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(D) Given the determinant equation: $\left| \begin{array}{ccc} x+1 & 3 & 5 \ 2 & x+2 & 5 \ 2 & 3 & x+4 \end{array} ight| = 0$. Applying the column

Vedclass · Accepted Answer

$1, -9$

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(D) Given the determinant equation: $\left| \begin{array}{ccc} x+1 & 3 & 5 \ 2 & x+2 & 5 \ 2 & 3 & x+4 \end{array} ight| = 0$. Applying the column operation $C_1 	o C_1 + C_2 + C_3$: $\left| \begin{array}{ccc} x+9 & 3 & 5 \ x+9 & x+2 & 5 \ x+9 & 3 & x+4 \end{array} ight| = 0$. Taking $(x+9)$ common from $C_1$: $(x+9) \left| \begin{array}{ccc} 1 & 3 & 5 \ 1 & x+2 & 5 \ 1 & 3 & x+4 \end{array} ight| = 0$. Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$: $(x+9) \left| \begin{array}{ccc} 1 & 3 & 5 \ 0 & x-1 & 0 \ 0 & 0 & x-1 \end{array} ight| = 0$. Expanding along $C_1$: $(x+9) \cdot 1 \cdot [(x-1)(x-1) - 0] = 0$. $(x+9)(x-1)^2 = 0$. Thus,$x = -9$ or $x = 1$. Therefore,the correct option is $D$.

If $\left| \begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array} \right| = 0$,then $x =$

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