MCQ
If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&3&5\\2&{x + 2}&5\\2&3&{x + 4}\end{array}\,} \right| = 0$, then $ x =$
- A$1, 9$
- B$-1, 9$
- C$-1, -9$
- ✓$1, -9$
✓
Answer
Correct option: D.
$1, -9$
d
(d) By ${C_1} \to {C_1} + {C_2} + {C_3}$,
(d) By ${C_1} \to {C_1} + {C_2} + {C_3}$,
we have $(9 + x)$ $\left| {\,\begin{array}{*{20}{c}}1&3&5\\1&{x + 2}&5\\1&3&{x + 4}\end{array}\,} \right|$ = 0
$ \Rightarrow $ $(x + 9)$ $\left| {\,\begin{array}{*{20}{c}}0&{1 - x}&0\\0&{ - (1 - x)}&{1 - x}\\1&3&{x + 4}\end{array}\,} \right| = 0$
$ \Rightarrow $ $(x + 9)$ ${(1 - x)^2}\left| {\,\begin{array}{*{20}{c}}0&1&0\\0&{ - 1}&1\\1&3&{x + 4}\end{array}\,} \right| = 0$
$ \Rightarrow $ $x = 1,\,1,\, - 9$,
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