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3 and 4 .Determinants and Matrices
medium
જો $\left| {\,\begin{array}{*{20}{c}}{x + 1}&3&5\\2&{x + 2}&5\\2&3&{x + 4}\end{array}\,} \right| = 0$ તો $x =$
A
$1, 9$
B
$-1, 9$
C
$-1, -9$
D
$1, -9$
Solution
(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 $ $x = 1,\,1,\, – 9$,
Standard 12
Mathematics
