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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Evaluate:
$\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
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Answer

$\triangle=\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&\lambda&\text{x}\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ $[$Applying $\left.C_1 \rightarrow C_1-C_2, C_2 \rightarrow C_2-C_3\right]$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&0&2\text{x}+\lambda\\0&-\lambda&\text{x}+\lambda\end{vmatrix} [$Applying $R_1→ R_2+ R_3]$
$=\lambda\begin{vmatrix}0&2\text{x}+\lambda\\-\lambda&\text{x}+\lambda\end{vmatrix}+\text{x}\begin{vmatrix}-\lambda&0\\0&-\lambda\end{vmatrix}$
$=\lambda\big[\lambda(2\text{x}+\lambda)\big]+\text{x}\lambda^2$
$=\lambda^2(2\text{x}+\lambda+\lambda^2\text{x})$
$=3\lambda^2\text{x}+\lambda^3$
$=\lambda^2(3\text{x}+\lambda)$

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