3 and 4 .Determinants and Matrices
easy

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&1&1\\2&{x + 2}&2\\3&3&{x + 3}\end{array}\,} \right| = 0,$ then $x$ is

A

$0, -6$

B

$0, 6$

C

$6$

D

$-6$

Solution

(a) From options put $x = 0,\,6$ and $-6$. Then only option $ (a)$ is satisfying.

Standard 12
Mathematics

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