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Home  >>  CBSE XII  >>  Math  >>  Determinants
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if $x=-9$ is a root of $\begin{vmatrix}x & 3 & 7\\2 & x & 2\\7 & 6 & x\end{vmatrix}=0$, then other roots are

$\begin{array}{1 1} -7,-2 \\ 7,-2 \\ -7,2 \\ 7,2 \end{array}$

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1 Answer

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Toolbox:
  • The value of a determinant of order $3\times 3$ can be found as :If $\Delta=\begin{vmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{vmatrix}$
  • $|\Delta|=a_{11}(a_{22}\times a_{33}-a_{23}\times a_{32})-a_{12}(a_{21}\times a_{33}-a_{23}\times a_{31})+a_{13}(a_{21}\times a_{32}-a_{22}\times a_{31})$
$\Delta=\begin{vmatrix}x & 3 & 7\\2 & x & 2\\7 & 6 & x\end{vmatrix}$
Now expanding along $R_1$ to find the value of the determinant,we get
$|\Delta|=x(x^2-12)-3(2x-14)+7(12-7x)$
$\quad=x^3-12x-6x+42+84-49x.$
$\quad=x^3-67x+126.$
It is given that x=-9 is one of the root.
Therefore (x+9) should be a factor.
To find the other factors let us follow the method of synthetic division:
$-9\mid1\; \;0\;\;-67\;\;126$
$\;\;\;\;\;\mid0 \;-9\;\;81\;-126$
____________________________
$\qquad 1\;-9\;14\;\;\; 0$
$x^3-67x+126=(x+9)(x^2-9x+14)$ should be the other factor
If we factorize this we get,
$x^2-9x+14=(x-7)(x-2)$
Hence the factors of $x^3-67x+126$ are (x+9),(x-7),(x-2).
Hence the roots of the equations are
x=7 and x=2.
answered Mar 27, 2013 by sreemathi.v
 

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