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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Using the properties of determinants:
$\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
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Answer

We have, $\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$ $=\begin{vmatrix}2\text{x}+4&2\text{x}+4&2\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$ $\big[\text{R}_1\rightarrow\text{R}_1+\text{R}_2\big]$$\begin{vmatrix}2\text{x}&2\text{x}&2\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}+\begin{vmatrix}4&4&4\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$=2\text{x}\begin{vmatrix}1&1&1\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}+4\begin{vmatrix}1&1&0\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$ [Taking 2x and 4 common from $R_1$]
$=2\text{x}\begin{vmatrix}0&0&1\\0&4&\text{x}\\-4&-4&\text{x}+4\end{vmatrix}+4\begin{vmatrix}0&1&0\\-4&\text{x}+4&\text{x}\\0&\text{x}&\text{x}+4\end{vmatrix}$ [$C_1 \rightarrow C_1 - C_3$ and $C_2 \rightarrow C_2 - C_3$ in first and $C_1 \rightarrow C_1 - C_2$ in second determinant]
Expanding along $C_1$, we get
$2\text{x}[-4(-4)]+4[4(\text{x}+4-0)]$
$=32\text{x}+16\text{x}+64$
$=16(3\text{x}+4)$

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