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Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Using properties of determinants prove that:
$\begin{vmatrix}\text{x}+4&2\text{x}&2\text{x}\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$
$=(5\text{x}+4)(4-\text{x})^2$
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Answer

$\text{L.H.S}=\begin{vmatrix}\text{x}+4&2\text{x}&2\text{x}\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}5\text{x}+4&5\text{x}+4&5\text{x}+4\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=5\text{x}+4\begin{vmatrix}1&1&1\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Take out 5x + 4 common from $R_1]$
$=5\text{x}+4\begin{vmatrix}1&0&0\\2\text{x}&4-\text{x}&0\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=5\text{x}+4(4-\text{x})^2$

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