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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Prove that:
$\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}=16(3\text{x}+4)$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}3\text{x}+4&3\text{x}+4&3\text{x}+4\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$[$Applying $R_1 → R_2 + R_2 + R_3]$
$=(3\text{x}+4)\begin{vmatrix}1&1&1\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$[$Taking out $(3x + 4)$ common from $R_1]$
$=(3\text{x}+4)\begin{vmatrix}1&0&0\\\text{x}&4&0\\\text{x}&0&4\end{vmatrix}$
$[$Applying $C_2 → C_2 - C_1$ and $C_3 → C_3 - C_1]$
$=(3\text{x}+4)(4)^2 [$Expanding along $R_1]$
$=16(3\text{x}+4)$
$=\text{R.H.S}$

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