mathrm{c}) & -(\mathrm{a}-\mathrm{c}) \\ 0 & -\mathrm{a}(\mathrm{b}-\mathrm{c}) & -(\mathrm{b}-\mathrm{c}) \\ 1 & \mathrm{ab} & \mathrm{a}+\mathrm{b} \end{array}\right| \end{array}
Taking (a-c) and (b-c) common from R_{1}, R_{2}
\rightarrow(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ccc} 0 & 1 & \mathrm{a}+\mathrm{c} \\ 0 & 1 & \mathrm{~b}+\mathrm{c} \\ 1 & \mathrm{c} & \mathrm{c}^{2} \end{array}\right|=(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ccc} 0 & -\mathrm{b} & -1 \\ 0 & -\mathrm{a} & -1 \\ 1 & \mathrm{ab} & \mathrm{a}+\mathrm{b} \end{array}\right|
Method 1:
If the determinants are equal, their difference must also be equal. (a-c) and (b-c) get cancelled.
\begin{array}{l} =\left|\begin{array}{ccc} 0 & 1 & a+c \\ 0 & 1 & b+c \\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{ccc} 0 & -b & -1 \\ 0 & -a & -1 \\ 1 & a b & a+b \end{array}\right| \\ =\left|\begin{array}{ccc} 0-0 & 1+b & a+c+1 \\ 0-0 & 1+a & b+c+1 \\ 1-1 & c-a b & c^{2}+a+b \end{array}\right| \\ =\left|\begin{array}{ccc} 0 & 1+b & a+c+1 \\ 0 & 1+a & b+c+1 \\ 0 & c-a b & c^{2}+a+b \end{array}\right| \end{array}
Since all elements of \mathrm{C}_{1} are 0 , by properties of determinants,
=0
\therefore The 2 determinants are equal.
Method 2:
Expanding with C_{1}
\begin{array}{l} \rightarrow(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})(\mathrm{b}+\mathrm{c}-\mathrm{a}-\mathrm{c})=(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})(\mathrm{b}-\mathrm{a}) \\ \rightarrow(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})(\mathrm{b}-\mathrm{a})=(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})(\mathrm{b}-\mathrm{a}) \end{array}
\therefore RHS and LHS are equal

i

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