10.6 Constant Coefficient Homogeneous Systems III

We now consider the system \({\bf y}'=A{\bf y}\), where \(A\) has a complex eigenvalue \(\lambda=\alpha+i\beta\) with \(\beta\ne0\). We continue to assume that \(A\) has real entries, so the characteristic polynomial of \(A\) has real coefficients. This implies that \(\overline\lambda=\alpha-i\beta\) is also an eigenvalue of \(A\).

An eigenvector \({\bf x}\) of \(A\) associated with \(\lambda=\alpha+i\beta\) will have complex entries, so we’ll write

where \({\bf u}\) and \({\bf v}\) have real entries; that is, \({\bf u}\) and \({\bf v}\) are the real and imaginary parts of \({\bf x}\). Since \(A{\bf x}=\lambda {\bf x}\),

Taking complex conjugates here and recalling that \(A\) has real entries yields

which shows that \({\bf x}={\bf u}-i{\bf v}\) is an eigenvector associated with \(\overline\lambda=\alpha-i\beta\). The complex conjugate eigenvalues \(\lambda\) and \(\overline\lambda\) can be separately associated with linearly independent solutions \({\bf y}'=A{\bf y}\); however, we won’t pursue this approach, since solutions obtained in this way turn out to be complex–valued. Instead, we’ll obtain solutions of \({\bf y}'=A{\bf y}\) in the form

where \(f_1\) and \(f_2\) are real–valued scalar functions. The next theorem shows how to do this.

Proof A function of the form (10.6.2) is a solution of \({\bf y}'=A{\bf y}\) if and only if

Carrying out the multiplication indicated on the right side of (10.6.1) and collecting the real and imaginary parts of the result yields

Equating real and imaginary parts on the two sides of this equation yields

We leave it to you (Exercise 25) to show from this that \({\bf u}\) and \({\bf v}\) are both nonzero. Substituting from these equations into (10.6.4) yields

This is true if

If we let \(f_1=g_1e^{\alpha t}\) and \(f_2=g_2e^{\alpha t}\), where \(g_1\) and \(g_2\) are to be determined, then the last two equations become

which implies that

so

The general solution of this equation is

Moreover, since \(g_2=g_1'/\beta\),

Multiplying \(g_1\) and \(g_2\) by \(e^{\alpha t}\) shows that

Substituting these into (10.6.2) shows that

is a solution of \({\bf y}'=A{\bf y}\) for any choice of the constants \(c_1\) and \(c_2\). In particular, by first taking \(c_1=1\) and \(c_2=0\) and then taking \(c_1=0\) and \(c_2=1\), we see that \({\bf y}_1\) and \({\bf y}_2\) are solutions of \( {\bf y}'=A{\bf y}\). We leave it to you to verify that they are, respectively, the real and imaginary parts of (10.6.3) (Exercise 26), and that they are linearly independent (Exercise 27).

Solution The characteristic polynomial of the coefficient matrix \(A\) in (10.6.6) is

Hence, \(\lambda=1+4i\) is an eigenvalue of \(A\). The associated eigenvectors satisfy \(\left(A-\left(1+4i\right)I\right){\bf x}={\bf 0}\). The augmented matrix of this system is

which is row equivalent to

Therefore \(x_1=(3+4i)x_2/5\). Taking \(x_2=5\) yields \(x_1=3+4i\), so

is an eigenvector. The real and imaginary parts of

are

which are linearly independent solutions of (10.6.6). The general solution of (10.6.6) is

Solution The characteristic polynomial of the coefficient matrix \(A\) in (10.6.7) is

Hence, \(\lambda=1+3i\) is an eigenvalue of \(A\). The associated eigenvectors satisfy \(\left(A-(1+3i)I\right){\bf x}={\bf 0}\). The augmented augmented matrix of this system is

which is row equivalent to

Therefore \(x_1=(5-i)/2\). Taking \(x_2=2\) yields \(x_1=5-i\), so

is an eigenvector. The real and imaginary parts of

are

which are linearly independent solutions of (10.6.7). The general solution of (10.6.7) is

Solution The characteristic polynomial of the coefficient matrix \(A\) in (10.6.8) is

Hence, the eigenvalues of \(A\) are \(\lambda_1=2\), \(\lambda_2=i\), and \(\lambda_3=-i\). The augmented matrix of \((A-2I){\bf x=0}\) is

which is row equivalent to

Therefore \(x_1=x_2=2x_3\). Taking \(x_3=1\) yields

so

is a solution of (10.6.8).

The augmented matrix of \((A-iI){\bf x=0}\) is

which is row equivalent to

Therefore \(x_1=ix_3\) and \(x_2=-(1-i)x_3\). Taking \(x_3=1\) yields the eigenvector

The real and imaginary parts of

are

which are solutions of (10.6.8). Since the Wronskian of \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) at \(t=0\) is

\(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions of (10.6.8). The general solution of (10.6.8) is

Solution The characteristic polynomial of the coefficient matrix \(A\) in (10.6.9) is

Hence, the eigenvalues of \(A\) are \(\lambda_1=2\), \(\lambda_2=2+2i\), and \(\lambda_3=2-2i\). The augmented matrix of \((A-2I){\bf x=0}\) is

which is row equivalent to

Therefore \(x_1=x_2=-x_3\). Taking \(x_3=1\) yields

so

is a solution of (10.6.9).

The augmented matrix of \(\left(A-(2+2i)I\right){\bf x=0}\) is

which is row equivalent to

Therefore \(x_1=ix_3\) and \(x_2=-ix_3\). Taking \(x_3=1\) yields the eigenvector

The real and imaginary parts of

are

which are solutions of (10.6.9). Since the Wronskian of \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) at \(t=0\) is

\(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions of (10.6.9). The general solution of (10.6.9) is

Geometric Properties of Solutions when \(n=2\)

We’ll now consider the geometric properties of solutions of a \(2\times2\) constant coefficient system

under the assumptions of this section; that is, when the matrix

has a complex eigenvalue \(\lambda=\alpha+i\beta\) (\(\beta\ne0\)) and \({\bf x}={\bf u}+i{\bf v}\) is an associated eigenvector, where \({\bf u}\) and \({\bf v}\) have real components. To describe the trajectories accurately it’s necessary to introduce a new rectangular coordinate system in the \(y_1\)-\(y_2\) plane. This raises a point that hasn’t come up before: It is always possible to choose \({\bf x}\) so that \(({\bf u},{\bf v})=0\). A special effort is required to do this, since not every eigenvector has this property. However, if we know an eigenvector that doesn’t, we can multiply it by a suitable complex constant to obtain one that does. To see this, note that if \({\bf x}\) is a \(\lambda\)-eigenvector of \(A\) and \(k\) is an arbitrary real number, then

is also a \(\lambda\)-eigenvector of \(A\), since

The real and imaginary parts of \({\bf x}_1\) are

so

Therefore \(({\bf u}_1,{\bf v}_1)=0\) if

If \(({\bf u},{\bf v})\ne0\) we can use the quadratic formula to find two real values of \(k\) such that \(({\bf u}_1,{\bf v}_1)=0\) (Exercise 28).

(The numbers don’t always work out as nicely as in this example. You’ll need a calculator or computer to do Exercises 29-40.)

Henceforth, we’ll assume that \(({\bf u},{\bf v})=0\). Let \({\bf U}\) and \({\bf V}\) be unit vectors in the directions of \({\bf u}\) and \({\bf v}\), respectively; that is, \({\bf U}={\bf u}/\|{\bf u}\|\) and \({\bf V}={\bf v}/\|{\bf v}\|\). The new rectangular coordinate system will have the same origin as the \(y_1\)-\(y_2\) system. The coordinates of a point in this system will be denoted by \((z_1,z_2)\), where \(z_1\) and \(z_2\) are the displacements in the directions of \({\bf U}\) and \({\bf V}\), respectively.

From (10.6.5), the solutions of (10.6.10) are given by

For convenience, let’s call the curve traversed by \(e^{-\alpha t}{\bf y}(t)\) a shadow trajectory of (10.6.10). Multiplying (10.6.13) by \(e^{-\alpha t}\) yields

where

Therefore

(verify!), which means that the shadow trajectories of (10.6.10) are ellipses centered at the origin, with axes of symmetry parallel to \({\bf U}\) and \({\bf V}\). Since

the vector from the origin to a point on the shadow ellipse rotates in the same direction that \({\bf V}\) would have to be rotated by \(\pi/2\) radians to bring it into coincidence with \({\bf U}\) (Figures 10.6.1 and 10.6.2).

If \(\alpha=0\), then any trajectory of (10.6.10) is a shadow trajectory of (10.6.10); therefore, if \(\lambda\) is purely imaginary, then the trajectories of (10.6.10) are ellipses traversed periodically as indicated in Figures 10.6.1 and 10.6.2.

If \(\alpha>0\), then

so the trajectory spirals away from the origin as \(t\) varies from \(-\infty\) to \(\infty\). The direction of the spiral depends upon the relative orientation of \({\bf U}\) and \({\bf V}\), as shown in Figures 10.6.3 and 10.6.4.

If \(\alpha<0\), then

so the trajectory spirals toward the origin as \(t\) varies from \(-\infty\) to \(\infty\). Again, the direction of the spiral depends upon the relative orientation of \({\bf U}\) and \({\bf V}\), as shown in Figures 10.6.5 and 10.6.6.

10.6 Exercises

In Exercises 1–16 find the general solution.

1. \(\dst{{\bf y}'=\twobytwo{-1}2{-5}5{\bf y} }\)

   Show answer

   \({\bf y}=\dst{c_1e^{2t}\ctwocol{3\cos t+\sin t}{5\cos t} +c_2e^{2t}\ctwocol{3\sin t-\cos t}{5\sin t}}\).

2. \(\dst{{\bf y}'=\twobytwo{-11}4{-26}9{\bf y} }\)

   Show answer

   \({\bf y}=\dst{c_1e^{-t}\ctwocol{5\cos2t+\sin2t}{13\cos2t} +c_2e^{-t}\ctwocol{5\sin2t-\cos2t}{13\sin2t}}\).

3. \(\dst{{\bf y}'=\twobytwo12{-4}5{\bf y}}\)

   Show answer

   \({\bf y}=\dst{c_1e^{3t}\ctwocol{\cos2t+\sin2t}{2\cos2t} +c_2e^{3t}\ctwocol{\sin2t-\cos2t}{2\sin2t}}\).

4. \(\dst{{\bf y}'=\twobytwo5{-6}3{-1}{\bf y}}\)

   Show answer

   \({\bf y}=\dst{c_1e^{2t}\ctwocol{\cos3t-\sin3t}{\cos3t} +c_2e^{2t}\ctwocol{\sin3t+\cos3t}{\sin3t}}\).

5. \(\dst{{\bf y}'=\threebythree3{-3}1022511{\bf y}}\)

   Show answer

   \(\dst{{\bf y}=c_1\threecol{-1}{-1}2e^{-2t}+c_2e^{4t}\cthreecol{\cos2t-\sin 2t}{\cos2t+\sin 2t}{2\cos2t}+c_3e^{4t}\cthreecol {\sin2t+\cos2t}{\sin2t-\cos2t}{2\sin2t}}\).

6. \(\dst{{\bf y}'=\threebythree{-3}311{-5}{-3}{-3}73{\bf y}}\)

   Show answer

   \(\dst{{\bf y}=c_1\threecol{-1}{-1}1e^{-t}+c_2e^{-2t}\cthreecol{\cos2t-\sin 2t}{-\cos2t-\sin 2t}{2\cos2t}+c_3e^{-2t}\cthreecol {\sin2t+\cos2t}{-\sin2t+\cos2t}{2\sin2t}}\)

7. \(\dst{{\bf y}'=\threebythree21{-1}011101{\bf y}}\)

   Show answer

   \(\dst{{\bf y}=c_1\threecol111e^{2t}+c_2e^t\threecol{-\sin t}{\sin t}{\cos t}+c_3e^t\threecol {\cos t}{-\cos t}{\sin t}}\)

8. \(\dst{{\bf y}'=\threebythree{-3}1{-3}4{-1}24{-2}3{\bf y}}\)

   Show answer

   \(\dst{{\bf y}=c_1\threecol{-1}11e^t+c_2e^{-t}\cthreecol{-\sin 2t-\cos2t}{2\cos2t}{2\cos2t}+c_3e^{-t}\cthreecol {\cos2t-\sin2t}{2\sin2t}{2\sin2t}}\)

9. \(\dst{{\bf y}'=\twobytwo5{-4}{10}1{\bf y}}\)

   Show answer

   \(\dst{{\bf y}=c_1e^{3t}\left[\begin{array}{c}\cos6t-3\sin6t \\5\cos6t\end{array}\right]+ c_2e^{3t}\left[\begin{array}{c} \sin6t+3\cos6t\\5\sin6t\end{array}\right]}\)

10. \(\dst{{\bf y}'={1\over3}\twobytwo7{-5}25{\bf y}}\)

    Show answer

    \(\dst{{\bf y}=c_1e^{2t}\left[\begin{array}{c}\cos t-3\sin t \\2\cos t\end{array}\right]+ c_2e^{2t}\left[\begin{array}{c} \sin t+3\cos t\\ 2\sin t\end{array}\right]}\)

11. \(\dst{{\bf y}'=\twobytwo32{-5}1{\bf y}}\)

    Show answer

    \(\dst{{\bf y}=c_1e^{2t}\left[\begin{array}{c}3\sin3t-\cos3t \\ 5\cos3t\end{array}\right]+ c_2e^{2t}\left[\begin{array}{c} -3\cos3t-\sin3t\\ 5\sin3t\end{array}\right]}\)

12. \(\dst{{\bf y}'=\twobytwo{34}{52}{-20}{-30}{\bf y}}\)

    Show answer

    \(\dst{{\bf y}=c_1e^{2t}\left[\begin{array}{c}\sin4t-8\cos4t \\ 5\cos4t\end{array}\right]+ c_2e^{2t}\left[\begin{array}{c} -\cos4t-8\sin4t\\ 5\sin4t\end{array}\right]}\)

13. \(\dst{{\bf y}' =\threebythree11210{-1}{-1}{-2}{-1}{\bf y}}\)

    Show answer

    \(\dst{{\bf y}=c_1\left[\begin{array}{r}-1\\1\\1 \end{array}\right]e^{-2t}+c_2e^t\left[\begin{array}{r}\sin t\\-\cos t\\\cos t \end{array}\right]+c_3e^t\left[\begin{array}{r}-\cos t \\-\sin t\\\sin t \end{array}\right]}\)

14. \(\dst{{\bf y}' =\threebythree3{-4}{-2}{-5}7{-8}{-10}{13}{-8}{\bf y}}\)

    Show answer

    \(\dst{{\bf y}=c_1\left[\begin{array}{c} 2\\ 2\\1 \end{array}\right]e^{-2t}+ c_2e^{2t}\left[\begin{array}{c}-\cos3t-\sin3t\\-\sin3t\\ \cos3t\end{array}\right]+c_3e^{2t}\left[\begin{array}{c} -\sin3t+\cos3t\\\cos3t\\\sin3t\end{array}\right]}\)

15. \(\dst{{\bf y}'=\threebythree60{-3}{-3}331{-2}6{\bf y}'}\)

    Show answer

    \(\dst{{\bf y}=c_1\left[\begin{array}{r}1\\2\\1 \end{array}\right]e^{3t}+c_2e^{6t}\left[\begin{array}{r}-\sin3t \\\sin3t\\\cos3t \end{array}\right]+ c_3e^{6t}\left[\begin{array}{r}\cos3t \\-\cos3t\\\sin3t\end{array}\right]}\)

16. \(\dst{{\bf y}'=\threebythree12{-2}02{-1}100{\bf y}'}\)

    Show answer

    \(\dst{{\bf y}=c_1\left[\begin{array}{c} 1\\1\\1 \end{array}\right]e^t+ c_2e^t\left[\begin{array}{c} 2\cos t-2\sin t \\\cos t-\sin t\\ 2\cos t\end{array}\right]+ c_3e^t\left[\begin{array}{ccc} 2\sin t+2\cos t\\\cos t+\sin t\\ 2\sin t\end{array}\right]}\)

17. \(\dst{{\bf y}'=\twobytwo4{-6}3{-2}{\bf y},\quad {\bf y}(0)=\twocol52}\)

    Show answer

    \(\dst{{\bf y}=e^t\left[\begin{array}{c} 5\cos3t+\sin3t\\2\cos3t+3\sin3t\end{array}\right]}\)

18. \(\dst{{\bf y}'=\twobytwo7{15}{-3}1{\bf y},\quad {\bf y}(0)=\twocol51}\)

    Show answer

    \(\dst{{\bf y}=e^{4t}\left[\begin{array}{c}5\cos6t+5\sin6t\\\cos6t-3\sin6t \end{array}\right]}\)

19. \(\dst{{\bf y}'=\twobytwo7{-15}3{-5}{\bf y},\quad {\bf y}(0)=\twocol{17}7}\)

    Show answer

    \(\dst{{\bf y}=e^t\left[\begin{array}{c}17\cos3t-\sin3t\\7\cos3t+3\sin3t \end{array}\right]}\)

20. \(\dst{{\bf y}'={1\over6}\twobytwo4{-2}52{\bf y},\quad {\bf y}(0)=\twocol1{-1}}\)

    Show answer

    \({\bf y}=\dst{e^{t/2}\left[\begin{array}{c}\cos(t/2)+ \sin(t/2)\\-\cos(t/2)+2\sin(t/2) \end{array}\right]}\)

21. \(\dst{{\bf y}'=\threebythree52{-1}{-3}22132{\bf y},\quad {\bf y}(0)=\threecol406}\)

    Show answer

    \(\dst{{\bf y}=\threecol1{-1}2e^t +e^{4t}\left[\begin{array}{c}3\cos t+\sin t\\\cos t-3\sin t\\4\cos t-2\sin t\end{array}\right]}\)

22. \(\dst{{\bf y}'=\threebythree4408{10}{-20}23{-2}{\bf y},\quad {\bf y}(0)=\threecol865}\)

    Show answer

    \(\dst{{\bf y}=\threecol442e^{8t} +e^{2t}\left[\begin{array}{c}4\cos2t+8\sin2t\\-6\sin2t+2\cos2t\\3\cos2t+\sin2t \end{array}\right]}\)

23. \(\dst{{\bf y}'=\threebythree1{15}{-15}{-6}{18}{-22}{-3}{11}{-15}{\bf y},\quad {\bf y}(0)=\threecol{15}{17}{10}}\)

    Show answer

    \(\dst{{\bf y}=\threecol033e^{-4t} +e^{4t}\left[\begin{array}{c}15\cos6t +10\sin6t\\14\cos6t-8\sin6t\\7\cos6t-4\sin6 t\end{array}\right]}\)

24. \(\dst{{\bf y}'=\threebythree4{-4}4{-10}3{15}2{-3}1{\bf y},\quad {\bf y}(0)=\threecol{16}{14}6}\)

    Show answer

    \(\dst{{\bf y}=\threecol6{-3}3e^{8t} +\left[\begin{array}{c}10\cos4t-4\sin4t\\17\cos4t-\sin4t\\3\cos4t-7\sin4t \end{array}\right]}\)

25. Suppose an \(n\times n\) matrix \(A\) with real entries has a complex eigenvalue \(\lambda=\alpha+i\beta\) (\(\beta\ne0\)) with associated eigenvector \({\bf x}={\bf u}+i{\bf v}\), where \({\bf u}\) and \({\bf v}\) have real components. Show that \({\bf u}\) and \({\bf v}\) are both nonzero.

26. Verify that

    \[ {\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \mbox{\quad and\quad} {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t), \]

    are the real and imaginary parts of

    \[ e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}). \]

27. Show that if the vectors \({\bf u}\) and \({\bf v}\) are not both \({\bf 0}\) and \(\beta\ne0\) then the vector functions

    \[ {\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)\quad \mbox{ and }\quad {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t) \]

    are linearly independent on every interval. Hint: There are two cases to consider: (i) \(\{{\bf u},{\bf v}\}\) linearly independent, and (ii) \(\{{\bf u},{\bf v}\}\) linearly dependent. In either case, exploit the the linear independence of \(\{\cos\beta t,\sin\beta t\}\) on every interval.

28. Suppose \({\bf u}=\dst{\twocol{u_1}{u_2}}\) and \({\bf v}=\dst{\twocol{v_1}{v_2}}\) are not orthogonal; that is, \(({\bf u},{\bf v})\ne0\).

    1. Show that the quadratic equation

       \[ ({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf v})=0 \]

       has a positive root \(k_1\) and a negative root \(k_2=-1/k_1\).

    2. Let \({\bf u}_1^{(1)}={\bf u}-k_1{\bf v}\), \({\bf v}_1^{(1)}={\bf v}+k_1{\bf u}\), \({\bf u}_1^{(2)}={\bf u}-k_2{\bf v}\), and \({\bf v}_1^{(2)}={\bf v}+k_2{\bf u}\), so that \(({\bf u}_1^{(1)},{\bf v}_1^{(1)}) =({\bf u}_1^{(2)},{\bf v}_1^{(2)})=0\), from the discussion given above. Show that

       \[ {\bf u}_1^{(2)}={{\bf v}_1^{(1)}\over k_1} \mbox{\quad and \quad} {\bf v}_1^{(2)}=-{{\bf u}_1^{(1)}\over k_1}. \]

    3. Let \({\bf U}_1\), \({\bf V}_1\), \({\bf U}_2\), and \({\bf V}_2\) be unit vectors in the directions of \({\bf u}_1^{(1)}\), \({\bf v}_1^{(1)}\), \({\bf u}_1^{(2)}\), and \({\bf v}_1^{(2)}\), respectively. Conclude from (a) that \({\bf U}_2={\bf V}_1\) and \({\bf V}_2=-{\bf U}_1\), and that therefore the counterclockwise angles from \({\bf U}_1\) to \({\bf V}_1\) and from \({\bf U}_2\) to \({\bf V}_2\) are both \(\pi/2\) or both \(-\pi/2\).

29. C/G   \(\dst{{\bf y}'=\twobytwo3{-5}5{-3}{\bf y}}\)

    Show answer

    \({\bf U}=\dst{{1\over\sqrt2}\twocol{-1}1}\), \({\bf V}=\dst{{1\over\sqrt2}\twocol11}\) 

30. C/G   \(\dst{{\bf y}'=\twobytwo{-15}{10}{-25}{15}{\bf y}}\)

    Show answer

    \({\bf U}\approx\dst{\twocol{.5257}{.8507}}\), \({\bf V}\approx\dst{\twocol{-.8507}{\phantom{-}.5257}}\)

31. C/G   \(\dst{{\bf y}'=\twobytwo{-4}8{-4}4{\bf y}}\)

    Show answer

    \({\bf U}\approx\dst{\twocol{.8507}{.5257}}\),

    \({\bf V}\approx\dst{\twocol{-.5257}{\phantom{-}.8507}}\) 

32. C/G   \(\dst{{\bf y}'=\twobytwo{-3}{-15}33{\bf y}}\)

    Show answer

    \({\bf U}\approx\dst{\twocol{-.9732}{\phantom{-}.2298}}\), \({\bf V}\approx\dst{\twocol{.2298}{.9732}}\)

33. C/G   \(\dst{{\bf y}'=\twobytwo{-5}6{-12}7{\bf y}}\)

    Show answer

    \({\bf U}\approx\dst{\twocol{\phantom{-}.5257}{.8507}}\), \({\bf V}\approx\dst{\twocol{-.8507}{.5257}}\) 

34. C/G   \(\dst{{\bf y}'=\twobytwo5{-12}6{-7}{\bf y}}\)

    Show answer

    \({\bf U}\approx\dst{\twocol{-.5257}{\phantom{-}.8507}}\), \({\bf V}\approx\dst{\twocol{.8507}{.5257}}\)

35. C/G   \(\dst{{\bf y}'=\twobytwo4{-5}9{-2}{\bf y}}\)

    Show answer

    \({\bf U}\approx\dst{\twocol{-.8817}{\phantom{-}.4719}}\), \({\bf V}\approx\dst{\twocol{.4719}{.8817}}\)  

36. C/G   \(\dst{{\bf y}'=\twobytwo{-4}9{-5}2{\bf y}}\)

    Show answer

    \({\bf U}\approx\dst{\twocol{.8817}{.4719}}\), \({\bf V}\approx\dst{\twocol{-.4719}{\phantom{-}.8817}}\)

37. C/G   \(\dst{{\bf y}'=\twobytwo{-1}{10}{-10}{-1}{\bf y}}\)

    Show answer

    \({\bf U}=\dst{\twocol01}\), \({\bf V}=\dst{\twocol{-1}0}\)           

38. C/G   \(\dst{{\bf y}'=\twobytwo{-1}{-5}{20}{-1}{\bf y}}\)

    Show answer

    \({\bf U}=\dst{\twocol01}\), \({\bf V}=\dst{\twocol10}\)

39. C/G   \(\dst{{\bf y}'=\twobytwo{-7}{10}{-10}9{\bf y}}\)

    Show answer

    \({\bf U}=\dst{{1\over\sqrt2}\twocol11}\), \({\bf V}=\dst{{1\over\sqrt2}\twocol{-1}{\phantom{-}1}}\)   

40. C/G   \(\dst{{\bf y}'=\twobytwo{-7}6{-12}5{\bf y}}\)

    Show answer

    \({\bf U}\approx\dst{\twocol{.5257}{.8507}}\), \({\bf V}\approx\dst{\twocol{-.8507}{\phantom{-}.5257}}\)