3 and 4 .Determinants and Matrices
easy

In order that the matrix $\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\\3&\lambda &5\end{array}} \right]$ be non-singular, $\lambda $ should not be equal to

A

$1$

B

$2$

C

$3$

D

$4$

Solution

(d) Matrix $\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\\3&\lambda &5\end{array}} \right]$be non singular,

only if $\left| {\,\begin{array}{*{20}{c}}1&2&3\\4&5&6\\3&\lambda &5\end{array}\,} \right| \ne 0$

==> $\,1(25 – 6\lambda ) – 2(20 – 18) + 3(4\lambda – 15) \ne 0$

==> $25 – 6\lambda – 4 + 12\lambda – 45 \ne 0$

==> $6\lambda – 24 \ne 0$ ==> $\lambda \ne 4$.

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

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