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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Prove that:
$\begin{vmatrix}\text{a}&\text{b}-\text{c}&\text{c}-\text{b}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{a}+\text{b}-\text{c})(\text{b}+\text{c}-\text{a})(\text{c}+\text{a}-\text{b})$
✓

Answer

$\text{L.H.S}=\begin{vmatrix}\text{a}&\text{b}-\text{c}&\text{c}-\text{b}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$R_1 \rightarrow R_1-R_2-R_3$
$=\begin{vmatrix}-\text{a}+\text{c}+\text{b}&-\text{b}-\text{c}+\text{a}&-\text{c}-\text{b}+\text{a}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{b}+\text{c}-\text{a})\begin{vmatrix}1&-1&1\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{b}+\text{c}-\text{a})\begin{vmatrix}1&0&0\\\text{a}-\text{c}&\text{b}+\text{a}-\text{c}&0\\\text{a}-\text{b}&0&\text{c}+\text{a}-\text{b}\end{vmatrix}$
$=(\text{a}+\text{b}-\text{c})(\text{b}+\text{c}-\text{a})(\text{c}+\text{a}-\text{b})$
$=\text{R.H.S}$

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