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Question One §

Question §

Find $x$ in the following equation…

$$\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\left[\begin{array}{cc}x& 3\\ 2& -1\end{array}\right]=\left[\begin{array}{cc}-6& 7\\ 4& 0\end{array}\right]$$

Solution §

$$\left[\begin{array}{cc}2x-2& 7\\ x+6& 0\end{array}\right]=\left[\begin{array}{cc}-6& 7\\ 4& 0\end{array}\right]$$

$$\{\begin{array}{l}2x-2=-6\\ x+6=4\end{array}⟹\boxed{x=-2}$$

Question Two §

Question §

Do the vectors $\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$, $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$, $\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]$ form a basis for ${\mathbb{R}}^3$?

Solution §

There are three vectors.

$$c_1\mathbf{u}+c_2\mathbf{v}+c_3\mathbf{w}=0⟹c_1=c_2=c_3=0:\text{linearly independent.}$$

$$\begin{array}{rl}\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& 1\end{array}\right]& =\left[\begin{array}{ccc}1& 1& 0\\ 0& -1& 0\\ 0& 1& 1\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 1& 0\\ 0& -1& 1\\ 0& 0& 2\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& -1\\ 0& 0& 1\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}⟹\text{linearly independent}$$

$$\boxed{\therefore \text{basic for }{\mathbb{R}}^3}$$

Question Three §

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Solution §

$$\begin{array}{rl}\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 3\\ 0& 1& -2\end{array}\right]& =\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 0& 1& -2\end{array}\right]\text{by }{R}_2+R_3\\ & \\ & =\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& -2\\ 0& 0& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 0& 3\\ 0& 1& -2\\ 0& 0& 0\end{array}\right]\\ & =\dots \end{array}$$

Question Four §

Question §

$a\ne 0$:

$F(x)=f(x)+a$ shifts up for a positive $a$, down for a negative $a$. $F(x)=f(x+a)$ shifts left for a positive $a$, right for a negative $a$. $F(x)=f(ax)$ stretches for $a<1$, compresses for $a>1$.

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$f:[0,1)\to [1,\infty )$

$f(x)=\frac{1}{\sqrt{1-x^2}}$

Use the aforementioned transformations to produce another bijection (mapping) $F:[2,4)\to [0,\infty )$.

Solution §

$F_1(x)=f(\frac{x}{2})=\frac{1}{\sqrt{1-\frac{1}{4}{x}^2}}$. $F_2(x)=F_1(x-2)=\frac{1}{\sqrt{1-\frac{1}{4}(x-2{)}^2}}$. $F_3(x)=f_2(x)-1=\frac{1}{\sqrt{1-\frac{1}{4}(x-2{)}^2}}-1$.

Therefore,

$$\boxed{F(x)=\frac{1}{\sqrt{1-\frac{1}{4}(x-2{)}^2}}-1}$$

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