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Determinants — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDeterminants5 Marks
Question
Prove that: $\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}\text{a}^2&-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C1]$
$=(-1)\begin{vmatrix}\text{a}^2&(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=-\begin{vmatrix}\text{a}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - 2C_1]$
$=-\begin{vmatrix}\text{a}^2+\text{b}^2+\text{c}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2+\text{c}^2+\text{a}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2+\text{a}^2+\text{b}&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 + C_2]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\1&\text{c}^2+\text{a}^2&\text{ca}\\1&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}^2-\text{b}^2&\text{c}(\text{a}-\text{b})\\0&\text{a}^2-\text{c}^2&\text{b}(\text{a}-\text{c})\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{a}-\text{c})\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}+\text{b}&\text{c}\\0&\text{a}+\text{c}&\text{b}\end{vmatrix} $
$[$Taking $(a - b)$ common from $R_2$ and $(a - c)$ common from $R_3]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\times\left\{1\times\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{a}+\text{c}&\text{b}\end{vmatrix}\right\}$
$[\because(\text{c}-\text{a})=-(\text{a}-\text{c})] [$Expanding along $C_1]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})(\text{ab}+\text{b}^2-\text{ac}-\text{c}^2)$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\{\text{a}(\text{b}-\text{c})+(\text{b}+\text{c})(\text{b}-\text{c})\}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
Hence prove.

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