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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants4 Marks
Question
Prove the following identities:
$\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
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Answer

$\text{L.H.S}=\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
$=\begin{vmatrix}\text{a}+\text{x}+\text{y}+\text{z}&\text{y}&\text{z}\\\text{a}+\text{x}+\text{y}+\text{z}&\text{a}+\text{y}&\text{z}\\\text{a}+\text{x}+\text{y}+\text{z}&\text{y}&\text{a}+\text{z}\end{vmatrix}$ [Applying C1 → C1 + C2 + C3]
$=(\text{a}+\text{x}+\text{y}+\text{z})\begin{vmatrix}1&\text{y}&\text{z}\\1&\text{a}+\text{y}&\text{z}\\1&\text{y}&\text{a}+\text{z}\end{vmatrix}$
$=(\text{a}+\text{x}+\text{y}+\text{z})\begin{vmatrix}1&\text{y}&\text{z}\\0&\text{a}&0\\0&0&\text{a}\end{vmatrix}$ [Applying R2 → R2 - R1 and R3 → R2 - R1]
$=(\text{a}+\text{x}+\text{y}+\text{z})\text{a}^2$ [Expanding along first column]
$=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$
$\therefore\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$

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