Let A = begin{bmatrix} a & b c & d end{bmatr | vedclass

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(A) The characteristic equation of matrix $A$ is given by $\det(A - \lambda I_2) = 0$. For a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \

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$(a+d)^2 < 4$

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(A) The characteristic equation of matrix $A$ is given by $\det(A - \lambda I_2) = 0$. For a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, the characteristic equation is $\lambda^2 - 	ext{tr}(A)\lambda + \det(A) = 0$. Given $\det(A) = 1$, the equation becomes $\lambda^2 - (a+d)\lambda + 1 = 0$. For this quadratic equation to have imaginary roots, its discriminant $D$ must be less than $0$. The discriminant $D$ is given by $D = [-(a+d)]^2 - 4(1)(1) = (a+d)^2 - 4$. Setting $D < 0$, we get $(a+d)^2 - 4 < 0$, which implies $(a+d)^2 < 4$.

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be a $2 \times 2$ real matrix with $\det(A) = 1$. If the equation $\det(A - \lambda I_2) = 0$ has imaginary roots (where $I_2$ is the identity matrix of order $2$), then:

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