If px^4 + qx^3 + rx^2 + sx + t equiv left| be | vedclass

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(C) Given the identity $px^4 + qx^3 + rx^2 + sx + t = \left| \begin{array}{ccc} x^2 + 3x & x - 1 & x + 3 \ x + 1 & 2 - x & x - 3 \ x - 3 & x + 4 & 3

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$21$

Vedclass · Answer

(C) Given the identity $px^4 + qx^3 + rx^2 + sx + t = \left| \begin{array}{ccc} x^2 + 3x & x - 1 & x + 3 \ x + 1 & 2 - x & x - 3 \ x - 3 & x + 4 & 3x \end{array} ight|$.To find the value of $t$,we set $x = 0$ in the identity.Substituting $x = 0$ on both sides:$p(0)^4 + q(0)^3 + r(0)^2 + s(0) + t = \left| \begin{array}{ccc} 0^2 + 3(0) & 0 - 1 & 0 + 3 \ 0 + 1 & 2 - 0 & 0 - 3 \ 0 - 3 & 0 + 4 & 3(0) \end{array} ight|$$t = \left| \begin{array}{ccc} 0 & -1 & 3 \ 1 & 2 & -3 \ -3 & 4 & 0 \end{array} ight|$Now,expand the determinant along the first row:$t = 0(2 	imes 0 - (-3) 	imes 4) - (-1)(1 	imes 0 - (-3) 	imes (-3)) + 3(1 	imes 4 - 2 	imes (-3))$$t = 0 - (-1)(0 - 9) + 3(4 + 6)$$t = 0 - (9) + 3(10)$$t = -9 + 30$$t = 21$Thus,the value of $t$ is $21$.

If $px^4 + qx^3 + rx^2 + sx + t \equiv \left| \begin{array}{ccc} x^2 + 3x & x - 1 & x + 3 \\ x + 1 & 2 - x & x - 3 \\ x - 3 & x + 4 & 3x \end{array} \right|$,then $t =$

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