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3 and 4 .Determinants and Matrices
easy
If the matrix $\left[ {\begin{array}{*{20}{c}}1&3&{\lambda + 2}\\2&4&8\\3&5&{10}\end{array}} \right]$ is singular, then $\lambda = $
A
$-2$
B
$4$
C
$2$
D
$-4$
Solution
(b) The matrix $\left[ {\begin{array}{*{20}{c}}1&3&{\lambda + 2}\\2&4&8\\3&5&{10}\end{array}} \right]$ is singular,
If $\,\left| {\,\begin{array}{*{20}{c}}1&3&{\lambda + 2}\\2&4&8\\3&5&{10}\end{array}\,} \right| = 0$
==> $1(40 – 40) – 3(20 – 24) + (\lambda + 2)(10 – 12) = 0$
==> $2(\lambda + 2) = 12 \Rightarrow \lambda = 4.$
Standard 12
Mathematics