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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