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Appendix A Answers to Selected Exercises
1 The Real and Complex Numbers 1.2 Real Number Arithmetic
Exercises
1.2.1. 1.2.3. Answer .
\(-\dfrac{3\sqrt[5]{3}}{8} = -\dfrac{3^{6/5}}{8}\)
1.2.5. 1.2.7. Answer .
\(\dfrac{-4 + \sqrt{2}}{7}\)
1.2.9. 1.2.11. 1.2.13. 1.2.15. 1.2.17. 1.2.19. 1.2.21. 1.2.23. 1.2.25. 1.2.27. 1.2.29. 1.2.31. 1.2.33.
1.3 The Cartesian Coordinate Plane
Exercises 1.3.1. Answer .
The required points
\(A(-3, -7)\text{,}\) \(B(1.3, -2)\text{,}\) \(C(\pi, \sqrt{10})\text{,}\) \(D(0, 8)\text{,}\) \(E(-5.5, 0)\text{,}\) \(F(-8, 4)\text{,}\) \(G(9.2, -7.8)\text{,}\) and
\(H(7, 5)\) are plotted in the Cartesian Coordinate Plane below.
1.3.3. Answer .
\(d = 4 \sqrt{10}\text{,}\) \(M = \left(1, -4 \right)\)
1.3.5. Answer .
\(d= \frac{\sqrt{37}}{2}\text{,}\) \(M = \left(\frac{5}{6}, \frac{7}{4} \right)\)
1.3.7. Answer .
\(d= 3\sqrt{5}\text{,}\) \(M = \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{3}}{2} \right)\)
1.3.9. Answer .
\(d = \sqrt{x^2 + y^2}\text{,}\) \(M = \left( \frac{x}{2}, \frac{y}{2}\right)\)
1.3.11. 1.3.13. Answer .
\(\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2} \right)\text{,}\) \(\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\)
1.4 Complex Numbers
Exercises
1.4.1. Answer .
\(\displaystyle z+w = 2+7i\)
\(\displaystyle zw = -12+8i\)
\(\displaystyle z^2 = -5 + 12i\)
\(\displaystyle \frac{1}{z} = \frac{2}{13} - \frac{3}{13} \, i\)
\(\displaystyle \frac{z}{w} = \frac{3}{4} - \frac{1}{2} \, i\)
\(\displaystyle \frac{w}{z} = \frac{12}{13} + \frac{8}{13} \,i\)
\(\displaystyle \overline{z} = 2-3i\)
\(\displaystyle z\overline{z} = 13\)
\(\displaystyle (\overline{z})^2 = -5-12i\)
1.4.3. Answer .
\(\displaystyle z+w = -1+3i\)
\(\displaystyle zw = -2-i\)
\(\displaystyle z^2 = -1\)
\(\displaystyle \frac{1}{z} = -i\)
\(\displaystyle \frac{z}{w} = \frac{2}{5} - \frac{1}{5} \, i\)
\(\displaystyle \frac{w}{z} = 2+i\)
\(\displaystyle \overline{z} = -i\)
\(\displaystyle z\overline{z} = 1\)
\(\displaystyle (\overline{z})^2 = -1\)
1.4.5. Answer .
\(\displaystyle z+w = 5+2i\)
\(\displaystyle zw = 41+11i\)
\(\displaystyle z^2 = -16-30i\)
\(\displaystyle \frac{1}{z} = \frac{3}{34} + \frac{5}{34} \,i\)
\(\displaystyle \frac{z}{w} = -\frac{29}{53} - \frac{31}{53} \, i\)
\(\displaystyle \frac{w}{z} = -\frac{29}{34} + \frac{31}{34} \,i\)
\(\displaystyle \overline{z} = 3+5i\)
\(\displaystyle z\overline{z} = 34\)
\(\displaystyle (\overline{z})^2 = -16+30i\)
1.4.7. Answer .
\(\displaystyle z+w = 2\sqrt{2}\)
\(\displaystyle z^2 = -4i\)
\(\displaystyle \frac{1}{z} = \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} \,i\)
\(\displaystyle \frac{z}{w} = -i\)
\(\displaystyle \frac{w}{z} = i\)
\(\displaystyle \overline{z} = \sqrt{2}+i\sqrt{2}\)
\(\displaystyle z\overline{z} = 4\)
\(\displaystyle (\overline{z})^2 = 4i\)
1.4.9. Answer .
\(\displaystyle z+w = i\sqrt{3}\)
\(\displaystyle zw = -1\)
\(\displaystyle z^2 = -\frac{1}{2} + \frac{\sqrt{3}}{2} \,i\)
\(\displaystyle \frac{1}{z} = \frac{1}{2} - \frac{\sqrt{3}}{2} \, i\)
\(\displaystyle \frac{z}{w} = \frac{1}{2} - \frac{\sqrt{3}}{2} \, i\)
\(\displaystyle \frac{w}{z} = \frac{1}{2} + \frac{\sqrt{3}}{2} \, i\)
\(\displaystyle \overline{z} = \frac{1}{2} - \frac{\sqrt{3}}{2} \, i\)
\(\displaystyle z\overline{z} = 1\)
\(\displaystyle (\overline{z})^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} \, i\)
1.4.11. 1.4.13. 1.4.15. 1.4.17.
1.4.19. Answer .
\(i^{5} = i^{4} \cdot i = 1 \cdot i = i\)
1.4.21. Answer .
\(i^{7} = i^{4} \cdot i^{3} = 1 \cdot (-i) = -i\)
1.4.23. Answer .
\(i^{15} = \left(i^{4}\right)^{3} \cdot i^{3} = 1 \cdot (-i) = -i\)
1.4.25. Answer .
\(i^{117} = \left(i^{4}\right)^{29} \cdot i = 1 \cdot i = i\)
1.4.27. Answer .
\(x = \dfrac{2 \pm i\sqrt{14}}{3}\)
1.4.29. 1.4.31. Answer .
\(y = \pm \dfrac{3i \sqrt{2}}{2}\)
1.4.33. Answer .
\(x = \dfrac{\sqrt{5} \pm i\sqrt{3}}{2}\)
1.4.35.
2 Vectors 2.1 Introduction to Cartesian Coordinates in Space 2.1.3 Exercises
2.1.3.1.
Answer 1 . Answer 2 . Answer 3 . Answer 4 .
2.2 An Introduction to Vectors
Exercises
2.2.1. 2.2.1.a 2.2.1.b Answer .
\(\boldsymbol{i}+6\boldsymbol{j}\)
2.2.3. 2.2.3.a 2.2.3.b Answer .
\(6\boldsymbol{i}-\boldsymbol{j}+6\boldsymbol{k}\)
2.2.5. 2.2.5.a Answer .
\(\vec u+\vec v = \bbm 2\\-1\ebm\text{;}\) \(\vec u -\vec v = \bbm 0\\-3\ebm\text{;}\) \(2\vec u-3\vec v = \bbm -1\\-7\ebm\text{.}\)
2.2.5.c Answer .
\(\vec x = \bbm 1/2\\2\ebm\text{.}\)
2.2.11.
Answer 1 . Answer 2 . Answer 3 . Answer 4 . 2.2.13.
Answer 1 . Answer 2 . Answer 3 . Answer 4 .
2.2.17. 2.2.19. Answer .
\(\left<\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right>\)
2.2.21. Answer .
\(\left<\frac{-1}{2},\frac{\sqrt{3}}{2}\right>\)
2.3 The Dot Product 2.3.2 Exercises
2.3.2.1. 2.3.2.3. 2.3.2.5. 2.3.2.7.
2.3.2.9. Answer .
\(\cos^{-1}\mathopen{}\left(\frac{3}{\sqrt{10}}\right)\)
2.3.2.11.
2.3.2.17. Answer .
\(\left<\frac{-5}{10},\frac{15}{10}\right>\)
2.3.2.19. Answer .
\(\left<\frac{-1}{2},\frac{-1}{2}\right>\)
2.3.2.21. Answer .
\(\left<\frac{14}{14},\frac{28}{14},\frac{42}{14}\right>\)
2.3.2.23.
Answer 1 .
\(\left<\frac{-5}{10},\frac{15}{10}\right>\)
Answer 2 .
\(\left<\frac{15}{10},\frac{5}{10}\right>\)
2.3.2.25.
Answer 1 .
\(\left<\frac{-1}{2},\frac{-1}{2}\right>\)
Answer 2 .
\(\left<\frac{-5}{2},\frac{5}{2}\right>\)
2.3.2.27.
Answer 1 .
\(\left<\frac{14}{14},\frac{28}{14},\frac{42}{14}\right>\)
Answer 2 .
\(\left<\frac{0}{14},\frac{42}{14},\frac{-28}{14}\right>\)
2.3.2.29. 2.3.2.31. 2.3.2.33. 2.3.2.35.
2.4 The Cross Product 2.4.3 Exercises
2.4.3.1. Answer .
\(\left<12,-15,3\right>\)
2.4.3.3. Answer .
\(\left<-5,-31,27\right>\)
2.4.3.5. 2.4.3.7. Answer .
\(\vec u\times \vec v = \bbm 0\\0\\ad-bc\ebm\)
2.4.3.9. 2.4.3.11.
2.4.3.29. Answer .
\(\left<0.408248,0.408248,-0.816497\right>\hbox{ or }\left<-0.408248,-0.408248,0.816497\right>\)
2.4.3.31. Answer .
\(\left<0,1,0\right>\hbox{ or }\left<0,-1,0\right>\)
2.4.3.33. 2.4.3.35. Answer .
\(200/3\approx 66.67\) ft–lb
2.5 Lines 2.5.4 Exercises
2.5.4.7.
Answer 1 .
\(\left(7,2,-1\right)+t\mathopen{}\left<1,-1,2\right>\)
Answer 2 .
\(x = 7+t, y = 2-t, z = -1+2t\)
Answer 3 .
\(x-7 = 2-y = \frac{z+1}{2}\)
2.6 Planes 2.6.2 Exercises
2.6.2.15. 2.6.2.17.
Answer 1 .
\(3\mathopen{}\left(x+4\right)+8\mathopen{}\left(y-7\right)-10\mathopen{}\left(z-2\right) = 0\)
Answer 2 .
2.7 Span and Linear Independence 2.7.4 Exercises
2.7.4.1. Answer .
\(\bbm -9\\-8\\16\\12\ebm\)
2.7.4.3. Answer .
\(\bbm -22\\2\\12\\39\ebm\)
2.7.4.9. Answer .
True. Suppose
\(S=\{\vec{v}_1,\vec{v}_2,\ldots, \vec{v}_n\}\) is linearly independent. Let
\(T\subseteq S\) be a subset. By re-ordering the vectors we can assume
\(T=\{\vec{v}_1,\vec{v}_2,\ldots, \vec{v}_m\}\) for some
\(m\leq n\text{.}\) It
\(T\) were linearly dependent, then there would exist scalars
\(c_1,\ldots, c_m\text{,}\) not all zero, such that
\(c_1\vec{v}_1+\cdots +c_m\vec{v}_m=\vec{0}\text{.}\) But if this is the case, then we would have
\(c_1\vec{v}_1+\cdots +c_m\vec{v}_m+0\vec{v}_{m+1}+\cdots +0\vec{v}_n=\vec{0}\text{,}\) which is impossible if
\(S\) is independent.
2.7.4.11. Answer .
True, since we can add any scalar multiple of the zero vector to a linear combination without affecting the value of that linear combination.
2.7.4.13. Answer .
False. The set
\(\left\{\bbm 1\\0\ebm, \bbm 0\\1\ebm, \bbm 0\\0\ebm\right\}\) is linearly dependent, since
\(\bbm 0\\0\ebm = 0\bbm 1\\0\ebm+0\bbm 0\\1\ebm\text{,}\) but
\(\bbm 1\\0\ebm\) does not belong to the span of the set
\(\left\{\bbm 0\\1\ebm, \bbm 0\\0\ebm\right\}\text{.}\)
3 Systems of Linear Equations 3.1 Introduction to Linear Equations
Exercises
3.1.1. 3.1.3. 3.1.5. 3.1.7. 3.1.9. 3.1.15. Answer .
35 blue, 40 green, 20 red, 5 yellow
3.1.17. Answer .
12 $0.30 trinkets, 8 $0.65 trinkets
3.2 Using Matrices To Solve Systems of Linear Equations
Exercises
3.2.1. Answer .
\(\left[\begin{array}{ccc|c} 3 \amp 4 \amp 5\amp 7\\-1\amp 1\amp -3\amp 1\\2\amp -2\amp 3\amp 5 \end{array}\right]\)
3.2.3. Answer .
\(\left[\begin{array}{cccc|c} 1 \amp 3 \amp -4\amp 5\amp 17\\-1\amp 0\amp 4\amp 8\amp 1\\ 2\amp 3\amp 4\amp 5\amp 6 \end{array}\right]\)
3.2.5. Answer .
\(\begin{array}{rl}
x_1+2x_2=\amp 3\\
-x_1+3x_2=\amp 9\\ \end{array}\)
3.2.7. Answer .
\(\begin{array}{rl}
x_1+x_2-x_3-x_4=\amp 2\\
2x_1+x_2+3x_3+5x_4=\amp 7\\ \end{array}\)
3.2.9. Answer .
\(\begin{array}{rl}
x_1=\amp 2\\
x_2=\amp -1\\
x_3=\amp 5\\
x_4=\amp 3\\ \end{array}\)
3.2.11. Answer .
\(\bbm -2 \amp 1 \amp -7\\0\amp 4\amp -2\\5\amp 0\amp 3\\ \ebm\)
3.2.13. Answer .
\(\bbm 2 \amp -1 \amp 7\\2\amp 3\amp 5\\5\amp 0\amp 3\\ \ebm\)
3.2.15. Answer .
\(\bbm 2\amp -1\amp 7\\0\amp 2\amp -1\\5\amp 0\amp 3\\ \ebm\)
3.2.17. 3.2.19. Answer .
\(2R_3+R_1\rightarrow R_1\)
3.2.21. Answer .
\(-R_2+R_3\leftrightarrow R_3\)
3.3 Elementary Row Operations and Gaussian Elimination
Exercises
3.3.5. Answer .
\(\bbm 1 \amp 0\\0 \amp 1\\ \ebm\)
3.3.7. Answer .
\(\bbm 1 \amp 3\\0 \amp 0\\ \ebm\)
3.3.9. Answer .
\(\bbm 1 \amp 0 \amp 3\\0\amp 1\amp 7\\ \ebm\)
3.3.11. Answer .
\(\bbm 1 \amp -1 \amp 2\\0\amp 0\amp 0\\ \ebm\)
3.3.13. Answer .
\(\bbm 1 \amp 0 \amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\\ \ebm\)
3.3.15. Answer .
\(\bbm 1 \amp 0 \amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\\ \ebm\)
3.3.17. Answer .
\(\bbm 1 \amp 0 \amp 0\amp 1\\0\amp 1\amp 1\amp 1\\0\amp 0\amp 0\amp 0\\ \ebm\)
3.3.19. Answer .
\(\bbm 1 \amp 0 \amp 1\amp 3\\0\amp 1\amp -2\amp 4\\0\amp 0\amp 0\amp 0\\ \ebm\)
3.3.21. Answer .
\(\bbm 1 \amp 1 \amp 0\amp 0\amp 0\amp 0\\0\amp 0\amp 1\amp 3\amp 1\amp 4\\ \ebm\)
3.4 Existence and Uniqueness of Solutions
Exercises
3.4.1. Answer .
\(x_1=1-2x_2\text{;}\) \(x_2\) is free. Possible solutions:
\(x_1=1\text{,}\) \(x_2=0\) and
\(x_1=-1\text{,}\) \(x_2=1\text{.}\) Geometrically, both equations describe the same line, and every point on this line is a solution to the system.
3.4.3. Answer .
No solution; the system is inconsistent. Geometrically, the system describes three planes, any two of which intersect along a line. However, there is no point common to all three.
3.4.5. Answer .
\(x_1=1-x_2-x_4\text{;}\) \(x_2\) is free;
\(x_3=1-2x_4\text{;}\) \(x_4\) is free. Possible solutions:
\(x_1 = 1\text{,}\) \(x_2 = 0\text{,}\) \(x_3 = 1\text{,}\) \(x_4 = 0\) and
\(x_1 = -2\text{,}\) \(x_2 = 1\text{,}\) \(x_3 = -3\text{,}\) \(x_4=2\text{.}\) Since there are four variables, a geometric description is more difficult (but see if you can come up with one!).
3.4.7. Answer .
\(x_1=3-x_3-2x_4\text{;}\) \(x_2=-3-5x_3-7x_4\text{;}\) \(x_3\) is free;
\(x_4\) is free. Possible solutions:
\(x_1 =3\text{,}\) \(x_2 = -3\text{,}\) \(x_3=0\text{,}\) \(x_4=0\) and
\(x_1 = 0\text{,}\) \(x_2 = -5\text{,}\) \(x_3 =-1\text{,}\) \(x_4=1\text{.}\) Since there are four variables, a geometric description is more difficult (but see if you can come up with one!).
3.4.9. Answer .
\(x_1=1\text{;}\) \(x_2=2\text{.}\) Geometrically, the system represents two lines intersecting in a single point.
3.4.11. Answer .
\(x_1=0\text{;}\) \(x_2=-1\text{.}\) Geometrically, the system represents two lines intersecting in a single point.
3.4.13. Answer .
\(x_1=\frac13-\frac43x_3\text{;}\) \(x_2=\frac13-\frac13x_3\text{;}\) \(x_3\) is free. Possible solutions:
\(x_1 = \frac13\text{,}\) \(x_2=\frac13\text{,}\) \(x_3=0\) and
\(x_1 = -1\text{,}\) \(x_2 = 0\text{,}\) \(x_3=1\text{.}\) Geometrically, the system represents three planes that all intersect along a common line.
3.4.15. Answer .
Exactly 1 solution if
\(k\neq 2\text{;}\) infinitely many solutions if
\(k=2\text{;}\) never no solution.
3.4.17. Answer .
Never exactly 1 solution; infinitely many solutions if
\(k=2\text{;}\) no solution if
\(k\neq 2\text{.}\)
3.5 Applications of Linear Systems
Exercises
3.5.1. 3.5.3. 3.5.5. 3.5.7. 3.5.9. 3.5.11. Answer .
Substitution yields the equations
\(2 = ae^{b}\) and
\(4 = ae^{2b}\text{;}\) these are not linear equations.
\(y = ae^{bx}\) implies that
\(\ln y = \ln (ae^{bx}) = \ln a + \ln e^{bx} = \ln a + bx\text{.}\)
Plugging in the points for \(x\) and \(y\) in the equation \(\ln y = \ln a + bx\text{,}\) we have equations
\begin{equation*}
\begin{array}{ccccc} \ln a \amp + \amp b \amp = \amp \ln 2 \\ \ln a \amp + \amp 2b \amp = \amp \ln 4\\ \end{array}\text{.}
\end{equation*}
To solve,
\begin{equation*}
\bbm 1 \amp 1\amp \ln 2 \\ 1 \amp 2 \amp \ln 4 \\ \ebm \overrightarrow{\text{rref}} \bbm 1\amp 0\amp 0\\0\amp 1\amp \ln 2\\ \ebm\text{.}
\end{equation*}
Therefore \(\ln a = 0\) and \(b = \ln 2\text{.}\)
Since
\(\ln a = 0\text{,}\) we know that
\(a = e^0 = 1\text{.}\) Thus our exponential function is
\(f(x) = e^{x\ln 2}\text{.}\)
3.5.13. Answer .
The augmented matrix from this system is \(\bbm 1 \amp 1 \amp 1\amp 1\amp 8\\6\amp 1\amp 2\amp 3\amp 29\\0\amp 1\amp -1\amp 0\amp 2\\ \ebm\text{.}\) From this we find the solution
\begin{align*}
t\amp =4-\frac13f\\
x\amp =3-\frac13f\\
w\amp =1-\frac13f\text{.}
\end{align*}
The only time each of these variables are nonnegative integers is when \(f=0\) or \(f=3\text{.}\) If \(f=0\text{,}\) then we have 4 touchdowns, 3 extra points and 1 two point conversions (no field goals). If \(f=3\text{,}\) then we have 3 touchdowns, 2 extra points and no two point conversions (and 3 field goals).
3.5.15. Answer .
Let \(x_1\text{,}\) \(x_2\) and \(x_3\) represent the number of free throws, 2 point and 3 point shots taken. The augmented matrix from this system is \(\bbm 1\amp 1\amp 1\amp 70\\1\amp 2\amp 3\amp 110 \ebm\text{.}\) From this we find the solution
\begin{align*}
x_1\amp =30+x_3\\
x_2\amp =40-2x_3\text{.}
\end{align*}
In order for \(x_2\) to be nonnegative, we need \(x_3\leq 20\text{.}\) Thus there are 21 different scenarios: the “first” is where 0 three point shots are taken (\(x_3=0\) ), 30 free throws and 40 two point shots; the “last” is where 20 three point shots are taken, 50 free throws, and no two point shots.
3.5.17. Answer .
Let
\(y = ax+b\text{;}\) all linear functions through (2,5) come in the form
\(y = (2.5-\frac12b)x+b\text{.}\) Examples:
\(b=1\) yields
\(y = 2x+1\text{;}\) \(b=-1\) yields
\(y=3x-1\text{.}\)
3.5.19. Answer .
Let
\(y = ax^2+bx+c\text{;}\) we find that
\(a = 2-\frac12 c\) and
\(b = -1+\frac12c\text{.}\) Examples:
\(c=0\) yields
\(y = 2x^2-x\text{;}\) \(c=-2\) yields
\(y=3x^2-2x-2\text{.}\)
3.5.21. 3.5.23. Answer .
Yes.
\(\vec{x} = -2\vec{w}_1+2\vec{w}_2\text{.}\)
3.6 Vector Solutions to Linear Systems
Exercises
3.6.1. Answer .
Multiply
\(A\vu\) and
\(A\vvv\) to verify.
3.6.3. Answer .
Multiply
\(A\vu\) and
\(A\vvv\) to verify.
3.6.5. Answer .
Multiply
\(A\vu\) and
\(A\vvv\) to verify.
3.6.7. Answer .
Multiply
\(A\vu\text{,}\) \(A\vvv\) and
\(A(\vu+\vvv)\) to verify.
3.6.9. Answer .
Multiply
\(A\vu\text{,}\) \(A\vvv\) and
\(A(\vu+\vvv)\) to verify.
3.6.11. Answer .
\(\displaystyle \vx=\bbm 0\\0\ebm\)
\(\displaystyle \vx=\bbm 2/5\\-13/5\ebm\)
3.6.13. Answer .
\(\displaystyle \vx=\bbm 0\\0\ebm\)
\(\displaystyle \vx=\bbm -2\\ -9/4\ebm\)
3.6.15. Answer .
\(\displaystyle \vx=x_3\bbm 5/4\\ 1\\ 1\ebm\)
\(\displaystyle \vx=\bbm 1\\ 0\\ 0\ebm+x_3\bbm 5/4\\ 1\\ 1\ebm\)
3.6.17. Answer .
\(\displaystyle \vx=x_3\bbm 6\\ -4\\ 1\ebm\)
\(\displaystyle \vx=\bbm -12\\ 8\\0\ebm +x_3\bbm 6\\ -4\\ 1\ebm\)
3.6.19. Answer .
\(\displaystyle \vx=x_3\bbm 2\\2/5\\1\\0\ebm+x_4\bbm-1\\2/5\\0\\1\ebm\)
\(\displaystyle \vx=\bbm-2\\2/5\\0\\0\ebm + x_3\bbm 2\\2/5\\1\\0\ebm +x_4\bbm-1\\2/5\\0\\1\ebm\)
3.6.21. Answer .
\(\displaystyle \vx=x_2\bbm-1/2\\1\\0\\0\\0\ebm+x_4\bbm 1/2\\0\\-1/2\\1\\0\ebm+x_5\bbm 13/2\\0\\-2\\0\\1\ebm\)
\(\displaystyle \vx=\bbm-5\\0\\ 3/2 \\0\\0\ebm + x_2\bbm-1/2\\1\\0\\0\\0\ebm+x_4\bbm 1/2 \\0\\-1/2\\1\\0\ebm+x_5\bbm 13/2\\0\\-2\\0\\1\ebm\)
3.6.23. Answer .
\(\displaystyle \vx=x_4\bbm 1\\ 13/9 \\-1/3 \\1\\0\ebm+x_5\bbm 0\\ -1 \\-1\\0\\1 \ebm\)
\(\displaystyle \vx=\bbm 1\\ 1/9 \\ 5/3 \\ 0\\ 0 \ebm + x_4\bbm 1\\ 13/9 \\ -1/3 \\1\\0\ebm+x_5\bbm 0\\ -1 \\-1\\0\\1 \ebm\)
3.6.25. Answer .
\(\vx = \bbm 0.5\\0\ebm + x_2\bbm 2.5\\1\ebm = \vec{x}_{p} + x_2\vvv\)
3.6.27. Answer .
\(\vx = x_2\bbm 2.5\\1\ebm = x_2\vvv\)
4 Matrix Algebra 4.1 Matrix Addition and Scalar Multiplication
Exercises
4.1.1. Answer .
\(\bbm -2 \amp -1\\12 \amp 13\ebm\)
4.1.3. Answer .
\(\bbm 2 \amp -2\\14 \amp 8\ebm\)
4.1.5. Answer .
\(\bbm 9 \amp -7\\11 \amp -6\ebm\)
4.1.11. Answer .
\(X = \bbm -5 \amp 9 \-1 \amp -14 \ebm\)
4.1.13. Answer .
\(X = \bbm -5 \amp -2 \-9/2 \amp -19/2 \ebm\)
4.1.15. Answer .
\(a = 2\text{,}\) \(b = 1\)
4.1.17. 4.1.19. 4.1.21.
4.2 Matrix Multiplication
Exercises
4.2.1. 4.2.3. 4.2.5. 4.2.7. 4.2.9. 4.2.11.
4.2.13. Answer .
\(AB=\bbm 8\amp 3\\10\amp -9\ebm\text{,}\) \(BA=\bbm -3\amp 24\\4\amp 2\ebm\)
4.2.15. Answer .
\(AB=\bbm -1\amp -2\amp 12\\10\amp 4\amp 32\ebm\text{,}\) \(BA\) is not possible.
4.2.17. Answer .
\(AB\) is not possible,
\(BA = \bbm 27\amp -33\amp 39\\-27\amp -3\amp -15\ebm\)
4.2.19. Answer .
\(AB =\bbm-32\amp 34\amp -24\\-32\amp 38\amp -8\\-16\amp 21\amp 4\ebm\text{,}\) \(BA = \bbm 22\amp -14\\-4\amp -12\ebm\)
4.2.21. Answer .
\(AB = \bbm -56\amp 2\amp -36\\ 20\amp 19\amp -30\\ -50\amp -13\amp 0\ebm\text{,}\) \(BA = \bbm -46\amp 40\\ 72\amp 9\ebm \)
4.2.23. Answer .
\(AB = \bbm -15\amp -22\amp -21\amp -1\\ 16\amp -53\amp -59\amp -31\ebm \text{,}\) \(BA\) is not possible.
4.2.25. Answer .
\(AB = \bbm 0\amp 0\amp 4\\ 6\amp 4\amp -2\\ 2\amp -4\amp -6\ebm\text{,}\) \(BA = \bbm 2\amp -2\amp 6\\ 2\amp 2\amp 4\\ 4\amp 0\amp -6\ebm\)
4.2.27. Answer .
\(AB = \bbm 21\amp -17\amp -5\\ 19\amp 5\amp 19\\ 5\amp 9\amp 4\ebm\text{,}\) \(BA = \bbm 19\amp 5\amp 23\\ 5\amp -7\amp -1\\ -14\amp 6\amp 18\ebm\)
4.2.29. Answer .
\(DA= \bbm 2\amp 2\amp 2\\ -6\amp -6\amp -6\\ -15\amp -15\amp -15\ebm\text{,}\) \(AD = \bbm 2\amp -3\amp 5\\ 4\amp -6\amp 10\\ -6\amp 9\amp -15\ebm\)
4.2.31. Answer .
\(DA= \bbm 4\amp -6\\ 4\amp -6\ebm\text{,}\) \(AD = \bbm 4\amp 8\\ -3\amp -6\ebm\)
4.2.33. Answer .
\(DA= \bbm d_1a\amp d_1b\amp d_1c\\ d_2d\amp d_2e\amp d_2f\\ d_3g\amp d_3h\amp d_3i\ebm\text{,}\) \(AD = \bbm d_1a\amp d_2b\amp d_3c\\ d_1d\amp d_2e\amp d_3f\\ d_1g\amp d_2h\amp d_3i\ebm\)
4.2.35. Answer .
\(A\vx= \bbm -6\\ 11\ebm\)
4.2.37. Answer .
\(A\vx= \bbm -5\\ 5\\ 21\ebm\)
4.2.39. Answer .
\(A\vx= \bbm2x_1-x_2\\ 4x_1+3x_2\ebm\)
4.2.41. Answer .
\(A^2 = \bbm 4 \amp 0\\0 \amp 9\ebm\text{;}\) \(A^3 = \bbm 8\amp 0\\0\amp 27\ebm\)
4.2.43. Answer .
\(A^2 = \bbm 0 \amp 0 \amp 1\\1\amp 0\amp 0\\0\amp 1\amp 0\ebm\text{;}\) \(A^3 = \bbm 1\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\ebm\)
4.2.45. Answer .
\(\displaystyle \bbm 0\amp -2\\-5\amp -1\ebm\)
\(\displaystyle \bbm 10\amp 2\\ 5\amp 11\ebm\)
\(\displaystyle \bbm -11 \amp -15\ 37 \amp 32\ebm\)
\((A+B)(A + B) = AA + AB + BA + BB = A^2+AB+BA+B^2\text{.}\)
4.3 Solving Matrix Equations \(AX=B\)
Exercises
4.3.1. Answer .
\(X=\bbm 0 \amp -2\ -8 \amp 6 \ebm \)
4.3.3. Answer .
\(X=\bbm -1 \amp 2 \amp -4\ -6\amp -2\amp 3 \ebm \)
4.3.5. Answer .
\(X=\bbm -5 \amp 2 \amp -3\ -4\amp -3\amp -2 \ebm \)
4.3.7. Answer .
\(X=\bbm 1 \amp -9\ -4 \amp -5 \ebm \)
4.3.9. Answer .
\(X=\bbm 3 \amp -3 \amp 3\ 2\amp -2\amp -3 \\ -3\amp -1\amp -2 \ebm \)
4.3.11. Answer .
\(X=\bbm -1/2 \amp -1/2 \amp 0\ -1/2 \amp -1 \amp 1/2 \\ -1/2 \amp -3/4 \amp 3/4 \ebm \)
4.4 The Matrix Inverse
Exercises 4.4.5. Answer .
\(A^{-1}\) does not exist.
4.4.21. Answer .
\(A^{-1}\) does not exist.
4.4.33. Answer .
\(\vx = \bbm 2\\ 3\ebm \)
4.5 Properties of the Matrix Inverse
Exercises
4.5.1. Answer .
\((AB)^{-1} = \bbm -2 \amp 3 \\1 \amp -1.4 \ebm\)
4.5.3. Answer .
\((AB)^{-1} = \bbm 29/5 \amp -18/5 \\-11/5 \amp 7/5 \ebm\)
4.5.5. Answer .
\(A^{-1} = \bbm -2 \amp -5\\-1 \amp -3 \ebm\text{,}\) \((A^{-1})^{-1} = \bbm -3 \amp 5 \\1 \amp -2 \ebm\)
4.5.7. Answer .
\(A^{-1} = \bbm -3 \amp 7\\ 1 \amp -2 \ebm\text{,}\) \((A^{-1})^{-1} = \bbm 2 \amp 7\\1 \amp 3 \ebm\)
4.5.9. 4.5.11. Answer .
Likely some entries that should be 0 will not be exactly 0, but rather very small values.
4.6 Elementary Matrices
Exercises
4.6.5. Answer .
\(\bbm 1 \amp -3\\0 \amp 1\ebm\)
4.6.7. Answer .
\(\bbm 1 \amp 0\\0 \amp 7\ebm\)
4.6.9. Answer .
\(\bbm 1 \amp 3\\0 \amp 1\ebm\) \(R_1+3R_2\to R_1\)
4.6.11. Answer .
\(\bbm 1 \amp 0 \amp 0\\4\amp 1\amp 0\\0\amp 0\amp 1\ebm\text{,}\) \(R_2+4R_1\to R_2\)
4.6.13. Answer .
Answers may vary. One possibility:
\begin{equation*}
A^{-1} = \bbm 1\amp 0\\-3\amp 1\ebm\bbm 0\amp 1\\1\amp 0\ebm
\end{equation*}
\begin{equation*}
A = \bbm 0\amp 1\\1\amp 0\ebm\bbm 1\amp 0\\3\amp 1\ebm
\end{equation*}
4.6.15. Answer .
Answers may vary. One possibility:
\begin{equation*}
A^{-1} = \bbm 1\amp 0\amp 0\\0\amp 1\amp 2\\0\amp 0\amp 1\ebm\bbm 1\amp 0\amp -1\\0\amp 1\amp 0\\0\amp 0\amp 1\ebm\bbm 1\amp 0\amp 0\\-3\amp 1\amp 0\\0\amp 0\amp 1\ebm\bbm \frac{1}{2}\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\ebm
\end{equation*}
\begin{equation*}
A = \bbm 2\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\ebm\bbm 1\amp 0\amp 0\\3\amp 1\amp 0\\0\amp 0\amp 1\ebm\bbm 1\amp 0\amp 1\\0\amp 1\amp 0\\0\amp 0\amp 1\ebm\bbm 1\amp 0\amp 0\\0\amp 1\amp -2\\0\amp 0\amp 1\ebm
\end{equation*}
5 Matrix Transformations 5.1 Matrix Transformations 5.1.4 Exercises
5.1.4.5. Answer .
\(A = \bbm 1\amp 2\\3\amp 4\ebm\)
5.1.4.7. Answer .
\(A = \bbm 1\amp 2\\1\amp 2\ebm\)
5.1.4.9. Answer .
\(A = \bbm 5\amp 2\\2\amp 1\ebm\)
5.1.4.11. Answer .
\(A = \bbm 0\amp 1\\3\amp 0\ebm\)
5.1.4.13. Answer .
\(A = \bbm 0\amp -1\\ -1\amp -1\ebm\)
5.1.4.15. Answer .
Yes, these are the same; the transformation matrix in each is
\(\bbm -1 \amp 0 \\0\amp -1\ebm\text{.}\)
5.1.4.17. Answer .
Yes, these are the same. Each produces the transformation matrix
\(\bbm 1/2 \amp 0 \\ 0 \amp 3\ebm\text{.}\)
5.2 Properties of Linear Transformations
Exercises
5.2.1. 5.2.3. Answer .
No; cannot add a constant.
5.2.5.
5.2.7. Answer .
\([\, T\, ] = \bbm 1 \amp 2\amp 3\amp 4\ebm\)
5.2.9. Answer .
\([\, T\, ] = \bbm 1 \amp 1\\1 \amp -1\ebm\)
5.2.11. Answer .
\([\, T\, ] = \bbm 1 \amp 0 \amp 3\\1\amp 0\amp -1\\1\amp 0\amp 1\ebm\)
5.3 Subspaces of \(\mathbb{R}^n\)
Exercises
5.3.1. Answer .
Not a subspace. The vector
\(\vec v = \bbm 2\\0\ebm\) belongs to
\(S\text{,}\) but
\(2\vec v\) does not.
5.3.3. Answer .
Subspace. If \(y=2x\text{,}\) we have
\begin{equation*}
\bbm x\\y\ebm = \bbm x\\2x\ebm = x\bbm 1\\2\ebm,
\end{equation*}
so \(U\) is equal to the span of the vector \(\bbm 1\\2\ebm\text{,}\) and therefore a subspace.
5.4 Null Space and Column Space 5.4.4 Exercises
5.4.4.1. Answer .
\(\operatorname{null}(A) = \operatorname{span}\left\{\bbm -2\\1\\0\\0\ebm\right\}\) has dimension 1;
\(\operatorname{col}(A) = \operatorname{span}\left\{\bbm 1\\2\\-3\ebm, \bbm 0\\-1\\2\ebm, \bbm 3\\6\\-4\ebm\right\}\) has dimension 3;
\(4 = 1+3\text{.}\)
5.4.4.3. Answer .
\(\operatorname{null}(A) = \operatorname{span}\left\{\bbm 1\\1\\1\ebm\right\}\) has dimension 1;
\(\operatorname{col}(A) = \operatorname{span}\left\{\bbm 1\\-2\\0\ebm, \bbm -2\\-4\\-8\ebm\right\}\) has dimension 2;
\(3 = 1+2\text{.}\)
5.4.4.5. Answer .
\(\operatorname{null}(A) = \operatorname{span}\left\{\bbm -6\\-4\\-1\\3\ebm\right\}\) has dimension 1;
\(\operatorname{col}(A) = \operatorname{span}\left\{\bbm 2\\1\\0\ebm, \bbm 0\\-1\\3\ebm, \bbm 3\\4\\-6\ebm\right\}\) has dimension 3;
\(4 = 1+3\text{.}\)
5.4.4.7. Answer .
\(\operatorname{null}(A) = \operatorname{span}\left\{\bbm 2\\1\\0\\0\ebm,\bbm -3\\0\\-8\\5\ebm\right\}\) has dimension 2;
\(\operatorname{col}(A) = \operatorname{span}\left\{\bbm 1\\2\\-1\ebm, \bbm 4\\3\\1\ebm\right\}\) has dimension 2;
\(4 = 2+2\text{.}\)
6 Operations on Matrices 6.1 The Matrix Transpose
Exercises
6.1.1. Answer .
\(A\) is skew symmetric.
\(\bbm 0\amp -6\amp 1\\ 6\amp 0\amp 4\\ -1\amp -4\amp 0\ebm\)
6.1.3. Answer .
\(\bbm -9\amp 6\amp -8\\ 4\amp -3\amp 1\\ 10\amp -7\amp -1\ebm\)
6.1.5. Answer .
\(A\) is diagonal, as is
\(A^T\text{.}\) \(\bbm 1\amp 0\amp 0\\ 0\amp 2\amp 0\\ 0\amp 0\amp -1\ebm\)
6.1.7. Answer .
\(A\) is symmetric.
\(\bbm 6\amp -4\amp -5\\ -4\amp 0\amp 2\\ -5\amp 2\amp -2\ebm\)
6.1.9. Answer .
\(\bbm -7\\ -8\\2\\ -3\ebm \)
6.1.11. Answer .
\(A\) is symmetric.
\(\bbm 13\amp -3\\ -3\amp 1\ebm\)
6.1.13. Answer .
\(\bbm 4\amp -9\\ -7\amp 6\\ -4\amp 3\\ -9\amp -9\ebm\)
6.1.15. Answer .
\(\bbm 2\amp 5\amp 7\\ -5\amp 5\amp -4\\ -3\amp -6\amp -10\ebm\)
6.1.17. Answer .
\(A\) is upper triangular;
\(A^T\) is lower triangular.
\(\bbm -3\amp 0\amp 0\\ -4\amp -3\amp 0\\ -5\amp 5\amp -3\ebm\)
6.1.19. Answer .
\(A\) is symmetric.
\(\bbm 4\amp 0\amp -2\\ 0\amp 2\amp 3\\ -2\amp 3\amp 6\ebm\)
6.1.21. Answer .
\(\bbm -5\amp 3\amp -10\\ -9\amp 1\amp -8\ebm\)
6.1.23. Answer .
\(A\) is skew symmetric.
\(\bbm 0\amp -1\amp 2\\ 1\amp 0\amp -4\\ -2\amp 4\amp 0\ebm\)
6.2 The Matrix Trace
Exercises
6.2.1. Answer .
Not defined; the matrix must be square.
6.2.3. 6.2.5. 6.2.7. 6.2.9. 6.2.11. Answer .
Not defined; the matrix must be square.
6.2.13. 6.2.15.
6.2.17. Answer .
\(\tr(A)=-1\text{;}\) \(\tr(B)=\text{;}\) \(\tr(A+B)=5\)
\(\displaystyle \tr(AB) = 201 = \tr(BA)\)
6.2.19. Answer .
\(\tr(A)=-5\text{;}\) \(\tr(B)=-4\text{;}\) \(\tr(A+B)=-9\)
\(\displaystyle \tr(AB) = 23 = \tr(BA)\)
6.3 The Determinant
Exercises
6.3.1. 6.3.3. 6.3.5. 6.3.7.
6.3.9. Answer .
The submatrices are
\(\bbm 7 \amp 6\ 6 \amp 10\ebm\text{,}\) \(\bbm 3 \amp 6\ 1 \amp 10\ebm\text{,}\) and
\(\bbm 3 \amp 7\ 1 \amp 6\ebm\text{,}\) respectively.
\(C_{1,2}=34\text{,}\) \(C_{1,2}=-24\text{,}\) \(C_{1,3}=11\)
6.3.11. Answer .
The submatrices are
\(\bbm 3\amp 10\\ 3\amp 9\ebm\text{,}\) \(\bbm -3 \amp 10\ -9 \amp 9\ebm\text{,}\) and
\(\bbm -3 \amp 3\ -9 \amp 3\ebm\text{,}\) respectively.
\(C_{1,2}=-3\text{,}\) \(C_{1,2}=-63\text{,}\) \(C_{1,3}=18\)
6.3.13. 6.3.15. 6.3.17. 6.3.19. 6.3.21. 6.3.23. 6.3.25. Answer .
Hint:
\(C_{1,1}= d\text{.}\)
6.4 Properties of the Determinant
Exercises
6.4.1. 6.4.3. 6.4.5. 6.4.7. 6.4.9. 6.4.11. 6.4.13.
6.4.15. Answer .
\(\det(A) = 90\text{;}\) \(2R_1\rightarrow R_1\)
\(\det(B) = 45\text{;}\) \(10R_1+R_3\rightarrow R_3\)
\(\det(C) = 45\text{;}\) \(C = A^T\)
6.4.17. Answer .
\(\det(A) = -16\text{;}\) \(R_1\leftrightarrow R_2\) then
\(R_1\leftrightarrow R_3\)
\(\det(B) = -16\text{;}\) \(-R_1\rightarrow R_1\) and
\(-R_2\rightarrow R_2\)
\(\det(C) = -432\text{;}\) \(C = 3*M\)
6.4.19. Answer .
\(\det(A)=4\text{,}\) \(\det(B)=4\text{,}\) \(\det(AB)=16\)
6.4.21. Answer .
\(\det(A)=-12\text{,}\) \(\det(B)=29\text{,}\) \(\det(AB)=-348\)
6.4.23. 6.4.25. 6.4.27. 6.4.29.
6.5 Applications of the Determinant 6.5.3 Exercises
6.5.3.1. Answer .
\(\det(A) = -123\text{,}\) \(\det(A_1) = -492\text{,}\) \(\det(A_2) = 123\text{,}\) \(\det(A_3) = 492\)
\(\displaystyle \vx = \bbm 4\\ -1\\ -4\ebm\)
6.5.3.3. Answer .
\(\det(A) = 96\text{,}\) \(\det(A_1) = -960\text{,}\) \(\det(A_2) = 768\text{,}\) \(\det(A_3) = 288\)
\(\displaystyle \vx = \bbm -10\\ 8\\ 3\ebm\)
6.5.3.5. Answer .
\(\det(A) = -43\text{,}\) \(\det(A_1) = 215\text{,}\) \(\det(A_2) = 0\)
\(\displaystyle \vx = \bbm -5\\ 0\ebm\)
6.5.3.7. Answer .
\(\det(A) = 16\text{,}\) \(\det(A_1) = -64\text{,}\) \(\det(A_2) = 80\)
\(\displaystyle \vx = \bbm -4\\ 5\ebm\)
6.5.3.9. Answer .
\(\det(A) = 0\text{,}\) \(\det(A_1) = 0\text{,}\) \(\det(A_2) = 0\)
Infinitely many solutions exist.
6.5.3.11. Answer .
\(\det(A) = 0\text{,}\) \(\det(A_1) = 0\text{,}\) \(\det(A_2) = 0\text{,}\) \(\det(A_3) = 0\)
Infinitely many solutions exist.
6.5.3.13. Answer .
\(A^{-1} = \frac{1}{8}\bbm -11 \amp 10 \amp 13\\6\amp -4\amp -2\\9\amp -6\amp -7\ebm\)
6.5.3.15. 6.5.3.17. Answer .
\(A^{-1} = \frac{1}{10}\bbm 9 \amp 5 \amp -12\amp 0\\15\amp 13\amp -26\amp -4\\19\amp 13\amp -28\amp -4\\2\amp 4\amp -4\amp -2\ebm\)
7 Eigenvalues and Eigenvectors 7.1 Eigenvalues and Eigenvectors
Exercises
7.1.7. Answer .
\(\vx = \bbm 3\\-7\\7\ebm\)
7.1.9. Answer .
\(\vx = \bbm -1\\1\\1\ebm\)
7.1.11.
7.1.13. Answer .
\(\lambda _1 = 2\) with
\(\vx[1] = \bbm 1\\0\\0\ebm\text{;}\) \(\lambda _2 = 3\) with
\(\vx[2] = \bbm-1\\1\\0\ebm\text{;}\) \(\lambda _3 = 7\) with
\(\vx[3] = \bbm-1\\15\\10\ebm\)
7.1.15. Answer .
\(\lambda _1 = -2\) with
\(\vx[1] = \bbm 0\\0\\1\ebm\text{;}\) \(\lambda _2 = 1\) with
\(\vx[2] = \bbm 0\\3\\5\ebm\text{;}\) \(\lambda _3 = 5\) with
\(\vx[3] = \bbm 28\\7\\1\ebm\)
7.1.17. Answer .
\(\lambda _1 = -4\) with
\(\vx[1] = \bbm-6\\1\\11\ebm\text{;}\) \(\lambda _2 = -1\) with
\(\vx[2] = \bbm 0\\0\\1\ebm\text{;}\) \(\lambda _3 = 5\) with
\(\vx[3] = \bbm 3\\1\\2\ebm\)
7.1.19. Answer .
\(\lambda _1 = -3\) with
\(\vx[1] = \bbm-2\\1\ebm\text{;}\) \(\lambda _2 = 5\) with
\(\vx[2] = \bbm 6\\1\ebm\)
7.1.21. Answer .
\(\lambda _1 = -5\) with
\(\vx[1] = \bbm 24\\13\\8\ebm\text{;}\) \(\lambda _2 = -2\) with
\(\vx[2] = \bbm 6\\5\\1\ebm\text{;}\) \(\lambda _3 = 3\) with
\(\vx[3] = \bbm 0\\1\\0\ebm\)
7.1.23. Answer .
\(\lambda _1 = -5\) with
\(\vx[1] = \bbm 1\\1\ebm\text{;}\) \(\lambda _2 = 2\) with
\(\vx[2] = \bbm-4\\3\ebm\)
7.1.25. Answer .
\(\lambda _1 = -5\) with
\(\vx[1] = \bbm-1\\5\ebm\text{;}\) \(\lambda _2 = 5\) with
\(\vx[2] = \bbm 1\\5\ebm\)
7.1.27. Answer .
\(\lambda _1 = 4\) with
\(\vx[1] = \bbm 9\\1\ebm\text{;}\) \(\lambda _2 = 5\) with
\(\vx[2] = \bbm 8\\1\ebm\)
7.2 Properties of Eigenvalues and Eigenvectors
Exercises
7.2.1. Answer .
\(\lambda _1 = 1\) with
\(\vx[1] = \bbm 4\\1\ebm\text{;}\) \(\lambda _2 = 4\) with
\(\vx[2] = \bbm 1\\1\ebm\)
\(\lambda _1 = 1\) with
\(\vx[1] = \bbm-1\\1\ebm\text{;}\) \(\lambda _2 = 4\) with
\(\vx[2] = \bbm-1\\4\ebm\)
\(\lambda _1 = 1/4\) with
\(\vx[1] = \bbm 1\\1\ebm\text{;}\) \(\lambda _2 = 1\) with
\(\vx[2] = \bbm 4\\1\ebm\)
7.2.3. Answer .
\(\lambda _1 = -1\) with
\(\vx[1] = \bbm-5\\1\ebm\text{;}\) \(\lambda _2 = 0\) with
\(\vx[2] = \bbm-6\\1\ebm\)
\(\lambda _1 = -1\) with
\(\vx[1] = \bbm 1\\6\ebm\text{;}\) \(\lambda _2 = 0\) with
\(\vx[2] = \bbm 1\\5\ebm\)
7.2.5. Answer .
\(\lambda _1 = -4\) with
\(\vx[1] = \bbm-7\\-7\\6\ebm\text{;}\) \(\lambda _2 = 3\) with
\(\vx[2] = \bbm 0\\0\\1\ebm\text{;}\) \(\lambda _3 = 4\) with
\(\vx[3] = \bbm 9\\1\\22\ebm\)
\(\lambda _1 = -4\) with
\(\vx[1] = \bbm-1\\9\\0\ebm\text{;}\) \(\lambda _2 = 3\) with
\(\vx[2] = \bbm-20\\26\\7\ebm\text{;}\) \(\lambda _3 = 4\) with
\(\vx[3] = \bbm-1\\1\\0\ebm\)
\(\lambda _1 = -1/4\) with
\(\vx[1] = \bbm-7\\-7\\6\ebm\text{;}\) \(\lambda _2 = 1/3\) with
\(\vx[2] = \bbm 0\\0\\1\ebm\text{;}\) \(\lambda _3 = 1/4\) with
\(\vx[3] = \bbm 9\\1\\22\ebm\)
7.3 Eigenvalues and Diagonalization
Exercises
7.3.1. Answer .
\(c_A(x) = (x-7)(x-5)(x+3)\)
Eigenvalue
\(\lambda_1=7\) has algebraic and geometric multiplicity 1.
Eigenvalue
\(\lambda_2=5\) has algebraic and geometric multiplicity 1.
Eigenvalue
\(\lambda_3=-3\) has algebraic and geometric multiplicity 1.
\(P = \bbm 1\amp 0\amp 0\\1\amp 0\amp 1\\1\amp 1\amp 1\ebm\text{.}\)
7.3.3. Answer .
Eigenvalue
\(\lambda_1=6\) has algebraic and geometric multiplicity 1.
Eigenvalue
\(\lambda_2=23\) has algebraic and geometric multiplicity 1.
\(P=\bbm 7\amp -1\\3\amp 2\ebm\text{.}\)
7.3.5. Answer .
\(c_A(x) = (x+1)^2(x-2)\)
Eigenvalue
\(\lambda_1=-1\) has algebraic and geometric multiplicity 2.
Eigenvalue
\(\lambda_2=2\) has algebraic and geometric multiplicity 1.
\(P=\bbm -1 \amp -1 \amp 1\\0\amp 1\amp 1\\1\amp 0\amp 1\ebm\text{.}\)
7.3.7. Answer .
\(c_A(x) = (x-1)^2(x-2)\)
Eigenvalue
\(\lambda_1=1\) has algebraic multiplicity 2 and geometric multiplicity 1.
Eigenvalue
\(\lambda_2=2\) has algebraic and geometric multiplicity 1.
Since the geometric multiplicity of
\(\lambda_1\) is less than its algebraic multiplicity, no such
\(P\) exists; the matrix cannot be diagonalized.
