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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If $f(x)$ = $\small\begin{vmatrix}(1+x)^{17} & (1+x)^{19} & (1+x)^{23}\\(1+x)^{23} & (1+x)^{29} & (1+x)^{34}\\(1+x)^{41} & (1+x)^{43} & (1+x)^{47}\end{vmatrix}$ = $A+Bx+Cx^2+.....$, then $A$ =

$\begin{array}{1 1} 1 \\ -1 \\ 0 \\ \pm 1 \end{array} $

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Toolbox:
  • If $\Delta=\begin{vmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{vmatrix}$
  • $|\Delta|=a_{11}(a_{22}\times a_{33}-a_{23}\times a_{32})-a_{12}(a_{21}\times a_{33}-a_{23}\times a_{31})+a_{13}(a_{21}\times a_{32}-a_{22}\times a_{31})$
  • If each element of a row or a column of a determinant is multiplied by a constant k,then the value gets multiplied by k.
Given:$\begin{vmatrix}(1+x)^{17} & (1+x)^{19} & (1+x)^{23}\\(1+x)^{23} & (1+x)^{29} & (1+x)^{34}\\(1+x)^{41} & (1+x)^{43} & (1+x)^{47}\end{vmatrix}$
Let us take $(1+x)^{17}$ as the common factor from $C_1,(1+x)^{19}$ from $C_3$
$\;\;=(1+x)^{17}(1+x)^{19}(1+x)^{23}\begin{vmatrix}1 &1 &1 \\(1+x)^{6} & (1+x)^{10} & (1+x)^{11}\\(1+x)^{24} & (1+x)^{24} & (1+x)^{24}\end{vmatrix}$
Taking $(1+x)^{24}$ as a common factor from $R_3$
$\;\;=(1+x)^{17}(1+x)^{19}(1+x)^{23}(1+x)^{24} \begin{vmatrix}1 &1 &1 \\(1+x)^{6} & (1+x)^{10} & (1+x)^{11}\\1& 1 & 1\end{vmatrix}$
Since two rows are identical the value of the determinant is 0.
Therefore $A+Bx+Cx^2+.......=0$
Since it is of the form of a binomial expansion of $(1+x)^n=1+nx+nC_1x^2+nC_2x^3+......$
The value of A=1.
answered Mar 27, 2013 by sreemathi.v
 

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