If A = begin{bmatrix} 3 & 2 1 & 1 end{bmatri | vedclass

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(A) Given $A = \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix}$.The characteristic equation of $A$ is given by $|A - \lambda I| = 0$.$|A - \lambda I| = \

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$(-4, 1)$

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(A) Given $A = \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix}$.The characteristic equation of $A$ is given by $|A - \lambda I| = 0$.$|A - \lambda I| = \begin{vmatrix} 3 - \lambda & 2 \ 1 & 1 - \lambda \end{vmatrix} = 0$.$(3 - \lambda)(1 - \lambda) - (2)(1) = 0$.$3 - 3\lambda - \lambda + \lambda^{2} - 2 = 0$.$\lambda^{2} - 4\lambda + 1 = 0$.By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation. Substituting $\lambda = A$,we get:$A^{2} - 4A + I = 0$.Comparing this with the given equation $A^{2} + xA + yI = 0$,we get:$x = -4$ and $y = 1$.Thus,$(x, y) = (-4, 1)$.

Market	$x, y, z$
$I$	$10,000, 2,000, 18,000$
$II$	$6,000, 20,000, 8,000$

If $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$,then $A^{2} + xA + yI = 0$ for $(x, y)$ is

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