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(C) Let the given determinants be $D_1, D_2$ and $D_3$ respectively. $D_1 = \left|\begin{array}{ccc}x & 4 & 6 \ 2 & 3 & -9 \ 5 & 6 & 1\end{array}i

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Let the given determinants be $D_1, D_2$ and $D_3$ respectively. $D_1 = \left|\begin{array}{ccc}x & 4 & 6 \ 2 & 3 & -9 \ 5 & 6 & 1\end{array}ight| = x(3+54) - 4(2+45) + 6(12-15) = 57x - 188 - 18 = 57x - 206$. $D_2 = \left|\begin{array}{ccc}5 & 6 & 1 \ 6 & 4 & 5 \ 2 & 3 & -9\end{array}ight| = 5(-36-15) - 6(-54-10) + 1(18-8) = 5(-51) - 6(-64) + 10 = -255 + 384 + 10 = 139$. $D_3 = \left|\begin{array}{ccc}2 & 3 & -9 \ 1-2x & -8 & -11 \ 5 & 6 & 1\end{array}ight| = 2(-8+66) - 3(1-2x+55) - 9(6-12x+40) = 2(58) - 3(56-2x) - 9(46-12x) = 116 - 168 + 6x - 414 + 108x = 114x - 466$. Given $D_1 + D_2 = D_3$,so $(57x - 206) + 139 = 114x - 466$. $57x - 67 = 114x - 466$. $466 - 67 = 114x - 57x$. $399 = 57x$. $x = \frac{399}{57} = 7$. Thus,the correct option is $C$.

If $\left|\begin{array}{ccc}x & 4 & 6 \\ 2 & 3 & -9 \\ 5 & 6 & 1\end{array}\right|+\left|\begin{array}{ccc}5 & 6 & 1 \\ 6 & 4 & 5 \\ 2 & 3 & -9\end{array}\right|=\left|\begin{array}{ccc}2 & 3 & -9 \\ 1-2 x & -8 & -11 \\ 5 & 6 & 1\end{array}\right|$,then $x=$ . . . . . .

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