If p{lambda ^4} + q{lambda ^3} + r{lambda ^2} | vedclass

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Question

(B) Given the identity $p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t = \left| {\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda }&{\lam

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) Given the identity $p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t = \left| {\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda }&{\lambda - 1}&{\lambda + 3}\{\lambda + 1}&{2 - \lambda }&{\lambda - 4}\{\lambda - 3}&{\lambda + 4}&{3\lambda }\end{array}} ight|$.Since this is an identity in $\lambda$,it holds true for all values of $\lambda$.To find the value of $t$,we substitute $\lambda = 0$ into the equation:$t = \left| {\begin{array}{*{20}{c}}0&{ - 1}&3\1&2&{ - 4}\{ - 3}&4&0\end{array}} ight|$.Expanding the determinant along the first row:$t = 0(0 - (-16)) - (-1)(0 - 12) + 3(4 - (-6))$$t = 0 + 1(-12) + 3(10)$$t = -12 + 30 = 18$.Thus,the value of $t$ is $18$.

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

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