. If $f(x) = \left| {\begin{array}{llllllllllllllllllll}{{{(1 + x)}^{17}}}&{{{(1 + x)}^{19}}}&{{{(1 + x)}^{23}}}\\{{{(1 + x)}^{23}}}&{{{(1 + x)}^{29}}}&{{{(1 + x)}^{34}}}\\{{{(1 + x)}^{41}}}&{{{(1 + x)}^{43}}}&{{{(1 + x)}^{47}}}\end{array}} \right|$
$= A + Bx + C{x^2} +$……….., then $A$ is equal to……………
Since,
$f(x) = {(1 + x)^{17}}{(1 + x)^{23}}{(1 + x)^{41}}\left| {\begin{array}{cccccccccccccccccccc}1&{{{(1 + x)}^2}}&{{{(1 + x)}^6}}\\1&{{{(1 + x)}^6}}&{{{(1 + x)}^{11}}}\\1&{{{(1 + x)}^2}}&{{{(1 + x)}^6}}\end{array}} \right| = 0$
[since, ${R_1}$ and ${R_3}$ are identical]
$\therefore$ $A = 0$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Determinants. Curated by Sachin Sharma. Free for all students.