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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
If $\begin{bmatrix}4-\text{x}&4+\text{x}&4+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0,$ then find values of $x$.
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Answer

Given, $\begin{bmatrix}4-\text{x}&4+\text{x}&4+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}12+\text{x}&12+\text{x}&12+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0$ $[\because\ \text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3]$
$\Rightarrow\ (12+\text{x})\begin{bmatrix}1&1&1\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0$ [Taking (12 + x) common from $R_1$]
$\Rightarrow\ (12+\text{x})\begin{vmatrix}0&0&0\\0&8&4+\text{x}\\2\text{x}&8&4-\text{x}\end{vmatrix}=0$ $\big[\because\text{C}_1\rightarrow\text{C}_ 1-\text{C}_ 3\text{ and }\text{C}_ 2\rightarrow\text{C}_ 2+\text{C}_ 3]$
$\Rightarrow\ (12+\text{x})\big[1.(-16\text{x})\big]=0$
$\Rightarrow\ (12+\text{x})(-16\text{x})=0$
$\therefore\ \text{x}=-12,0$

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