The solutions of operatorname{det}(A-lambda I | vedclass

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Question

(D) The characteristic equation is given by $\operatorname{det}(A-\lambda I_2)=0$.$\left|\begin{array}{cc} 2-\lambda & 3 \ x & y-\lambda \end{array}\

Vedclass · Accepted Answer

$x=-4, y=10$

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(D) The characteristic equation is given by $\operatorname{det}(A-\lambda I_2)=0$.$\left|\begin{array}{cc} 2-\lambda & 3 \ x & y-\lambda \end{array}ight|=0$$(2-\lambda)(y-\lambda)-3x=0$$\lambda^2 - (y+2)\lambda + 2y - 3x = 0$Since the roots are $4$ and $8$,the sum of roots is $4+8=12$ and the product of roots is $4 	imes 8 = 32$.Comparing with $\lambda^2 - (y+2)\lambda + (2y-3x) = 0$:$y+2 = 12 \Rightarrow y=10$.$2y-3x = 32 \Rightarrow 2(10)-3x=32 \Rightarrow 20-3x=32 \Rightarrow -3x=12 \Rightarrow x=-4$.Thus,$x=-4$ and $y=10$.

The solutions of $\operatorname{det}(A-\lambda I_2)=0$ are $4$ and $8$,where $A=\begin{bmatrix} 2 & 3 \\ x & y \end{bmatrix}$. Then:

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