If ax^4+bx^3+cx^2+50x+d = begin{vmatrix} x^3- | vedclass

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(A) To find $\lambda$,we expand the determinant along the third row: $D = -3 \begin{vmatrix} -x & 3x+\lambda \ 3x & x-4 \end{vmatrix} - 4 \begin{vmat

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$2$

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(A) To find $\lambda$,we expand the determinant along the third row: $D = -3 \begin{vmatrix} -x & 3x+\lambda \ 3x & x-4 \end{vmatrix} - 4 \begin{vmatrix} x^3-14x^2 & 3x+\lambda \ 4x+1 & x-4 \end{vmatrix} + 0 \begin{vmatrix} x^3-14x^2 & -x \ 4x+1 & 3x \end{vmatrix}$ $= -3[-x(x-4) - 3x(3x+\lambda)] - 4[(x^3-14x^2)(x-4) - (4x+1)(3x+\lambda)]$ $= -3[-x^2+4x - 9x^2 - 3x\lambda] - 4[x^4-4x^3-14x^3+56x^2 - (12x^2+4x\lambda+3x+\lambda)]$ $= -3[-10x^2+4x-3x\lambda] - 4[x^4-18x^3+44x^2-4x\lambda-3x-\lambda]$ $= 30x^2-12x+9x\lambda - 4x^4+72x^3-176x^2+16x\lambda+12x+4\lambda$ $= -4x^4+72x^3-146x^2+(25\lambda)x+4\lambda$ Comparing this with $ax^4+bx^3+cx^2+50x+d$,we equate the coefficients of $x$: $25\lambda = 50$ $\lambda = 2$.

If $ax^4+bx^3+cx^2+50x+d = \begin{vmatrix} x^3-14x^2 & -x & 3x+\lambda \\ 4x+1 & 3x & x-4 \\ -3 & 4 & 0 \end{vmatrix}$,then find $\lambda$.

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