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# Given that $x=-9$ is a root of $\begin{vmatrix}x&3&7\\2 &x&2\\7&6&x\end{vmatrix}$ = 0, the other two roots are

$\begin{array}{1 1}(A)1,7 \\ (B) 3,7 \\ (C) 2,7 \\(D) 7,0 \end{array}$

Given :
$\begin{vmatrix}x&3&7\\2 &x&2\\7&6&x\end{vmatrix}=0$
Operating $R_1\rightarrow R_1+R_2+R_3$ we get,
$\begin{vmatrix}x+9&x+9&x+9\\2&x&2\\7 &6&x\end{vmatrix}=0$
$\Rightarrow (x+9)\begin{vmatrix}1&1&1\\2&x&2\\7&6&x\end{vmatrix}$=0
Operating $C_2\rightarrow C_2-C_1$
$C_3\rightarrow C_3-C_1$
$\Rightarrow (x+9)\begin{vmatrix}1 &0&0\\2 &x-2&0\\7 &-1&x-7\end{vmatrix}=0$
Expanding along $R_1$
$\Rightarrow (x+9)(x-2)(x-7)=0$
$\Rightarrow x=-9,2,7$
Hence the other two roots are 2 and 7.
Hence (c) is the correct answer.