Linear Algebra with Applications, 10th Edition

7.5

1. 1. 

   1. (a) $\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$;

   2. (b) $\left[\begin{array}{cc}\frac{\sqrt{3}}{\sqrt{2}}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$;

   3. (c) $\left[\begin{array}{cc}-\frac{4}{5}& \frac{3}{5}\\ -\frac{3}{5}& -\frac{4}{5}\end{array}\right]$

2. 2. 

   1. (a) $\left[\begin{array}{ccc}\frac{3}{5}& 0& \frac{4}{5}\\ 0& 1& 0\\ \frac{4}{5}& 0& -\frac{3}{5}\end{array}\right]$;

   2. (b) $\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}& 0\\ -\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}& 0\\ 0& 0& 1\end{array}\right]$;

   3. (c) $\left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{2}& \frac{\sqrt{3}}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right]$;

   4. (d) $\left[\begin{array}{ccc}1& 0& 0\\ 0& -\frac{\sqrt{3}}{2}& \frac{1}{2}\\ 0& \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$

3. 3. $H=I-\frac{1}{\beta }{\mathbf{\text{vv}}}^T$ for the given $\beta$ and v. 

   1. (a) $\beta =90,\mathbf{v}=(-10,8,-4{)}^T;$

   2. (b) $\beta =70,\mathbf{v}=(10,6,2{)}^T;$

   3. (c) $\beta =15,\mathbf{v}=(-5,-3,4{)}^T$

4. 4. 

   1. (a) $\beta =90,\mathbf{v}=(0,10,4,6{)}^T;$

   2. (b) $\beta =15,\mathbf{v}=(0,0,-5,-1,2{)}^T$

5. 6. 

   1. (a) $H_2{H}_{1}A=R$, where $H_i=I-\frac{{\displaystyle 1}}{{\displaystyle {\beta }_i}}{\mathbf{v}}_i{\mathbf{v}}_i^T,i=1,2$, and ${\beta }_1=12,{\beta }_2=45$.

      $$\begin{array}{c}\begin{array}{cc}{\mathbf{v}}_1=\left[\begin{array}{c}\hfill -4\\ \hfill 2\\ \hfill -2\end{array}\right],\hfill & {\mathbf{v}}_2=\left[\begin{array}{c}\hfill 0\\ \hfill 9\\ \hfill -3\end{array}\right],\hfill \end{array}\hfill \\ R=\left[\begin{array}{ccc}3& \frac{19}{2}& \frac{9}{2}\\ 0& -5& -3\\ 0& 0& 6\end{array}\right],\hfill \\ \mathbf{c}=H_2{H}_1\mathbf{b}=\left[\begin{array}{r}-\frac{5}{2}\\ -5\\ 0\end{array}\right]\mathrm{;}\hfill \end{array}$$

   2. (b) $\mathbf{x}=(-4,1,0{)}^T$

6. 7. 

   1. (a) $\begin{array}{cc}G=\left[\begin{array}{cc}\frac{{\displaystyle 3}}{{\displaystyle 5}}& \hfill \frac{{\displaystyle \begin{array}{c}4\end{array}}}{{\displaystyle 5}}\\ \frac{{\displaystyle \begin{array}{c}4\end{array}}}{{\displaystyle 5}}& \hfill -\frac{{\displaystyle 3}}{{\displaystyle 5}}\end{array}\right],& \mathbf{x}=\left[\begin{array}{r}-1\\ 1\end{array}\right]\end{array}$

7. 8. It takes three multiplications, two additions, and one square root to determine H. It takes four multiplications/divisions, one addition, and one square root to determine G. The calculation of GA requires 4n multiplications and 2n additions, while the calculation ...