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3 and 4 .Determinants and Matrices
hard
If $p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t = $ $\left| {\,\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda }&{\lambda - 1}&{\lambda + 3}\\{\lambda + 1}&{2 - \lambda }&{\lambda - 4}\\{\lambda - 3}&{\lambda + 4}&{3\lambda }\end{array}\,} \right|,$ the value of $t$ is
A
$16$
B
$18$
C
$17$
D
$19$
(IIT-1981)
Solution
(b) Since it is an identity in $\lambda $ so satisfied by every value of $\lambda $.
Now put $\lambda = 0 $ in the given equation, we have $t = \left| {\,\begin{array}{*{20}{c}}0&{ – 1}&3\\1&2&{ – 4}\\{ – 3}&4&0\end{array}\,} \right|\, = – 12 + 30 = 18$.
Standard 12
Mathematics