# Let matrix $[A]=\left[\begin{array}{cc}2 & -6 \\ 0 & 2\end{array}\right]$. Then for any non-trivial vector $\{x\}=\left\{\begin{array}{l}x_{1} \\ x_{2}\end{array}\right\}$, which of the following is true for the value of $K=\{x\}^{T}[A]\{x\}$ : (A)K is always less than zero (B) $\mathrm{K}$ is always greater thanzero (C) $\mathrm{K}$ is non-negative (D) $\mathrm{K}$ can be anything

## Question ID - 155576 :- Let matrix $[A]=\left[\begin{array}{cc}2 & -6 \\ 0 & 2\end{array}\right]$. Then for any non-trivial vector $\{x\}=\left\{\begin{array}{l}x_{1} \\ x_{2}\end{array}\right\}$, which of the following is true for the value of $K=\{x\}^{T}[A]\{x\}$ : (A)K is always less than zero (B) $\mathrm{K}$ is always greater thanzero (C) $\mathrm{K}$ is non-negative (D) $\mathrm{K}$ can be anything

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(D) $\mathrm{K}$ can be anything

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Consider the initial value problem: $\frac{d^{2} y}{d t^{2}}+4 \frac{d y}{d t}+6 y=f(t) ; y(0)=2,\left(\frac{d y}{d t}\right)_{t=0}=1$
If $Y(s)=\int_{0}^{\infty} y(t) e^{-s t} d t$ and $F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t$ are the Laplace transforms of $y(t)$ and $f(t)$ respectively, then $Y(s)$ is given by:
$\begin{array}{ll} \text { (A) } \frac{F(s)}{\left(s^{2}+4 s+6\right)} & \text { (B) } \frac{F(s)+2 s+9}{\left(s^{2}+4 s+6\right)}\\ \text { (C) } \frac{F(s)}{\left(-s^{2}+4 s+6\right)} & \text { (D) } \frac{F(s)-2 s+9}{\left(s^{2}+4 s+6\right)} \end{array}$ 