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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS1 Mark
Question
Fill in the blanks: If $\text{f(x)}=\begin{vmatrix}(1+\text{x})^{17}&(1+\text{x})^{19}&(1+\text{x})^{23}\\(1+\text{x})^{23}&(1+\text{x})^{29}&(1+\text{x})^{34}\\(1+\text{x})^{41}&(1+\text{x})^{43}&(1+\text{x})^{47}\end{vmatrix}=\text{A}+\text{Bx}+\text{Cx}^2+\ \dots,$ then A = ________.
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Answer

If $\text{f(x)}=\begin{vmatrix}(1+\text{x})^{17}&(1+\text{x})^{19}&(1+\text{x})^{23}\\(1+\text{x})^{23}&(1+\text{x})^{29}&(1+\text{x})^{34}\\(1+\text{x})^{41}&(1+\text{x})^{43}&(1+\text{x})^{47}\end{vmatrix}=\text{A}+\text{Bx}+\text{Cx}^2+\ \dots,$ then A = 0. Solution: We have, $\text{f(x)}=\begin{vmatrix}(1+\text{x})^{17}&(1+\text{x})^{19}&(1+\text{x})^{23}\\(1+\text{x})^{23}&(1+\text{x})^{29}&(1+\text{x})^{34}\\(1+\text{x})^{41}&(1+\text{x})^{43}&(1+\text{x})^{47}\end{vmatrix}$ [Taking $(1+x)^{17},(1+x)^{23},(1+x)^{41}$ common from $R _1, R _2, R _3$ respectively]
$=(1+\text{x})^{17}(1+\text{x})^{23},(1+\text{x})^{41}\begin{vmatrix}1&(1+\text{x})^{2}&(1+\text{x})^{6}\\1&(1+\text{x})^{6}&(1+\text{x})^{11}\\1&(1+\text{x})^{2}&(1+\text{x})^{6}\end{vmatrix}=0$ $\big[\because\ \text{R}_1\text{ and R}_3\text{ are identical}\big]$ $\therefore\ \text{A}=0$

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