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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants4 Marks
Question
Prove the following identities:
$\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(5\text{x}+\lambda)(\lambda-\text{x})^2$
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Answer

$\text{L.H.S}=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}-\text{x}-\lambda&\text{x}+\lambda-2\text{x}&0\\2\text{x}-\text{x}-\lambda&0&\text{x}+\lambda-2\text{x}\end{vmatrix}$ [Applying R2 → R2 - R1 and R3 → R3 - R1]
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-(\lambda-\text{x})&\lambda-\text{x}&0\\-(\lambda-\text{x})&0&\lambda-\text{x}\end{vmatrix}$
$=(\lambda-\text{x})^2\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-1&1&0\\-1&0&1\end{vmatrix}$ [Taking $(\lambda-\text{x})$ common from R2 and $(\lambda-\text{x})$ common from R3]
$=(\lambda-\text{x})^2[-1(-2\text{x})+1(\text{x}+\lambda+2\text{x})]$ [Expanding along last row]
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
$=\text{R.H.S}$
$\because\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(\lambda-\text{x})^2(\lambda+5\text{x})$

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