Let A be a symmetric matrix such that |A|=2 a | vedclass

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(A) Let $A = \begin{bmatrix} a & b \ b & c \end{bmatrix}$ since it is a symmetric matrix.Given $|A| = ac - b^2 = 2$.From the matrix equation $\begin{

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$5$

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(A) Let $A = \begin{bmatrix} a & b \ b & c \end{bmatrix}$ since it is a symmetric matrix.Given $|A| = ac - b^2 = 2$.From the matrix equation $\begin{bmatrix} 2 & 1 \ 3 & \frac{3}{2} \end{bmatrix} \begin{bmatrix} a & b \ b & c \end{bmatrix} = \begin{bmatrix} 1 & 2 \ \alpha & \beta \end{bmatrix}$,we get:$2a + b = 1 \Rightarrow b = 1 - 2a$$2b + c = 2 \Rightarrow c = 2 - 2b = 2 - 2(1 - 2a) = 4a$Substitute $b$ and $c$ into $ac - b^2 = 2$:$a(4a) - (1 - 2a)^2 = 2$$4a^2 - (1 - 4a + 4a^2) = 2$$4a^2 - 1 + 4a - 4a^2 = 2$$4a = 3 \Rightarrow a = \frac{3}{4}$Then $b = 1 - 2(\frac{3}{4}) = 1 - \frac{3}{2} = -\frac{1}{2}$ and $c = 4(\frac{3}{4}) = 3$.Now,calculate $\alpha$ and $\beta$:$\alpha = 3a + \frac{3}{2}b = 3(\frac{3}{4}) + \frac{3}{2}(-\frac{1}{2}) = \frac{9}{4} - \frac{3}{4} = \frac{6}{4} = \frac{3}{2}$$\beta = 3b + \frac{3}{2}c = 3(-\frac{1}{2}) + \frac{3}{2}(3) = -\frac{3}{2} + \frac{9}{2} = \frac{6}{2} = 3$Sum of diagonal elements $s = a + c = \frac{3}{4} + 3 = \frac{15}{4}$.Finally,$\frac{\beta s}{\alpha^2} = \frac{3 	imes \frac{15}{4}}{(\frac{3}{2})^2} = \frac{\frac{45}{4}}{\frac{9}{4}} = 5$.

Let $A$ be a symmetric matrix such that $|A|=2$ and $\begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix} A = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$. If the sum of the diagonal elements of $A$ is $s$,then $\frac{\beta s}{\alpha^2}$ is equal to $..........$.

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