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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Prove that:
$\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}=4\text{abc}$
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Answer

$\text{L.H.S}=\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2&\text{c}^2\\\text{a}^2&\text{b}^2+\text{c}^2&\text{a}^2\\\text{b}^2&\text{b}^2&\text{c}^2+\text{a}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}0&-2\text{b}^2&-2\text{a}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}(-\text{a}^2)\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\\text{b}^2+\text{c}^2-\text{a}^2&0\end{vmatrix}$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2(\text{c}^2+\text{a}^2-\text{b}^2)\}+\text{b}^2\{0+2\text{a}^2(\text{b}^2+\text{c}^2-\text{a}^2)\}\big]$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2\text{c}^2-2\text{b}^2\text{a}^2+2\text{b}^4\}+\text{b}^2\{2\text{a}^2\text{b}^2+2\text{a}^2\text{c}^2-2\text{a}^4\}\big]$
$=\frac{1}{\text{abc}}\big[2\text{a}^2\text{b}^2\text{c}^2+2\text{a}^4\text{b}^2-2\text{a}^4\text{b}^4+2\text{a}^2\text{b}^4+2\text{a}^2\text{b}^2\text{c}^2-2\text{a}^4\text{b}^2\big]$
$=\frac{1}{\text{abc}}4\text{a}^2\text{b}^2\text{c}^2$
$=4\text{abc}$
$=\text{R.H.S}$

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