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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
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 ${C_1} \to {C_1} + {C_2} + {C_3}$,

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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