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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.