1.2.1.
Answer.
\(\dfrac{23}{9}\)
Appendix A Answers to Selected Exercises
1 The Real and Complex Numbers
1.2 Real Number Arithmetic
Exercises
1.2.3.
Answer.
\(-\dfrac{3\sqrt[5]{3}}{8} = -\dfrac{3^{6/5}}{8}\)
1.2.5.
Answer.
\(\dfrac{9}{22}\)
1.2.7.
Answer.
\(\dfrac{-4 + \sqrt{2}}{7}\)
1.2.9.
Answer.
\(\dfrac{\sqrt{61}}{2}\)
1.2.11.
Answer.
\(-\dfrac{24}{7}\)
1.2.13.
Answer.
\(6\)
1.2.15.
Answer.
\(18\)
1.2.17.
Answer.
undefined
1.2.19.
Answer.
\(\dfrac{15}{16}\)
1.2.21.
Answer.
\(\dfrac{3}{5}\)
1.2.23.
Answer.
\(-1\)
1.2.25.
Answer.
\(5\)
1.2.27.
Answer.
\(\sqrt{10}\)
1.2.29.
Answer.
\(-1\)
1.2.31.
Answer.
\(-\dfrac{1}{3}\)
1.2.33.
Answer.
\(\dfrac{25}{4}\)
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.
Answer.
\((0,3)\)
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.
- \(\displaystyle z+w = 2\sqrt{2}\)
- \(\displaystyle zw = 4\)
- \(\displaystyle z^2 = -4i\)
- \(\displaystyle \frac{1}{z} = \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} \,i\)
- \(\displaystyle \frac{z}{w} = -i\)
- \(\displaystyle \frac{w}{z} = i\)
- \(\displaystyle \overline{z} = \sqrt{2}+i\sqrt{2}\)
- \(\displaystyle z\overline{z} = 4\)
- \(\displaystyle (\overline{z})^2 = 4i\)
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.
Answer.
\(7i\)
1.4.13.
Answer.
\(-10\)
1.4.15.
Answer.
\(-12\)
1.4.17.
Answer.
\(3\)
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.
Answer.
\(y = \pm 2, \pm i\)
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.
Answer.
\(z = \pm 2, \pm 2i\)
2 Vectors
2.1 Introduction to Cartesian Coordinates in Space
2.1.3 Exercises
2.1.3.1.
Answer 1.
\(\sqrt{6}\)
Answer 2.
\(\sqrt{17}\)
Answer 3.
\(\sqrt{11}\)
Answer 4.
\(\text{do}\)
2.2 An Introduction to Vectors
Exercises
2.2.1.
Answer.
\(\left<1,6\right>\)
Answer.
\(\boldsymbol{i}+6\boldsymbol{j}\)
2.2.3.
Answer.
\(\left<6,-1,6\right>\)
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.
\(\sqrt{5}\)
Answer 2.
\(\sqrt{13}\)
Answer 3.
\(\sqrt{26}\)
Answer 4.
\(\sqrt{10}\)
2.2.13.
Answer 1.
\(\sqrt{5}\)
Answer 2.
\(3\sqrt{5}\)
Answer 3.
\(2\sqrt{5}\)
Answer 4.
\(4\sqrt{5}\)
2.2.17.
Answer.
\(\left<0.6,0.8\right>\)
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.
Answer.
\(-22\)
2.3.2.3.
Answer.
\(3\)
2.3.2.5.
Answer.
not defined
2.3.2.7.
Answer.
Answers will vary.
2.3.2.9.
Answer.
\(\cos^{-1}\mathopen{}\left(\frac{3}{\sqrt{10}}\right)\)
2.3.2.11.
Answer.
\(\frac{\pi }{4}\)
2.3.2.13.
Answer 1.
\(\left<-7,4\right>\)
Answer 2.
\(\left<4,7\right>\)
2.3.2.15.
Answer 1.
\(\left<1,0,-1\right>\)
Answer 2.
\(\left<1,1,1\right>\)
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.
Answer.
1.96lb
2.3.2.31.
Answer.
\(141.42\)ft–lb
2.3.2.33.
Answer.
\(500\)ft–lb
2.3.2.35.
Answer.
\(500\)ft–lb
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.
Answer.
\(\left<0,-2,0\right>\)
2.4.3.7.
Answer.
\(\vec u\times \vec v = \bbm 0\\0\\ad-bc\ebm\)
2.4.3.9.
Answer.
\(-\boldsymbol{j}\)
2.4.3.11.
Answer.
Answers will vary.
2.4.3.13.
Answer.
\(5\)
2.4.3.15.
Answer.
\(0\)
2.4.3.17.
Answer.
\(\sqrt{14}\)
2.4.3.19.
Answer.
\(3\)
2.4.3.21.
Answer.
\(\frac{5\sqrt{2}}{2}\)
2.4.3.23.
Answer.
\(1\)
2.4.3.25.
Answer.
\(7\)
2.4.3.27.
Answer.
\(2\)
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.
Answer.
\(87.5\)ft–lb
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.
\(x-7 = 2-y = \frac{z+1}{2}\)
2.5.4.11.
Answer.
\(\text{parallel}\)
2.5.4.15.
Answer.
\(\text{skew}\)
2.5.4.19.
Answer.
\(\sqrt{41}/3\)
2.5.4.21.
Answer.
\(5\sqrt{2}/2\)
2.5.4.23.
Answer.
\(3/\sqrt{2}\)
2.6 Planes
2.6.2 Exercises
2.6.2.1.
Answer.
Answers will vary.
2.6.2.3.
Answer.
Answers will vary.
2.6.2.15.
Answer 1.
\(x-5+y-7+z-3 = 0\)
Answer 2.
\(x+y+z = 15\)
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.
\(3x+8y-10z = 24\)
2.6.2.25.
Answer.
\(\sqrt{5/7}\)
2.6.2.27.
Answer.
\(1/\sqrt{3}\)
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.
\(\bbm -22\\2\\12\\39\ebm\)
2.7.4.5.
Answer.
\(\sqrt{71}\)
2.7.4.7.
Answer.
\(15\)
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.
Answer.
Yes
3.1.3.
Answer.
Yes
3.1.5.
Answer.
No
3.1.7.
Answer.
Yes
3.1.9.
Answer.
Yes
3.1.11.
Answer.
\(x = 1, y=-2\)
3.1.13.
Answer.
\(x = -1, y=0,z=2\)
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.
Answer.
\(2R_2\rightarrow R_2\)
3.2.19.
Answer.
\(2R_3+R_1\rightarrow R_1\)
3.2.21.
Answer.
\(-R_2+R_3\leftrightarrow R_3\)
3.2.23.
Answer.
\(x=-1,y=3\)
3.2.25.
Answer.
\(x=\frac12,y=\frac13\)
3.2.27.
Answer.
\(x_1=1,x_2=5,x_3=7\)
3.3 Elementary Row Operations and Gaussian Elimination
Exercises
3.3.1.
Answer.
- yes
- no
- no
- yes
3.3.3.
Answer.
- no
- yes
- yes
- yes
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.
Answer.
\(f(x) = 6x-3\)
3.5.3.
Answer.
\(f(x) = -x^2+x+5\)
3.5.5.
Answer.
\(f(x) = -x^3+x^2-x+1\)
3.5.7.
Answer.
\(f(x) = 3x-5\)
3.5.9.
Answer.
\(f(x) = x^2+1\)
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.
Answer.
No.
3.5.23.
Answer.
Yes. \(\vec{x} = -2\vec{w}_1+2\vec{w}_2\text{.}\)
3.5.25.
Answer.
Yes.
3.5.27.
Answer.
No.
3.6 Vector Solutions to Linear Systems
Exercises
3.6.1.
Answer.
Multiply \(\tta\vu\) and \(\tta\vvv\) to verify.
3.6.3.
Answer.
Multiply \(\tta\vu\) and \(\tta\vvv\) to verify.
3.6.5.
Answer.
Multiply \(\tta\vu\) and \(\tta\vvv\) to verify.
3.6.7.
Answer.
Multiply \(\tta\vu\text{,}\) \(\tta\vvv\) and \(\tta(\vu+\vvv)\) to verify.
3.6.9.
Answer.
Multiply \(\tta\vu\text{,}\) \(\tta\vvv\) and \(\tta(\vu+\vvv)\) 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.
\(\bbm 9 \amp -7\\11 \amp -6\ebm\)
4.1.7.
Answer.
\(\bbm -14\\6 \ebm\)
4.1.9.
Answer.
\(\bbm -15\\-25 \ebm\)
4.1.11.
Answer.
\(\ttx = \bbm -5 \amp 9 \-1 \amp -14 \ebm\)
4.1.13.
Answer.
\(\ttx = \bbm -5 \amp -2 \-9/2 \amp -19/2 \ebm\)
4.1.15.
Answer.
\(a = 2\text{,}\) \(b = 1\)
4.1.17.
Answer.
\(a = 5/2 + 3/2b\)
4.1.19.
Answer.
No solution.
4.1.21.
Answer.
No solution.
4.2 Matrix Multiplication
Exercises
4.2.1.
Answer.
\(-22\)
4.2.3.
Answer.
\(0\)
4.2.5.
Answer.
\(5\)
4.2.7.
Answer.
\(15\)
4.2.9.
Answer.
\(-2\)
4.2.11.
Answer.
Not possible.
4.2.13.
Answer.
\(\tta\ttb=\bbm 8\amp 3\\10\amp -9\ebm\text{,}\) \(\ttb\tta=\bbm -3\amp 24\\4\amp 2\ebm\)
4.2.15.
Answer.
\(\tta\ttb=\bbm -1\amp -2\amp 12\\10\amp 4\amp 32\ebm\text{,}\) \(\ttb\tta\) is not possible.
4.2.17.
Answer.
\(\tta\ttb\) is not possible, \(\ttb\tta = \bbm 27\amp -33\amp 39\\-27\amp -3\amp -15\ebm\)
4.2.19.
Answer.
\(\tta\ttb =\bbm-32\amp 34\amp -24\\-32\amp 38\amp -8\\-16\amp 21\amp 4\ebm\text{,}\) \(\ttb\tta = \bbm 22\amp -14\\-4\amp -12\ebm\)
4.2.21.
Answer.
\(\tta\ttb = \bbm -56\amp 2\amp -36\\ 20\amp 19\amp -30\\ -50\amp -13\amp 0\ebm\text{,}\) \(\ttb\tta = \bbm -46\amp 40\\ 72\amp 9\ebm \)
4.2.23.
Answer.
\(\tta\ttb = \bbm -15\amp -22\amp -21\amp -1\\ 16\amp -53\amp -59\amp -31\ebm \text{,}\) \(\ttb\tta\) is not possible.
4.2.25.
Answer.
\(\tta\ttb = \bbm 0\amp 0\amp 4\\ 6\amp 4\amp -2\\ 2\amp -4\amp -6\ebm\text{,}\) \(\ttb\tta = \bbm 2\amp -2\amp 6\\ 2\amp 2\amp 4\\ 4\amp 0\amp -6\ebm\)
4.2.27.
Answer.
\(\tta\ttb = \bbm 21\amp -17\amp -5\\ 19\amp 5\amp 19\\ 5\amp 9\amp 4\ebm\text{,}\) \(\ttb\tta = \bbm 19\amp 5\amp 23\\ 5\amp -7\amp -1\\ -14\amp 6\amp 18\ebm\)
4.2.29.
Answer.
\(\ttd\tta= \bbm 2\amp 2\amp 2\\ -6\amp -6\amp -6\\ -15\amp -15\amp -15\ebm\text{,}\) \(\tta\ttd = \bbm 2\amp -3\amp 5\\ 4\amp -6\amp 10\\ -6\amp 9\amp -15\ebm\)
4.2.31.
Answer.
\(\ttd\tta= \bbm 4\amp -6\\ 4\amp -6\ebm\text{,}\) \(\tta\ttd = \bbm 4\amp 8\\ -3\amp -6\ebm\)
4.2.33.
Answer.
\(\ttd\tta= \bbm d_1a\amp d_1b\amp d_1c\\ d_2d\amp d_2e\amp d_2f\\ d_3g\amp d_3h\amp d_3i\ebm\text{,}\) \(\tta\ttd = \bbm d_1a\amp d_2b\amp d_3c\\ d_1d\amp d_2e\amp d_3f\\ d_1g\amp d_2h\amp d_3i\ebm\)
4.2.35.
Answer.
\(\tta\vx= \bbm -6\\ 11\ebm\)
4.2.37.
Answer.
\(\tta\vx= \bbm -5\\ 5\\ 21\ebm\)
4.2.39.
Answer.
\(\tta\vx= \bbm2x_1-x_2\\ 4x_1+3x_2\ebm\)
4.2.41.
Answer.
\(\tta^2 = \bbm 4 \amp 0\\0 \amp 9\ebm\text{;}\) \(\tta^3 = \bbm 8\amp 0\\0\amp 27\ebm\)
4.2.43.
Answer.
\(\tta^2 = \bbm 0 \amp 0 \amp 1\\1\amp 0\amp 0\\0\amp 1\amp 0\ebm\text{;}\) \(\tta^3 = \bbm 1\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\ebm\)
4.2.45.
Answer.
- \(\displaystyle \bbm 0\amp -2\\-5\amp -1\ebm\)
- \(\displaystyle \bbm 10\amp 2\\ 5\amp 11\ebm\)
- \(\displaystyle \bbm -11 \amp -15\ 37 \amp 32\ebm\)
- No.
- \((\tta+\ttb)(\tta + \ttb) = \tta\tta + \tta\ttb + \ttb\tta + \ttb\ttb = \tta^2+\tta\ttb+\ttb\tta+\ttb^2\text{.}\)
4.3 Solving Matrix Equations \(\tta\ttx=\ttb\)
Exercises
4.3.1.
Answer.
\(\ttx=\bbm 0 \amp -2\ -8 \amp 6 \ebm \)
4.3.3.
Answer.
\(\ttx=\bbm -1 \amp 2 \amp -4\ -6\amp -2\amp 3 \ebm \)
4.3.5.
Answer.
\(\ttx=\bbm -5 \amp 2 \amp -3\ -4\amp -3\amp -2 \ebm \)
4.3.7.
Answer.
\(\ttx=\bbm 1 \amp -9\ -4 \amp -5 \ebm \)
4.3.9.
Answer.
\(\ttx=\bbm 3 \amp -3 \amp 3\ 2\amp -2\amp -3 \\ -3\amp -1\amp -2 \ebm \)
4.3.11.
Answer.
\(\ttx=\bbm -1/2 \amp -1/2 \amp 0\ -1/2 \amp -1 \amp 1/2 \\ -1/2 \amp -3/4 \amp 3/4 \ebm \)
4.4 The Matrix Inverse
Exercises
4.4.5.
Answer.
\(\ttai\) does not exist.
4.4.21.
Answer.
\(\ttai\) does not exist.
4.4.33.
Answer.
\(\vx = \bbm 2\\ 3\ebm \)
4.5 Properties of the Matrix Inverse
Exercises
4.5.1.
Answer.
\((\tta\ttb)^{-1} = \bbm -2 \amp 3 \\1 \amp -1.4 \ebm\)
4.5.3.
Answer.
\((\tta\ttb)^{-1} = \bbm 29/5 \amp -18/5 \\-11/5 \amp 7/5 \ebm\)
4.5.5.
Answer.
\(\ttai = \bbm -2 \amp -5\\-1 \amp -3 \ebm$, $(\ttai)^{-1} = \bbm -3 \amp 5 \\1 \amp -2 \ebm\)
4.5.7.
Answer.
\(\ttai = \bbm -3 \amp 7\\ 1 \amp -2 \ebm$, $(\ttai)^{-1} = \bbm 2 \amp 7\\1 \amp 3 \ebm\)
4.5.9.
Answer.
Solutions will vary.
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
4.6.1.
Answer.
\(R_1-3R_2\to R_1\)
4.6.3.
Answer.
\(R_2-4R_1\to R_2\)
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
5.1.4.1.
Answer.
5.1.4.3.
Answer.
5.1.4.5.
Answer.
\(\tta = \bbm 1\amp 2\\3\amp 4\ebm\)
5.1.4.7.
Answer.
\(\tta = \bbm 1\amp 2\\1\amp 2\ebm\)
5.1.4.9.
Answer.
\(\tta = \bbm 5\amp 2\\2\amp 1\ebm\)
5.1.4.11.
Answer.
\(\tta = \bbm 0\amp 1\\3\amp 0\ebm\)
5.1.4.13.
Answer.
\(\tta = \bbm 0\amp -1\\ -1\amp -1\ebm\)
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.
Answer.
Yes
5.2.3.
Answer.
No; cannot add a constant.
5.2.5.
Answer.
Yes.
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.
\(\tta\) 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.
\(\tta\) is diagonal, as is \(\ttat\text{.}\) \(\bbm 1\amp 0\amp 0\\ 0\amp 2\amp 0\\ 0\amp 0\amp -1\ebm\)
6.1.7.
Answer.
\(\tta\) 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.
\(\tta\) 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.
\(\tta\) is upper triangular; \(\ttat\) is lower triangular. \(\bbm -3\amp 0\amp 0\\ -4\amp -3\amp 0\\ -5\amp 5\amp -3\ebm\)
6.1.19.
Answer.
\(\tta\) 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.
\(\tta\) 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.
Answer.
\(3\)
6.2.5.
Answer.
\(-9\)
6.2.7.
Answer.
\(-5\)
6.2.9.
Answer.
\(1\)
6.2.11.
Answer.
Not defined; the matrix must be square.
6.2.13.
Answer.
\(0\)
6.2.15.
Answer.
\(n\)
6.2.17.
Answer.
- \(\tr(\tta)=-1\text{;}\) \(\tr(\ttb)=\text{;}\) \(\tr(\tta+\ttb)=5\)
- \(\displaystyle \tr(\tta\ttb) = 201 = \tr(\ttb\tta)\)
6.2.19.
Answer.
- \(\tr(\tta)=-5\text{;}\) \(\tr(\ttb)=-4\text{;}\) \(\tr(\tta+\ttb)=-9\)
- \(\displaystyle \tr(\tta\ttb) = 23 = \tr(\ttb\tta)\)
6.3 The Determinant
Exercises
6.3.1.
Answer.
\(34\)
6.3.3.
Answer.
\(-44\)
6.3.5.
Answer.
\(-44\)
6.3.7.
Answer.
\(28\)
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.
Answer.
\(-59\)
6.3.15.
Answer.
\(15\)
6.3.17.
Answer.
\(3\)
6.3.19.
Answer.
\(0\)
6.3.21.
Answer.
\(0\)
6.3.23.
Answer.
\(-113\)
6.3.25.
Answer.
Hint: \(C_{1,1}= d\text{.}\)
6.4 Properties of the Determinant
Exercises
6.4.1.
Answer.
\(84\)
6.4.3.
Answer.
\(0\)
6.4.5.
Answer.
\(10\)
6.4.7.
Answer.
\(24\)
6.4.9.
Answer.
\(175\)
6.4.11.
Answer.
\(-200\)
6.4.13.
Answer.
\(34\)
6.4.15.
Answer.
- \(\det(\tta) = 90\text{;}\) \(2R_1\rightarrow R_1\)
- \(\det(\ttb) = 45\text{;}\) \(10R_1+R_3\rightarrow R_3\)
- \(\det(\ttc) = 45\text{;}\) \(\ttc = \ttat\)
6.4.17.
Answer.
- \(\det(\tta) = -16\text{;}\) \(R_1\leftrightarrow R_2\) then \(R_1\leftrightarrow R_3\)
- \(\det(\ttb) = -16\text{;}\) \(-R_1\rightarrow R_1\) and \(-R_2\rightarrow R_2\)
- \(\det(\ttc) = -432\text{;}\) \(\ttc = 3*\ttm\)
6.4.19.
Answer.
\(\det(\tta)=4\text{,}\) \(\det(\ttb)=4\text{,}\) \(\det(\tta\ttb)=16\)
6.4.21.
Answer.
\(\det(\tta)=-12\text{,}\) \(\det(\ttb)=29\text{,}\) \(\det(\tta\ttb)=-348\)
6.4.23.
Answer.
\(-59\)
6.4.25.
Answer.
\(15\)
6.4.27.
Answer.
\(3\)
6.4.29.
Answer.
\(0\)
6.5 Applications of the Determinant
6.5.3 Exercises
6.5.3.1.
Answer.
- \(\det(\tta) = -123\text{,}\) \(\det(\tta_1) = -492\text{,}\) \(\det(\tta_2) = 123\text{,}\) \(\det(\tta_3) = 492\)
- \(\displaystyle \vx = \bbm 4\\ -1\\ -4\ebm\)
6.5.3.3.
Answer.
- \(\det(\tta) = 96\text{,}\) \(\det(\tta_1) = -960\text{,}\) \(\det(\tta_2) = 768\text{,}\) \(\det(\tta_3) = 288\)
- \(\displaystyle \vx = \bbm -10\\ 8\\ 3\ebm\)
6.5.3.5.
Answer.
- \(\det(\tta) = -43\text{,}\) \(\det(\tta_1) = 215\text{,}\) \(\det(\tta_2) = 0\)
- \(\displaystyle \vx = \bbm -5\\ 0\ebm\)
6.5.3.7.
Answer.
- \(\det(\tta) = 16\text{,}\) \(\det(\tta_1) = -64\text{,}\) \(\det(\tta_2) = 80\)
- \(\displaystyle \vx = \bbm -4\\ 5\ebm\)
6.5.3.9.
Answer.
- \(\det(\tta) = 0\text{,}\) \(\det(\tta_1) = 0\text{,}\) \(\det(\tta_2) = 0\)
- Infinite solutions exist.
6.5.3.11.
Answer.
- \(\det(\tta) = 0\text{,}\) \(\det(\tta_1) = 0\text{,}\) \(\det(\tta_2) = 0\text{,}\) \(\det(\tta_3) = 0\)
- Infinite 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.
Answer.
\(A\) is not invertible.
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
7.1.1.
Answer.
\(\lambda = 3\)
7.1.3.
Answer.
\(\lambda = -5\)
7.1.5.
Answer.
\(\lambda = -2\)
7.1.7.
Answer.
\(\vx = \bbm 3\\-7\\7\ebm\)
7.1.9.
Answer.
\(\vx = \bbm -1\\1\\1\ebm\)
7.1.11.
Answer.
\(\vx = \bbm 2\\3\ebm\)
7.1.13.
Answer.
\(\lda_1 = 2\) with \(\vx[1] = \bbm 1\\0\\0\ebm\text{;}\) \(\lda_2 = 3\) with \(\vx[2] = \bbm-1\\1\\0\ebm\text{;}\) \(\lda_3 = 7\) with \(\vx[3] = \bbm-1\\15\\10\ebm\)
7.1.15.
Answer.
\(\lda_1 = -2\) with \(\vx[1] = \bbm 0\\0\\1\ebm\text{;}\) \(\lda_2 = 1\) with \(\vx[2] = \bbm 0\\3\\5\ebm\text{;}\) \(\lda_3 = 5\) with \(\vx[3] = \bbm 28\\7\\1\ebm\)
7.1.17.
Answer.
\(\lda_1 = -4\) with \(\vx[1] = \bbm-6\\1\\11\ebm\text{;}\) \(\lda_2 = -1\) with \(\vx[2] = \bbm 0\\0\\1\ebm\text{;}\) \(\lda_3 = 5\) with \(\vx[3] = \bbm 3\\1\\2\ebm\)
7.1.19.
Answer.
\(\lda_1 = -3\) with \(\vx[1] = \bbm-2\\1\ebm\text{;}\) \(\lda_2 = 5\) with \(\vx[2] = \bbm 6\\1\ebm\)
7.1.21.
Answer.
\(\lda_1 = -5\) with \(\vx[1] = \bbm 24\\13\\8\ebm\text{;}\) \(\lda_2 = -2\) with \(\vx[2] = \bbm 6\\5\\1\ebm\text{;}\) \(\lda_3 = 3\) with \(\vx[3] = \bbm 0\\1\\0\ebm\)
7.1.23.
Answer.
\(\lda_1 = -5\) with \(\vx[1] = \bbm 1\\1\ebm\text{;}\) \(\lda_2 = 2\) with \(\vx[2] = \bbm-4\\3\ebm\)
7.1.25.
Answer.
\(\lda_1 = -5\) with \(\vx[1] = \bbm-1\\5\ebm\text{;}\) \(\lda_2 = 5\) with \(\vx[2] = \bbm 1\\5\ebm\)
7.1.27.
Answer.
\(\lda_1 = 4\) with \(\vx[1] = \bbm 9\\1\ebm\text{;}\) \(\lda_2 = 5\) with \(\vx[2] = \bbm 8\\1\ebm\)
7.2 Properties of Eigenvalues and Eigenvectors
Exercises
7.2.1.
Answer.
- \(\lda_1 = 1\) with \(\vx[1] = \bbm 4\\1\ebm\text{;}\) \(\lda_2 = 4\) with \(\vx[2] = \bbm 1\\1\ebm\)
- \(\lda_1 = 1\) with \(\vx[1] = \bbm-1\\1\ebm\text{;}\) \(\lda_2 = 4\) with \(\vx[2] = \bbm-1\\4\ebm\)
- \(\lda_1 = 1/4\) with \(\vx[1] = \bbm 1\\1\ebm\text{;}\) \(\lda_2 = 1\) with \(\vx[2] = \bbm 4\\1\ebm\)
- 5
- 4
7.2.3.
Answer.
- \(\lda_1 = -1\) with \(\vx[1] = \bbm-5\\1\ebm\text{;}\) \(\lda_2 = 0\) with \(\vx[2] = \bbm-6\\1\ebm\)
- \(\lda_1 = -1\) with \(\vx[1] = \bbm 1\\6\ebm\text{;}\) \(\lda_2 = 0\) with \(\vx[2] = \bbm 1\\5\ebm\)
- \(\tta\) is not invertible.
- \(\displaystyle -1\)
- \(\displaystyle 0\)
7.2.5.
Answer.
- \(\lda_1 = -4\) with \(\vx[1] = \bbm-7\\-7\\6\ebm\text{;}\) \(\lda_2 = 3\) with \(\vx[2] = \bbm 0\\0\\1\ebm\text{;}\) \(\lda_3 = 4\) with \(\vx[3] = \bbm 9\\1\\22\ebm\)
- \(\lda_1 = -4\) with \(\vx[1] = \bbm-1\\9\\0\ebm\text{;}\) \(\lda_2 = 3\) with \(\vx[2] = \bbm-20\\26\\7\ebm\text{;}\) \(\lda_3 = 4\) with \(\vx[3] = \bbm-1\\1\\0\ebm\)
- \(\lda_1 = -1/4\) with \(\vx[1] = \bbm-7\\-7\\6\ebm\text{;}\) \(\lda_2 = 1/3\) with \(\vx[2] = \bbm 0\\0\\1\ebm\text{;}\) \(\lda_3 = 1/4\) with \(\vx[3] = \bbm 9\\1\\22\ebm\)
- 3
- \(\displaystyle -48\)
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.
\(c_A(x) = (x-6)(x-23)\)
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.
