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Elementary Linear Algebra For University of Lethbridge Math 1410

Appendix A Answers to Selected Exercises

1 The Real and Complex Numbers
1.2 Real Number Arithmetic

Exercises

1.2.3.
Answer.
\(-\dfrac{3\sqrt[5]{3}}{8} = -\dfrac{3^{6/5}}{8}\)

1.3 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.
Cartesian plane showing the location of the points to be plotted in this exercise.
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.
\(d = \sqrt{x^2 + y^2}\text{,}\) \(M = \left( \frac{x}{2}, \frac{y}{2}\right)\)
1.3.13.
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.
1.4.3.
Answer.
1.4.5.
Answer.
1.4.7.
Answer.
1.4.9.
Answer.
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.
\(i^{117} = \left(i^{4}\right)^{29} \cdot i = 1 \cdot i = i\)

2 Vectors
2.1 Introduction to Cartesian Coordinates in Space
2.1.3 Exercises

2.2 An Introduction to Vectors

Exercises

2.2.5.
2.2.5.a
Answer.
\(\vec u+\vec v = \bbm 2\\-1\ebm\text{;}\) \(\vec u -\vec v = \bbm 0\\-3\ebm\text{;}\) \(2\vec u-3\vec v = \bbm -1\\-7\ebm\text{.}\)
2.2.19.
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

2.3.2.9.

Answer.
\(\cos^{-1}\mathopen{}\left(\frac{3}{\sqrt{10}}\right)\)

2.3.2.17.

Answer.
\(\left<\frac{-5}{10},\frac{15}{10}\right>\)

2.3.2.21.

Answer.
\(\left<\frac{14}{14},\frac{28}{14},\frac{42}{14}\right>\)

2.3.2.27.

Answer 1.
\(\left<\frac{14}{14},\frac{28}{14},\frac{42}{14}\right>\)
Answer 2.
\(\left<\frac{0}{14},\frac{42}{14},\frac{-28}{14}\right>\)

2.4 The Cross Product
2.4.3 Exercises

2.4.3.7.

Answer.
\(\vec u\times \vec v = \bbm 0\\0\\ad-bc\ebm\)

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.
\(\left<0,1,0\right>\hbox{ or }\left<0,-1,0\right>\)

2.5 Lines
2.5.4 Exercises

2.6 Planes
2.6.2 Exercises

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.
\(3x+8y-10z = 24\)

2.7 Span and Linear Independence
2.7.4 Exercises

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

3.2 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.
\(\bbm 2\amp -1\amp 7\\0\amp 2\amp -1\\5\amp 0\amp 3\\ \ebm\)

3.3 Elementary Row Operations and Gaussian Elimination

Exercises

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

3.5.11.
Answer.
  1. Substitution yields the equations \(2 = ae^{b}\) and \(4 = ae^{2b}\text{;}\) these are not linear equations.
  2. \(y = ae^{bx}\) implies that \(\ln y = \ln (ae^{bx}) = \ln a + \ln e^{bx} = \ln a + bx\text{.}\)
  3. 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 \(b = \ln 2\text{.}\)
  4. 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 β€œfirst” is where 0 three point shots are taken (\(x_3=0\)), 30 free throws and 40 two point shots; the β€œlast” 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 \(y=3x^2-2x-2\text{.}\)
3.5.23.
Answer.
Yes. \(\vec{x} = -2\vec{w}_1+2\vec{w}_2\text{.}\)

3.6 Vector Solutions to Linear Systems

Exercises

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.
3.6.15.
Answer.
  1. \(\displaystyle \vx=x_3\bbm 5/4\\ 1\\ 1\ebm\)
  2. \(\displaystyle \vx=\bbm 1\\ 0\\ 0\ebm+x_3\bbm 5/4\\ 1\\ 1\ebm\)
3.6.17.
Answer.
  1. \(\displaystyle \vx=x_3\bbm 6\\ -4\\ 1\ebm\)
  2. \(\displaystyle \vx=\bbm -12\\ 8\\0\ebm +x_3\bbm 6\\ -4\\ 1\ebm\)
3.6.19.
Answer.
  1. \(\displaystyle \vx=x_3\bbm 2\\2/5\\1\\0\ebm+x_4\bbm-1\\2/5\\0\\1\ebm\)
  2. \(\displaystyle \vx=\bbm-2\\2/5\\0\\0\ebm + x_3\bbm 2\\2/5\\1\\0\ebm +x_4\bbm-1\\2/5\\0\\1\ebm\)
3.6.21.
Answer.
  1. \(\displaystyle \vx=x_2\bbm-1/2\\1\\0\\0\\0\ebm+x_4\bbm 1/2\\0\\-1/2\\1\\0\ebm+x_5\bbm 13/2\\0\\-2\\0\\1\ebm\)
  2. \(\displaystyle \vx=\bbm-5\\0\\ 3/2 \\0\\0\ebm + x_2\bbm-1/2\\1\\0\\0\\0\ebm+x_4\bbm 1/2 \\0\\-1/2\\1\\0\ebm+x_5\bbm 13/2\\0\\-2\\0\\1\ebm\)
3.6.23.
Answer.
  1. \(\displaystyle \vx=x_4\bbm 1\\ 13/9 \\-1/3 \\1\\0\ebm+x_5\bbm 0\\ -1 \\-1\\0\\1 \ebm\)
  2. \(\displaystyle \vx=\bbm 1\\ 1/9 \\ 5/3 \\ 0\\ 0 \ebm + x_4\bbm 1\\ 13/9 \\ -1/3 \\1\\0\ebm+x_5\bbm 0\\ -1 \\-1\\0\\1 \ebm\)
3.6.25.
Answer.
\(\vx = \bbm 0.5\\0\ebm + x_2\bbm 2.5\\1\ebm = \vec{x}_{p} + x_2\vvv\)
A line in the plane

4 Matrix Algebra
4.1 Matrix Addition and Scalar Multiplication

Exercises

4.2 Matrix Multiplication

Exercises

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.
\(DA= \bbm d_1a\amp d_1b\amp d_1c\\ d_2d\amp d_2e\amp d_2f\\ d_3g\amp d_3h\amp d_3i\ebm\text{,}\) \(AD = \bbm d_1a\amp d_2b\amp d_3c\\ d_1d\amp d_2e\amp d_3f\\ d_1g\amp d_2h\amp d_3i\ebm\)
4.2.41.
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.
\(A^2 = \bbm 0 \amp 0 \amp 1\\1\amp 0\amp 0\\0\amp 1\amp 0\ebm\text{;}\) \(A^3 = \bbm 1\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\ebm\)

4.3 Solving Matrix Equations \(AX=B\)

Exercises

4.3.3.
Answer.
\(X=\bbm -1 \amp 2 \amp -4\ -6\amp -2\amp 3 \ebm \)
4.3.5.
Answer.
\(X=\bbm -5 \amp 2 \amp -3\ -4\amp -3\amp -2 \ebm \)
4.3.9.
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

4.5 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.
\(A^{-1} = \bbm -3 \amp 7\\ 1 \amp -2 \ebm\text{,}\) \((A^{-1})^{-1} = \bbm 2 \amp 7\\1 \amp 3 \ebm\)
4.5.11.
Answer.
Likely some entries that should be 0 will not be exactly 0, but rather very small values.

4.6 Elementary Matrices

Exercises

4.6.9.
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

5.1.4.15.

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

5.2.11.
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. \(\bbm 6\amp -4\amp -5\\ -4\amp 0\amp 2\\ -5\amp 2\amp -2\ebm\)
6.1.11.
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

6.3 The Determinant

Exercises

6.3.9.
Answer.
  1. 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.
  2. \(C_{1,2}=34\text{,}\) \(C_{1,2}=-24\text{,}\) \(C_{1,3}=11\)
6.3.11.
Answer.
  1. 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.
  2. \(C_{1,2}=-3\text{,}\) \(C_{1,2}=-63\text{,}\) \(C_{1,3}=18\)

6.4 Properties of the Determinant

Exercises

6.4.19.
Answer.
\(\det(A)=4\text{,}\) \(\det(B)=4\text{,}\) \(\det(AB)=16\)
6.4.21.
Answer.
\(\det(A)=-12\text{,}\) \(\det(B)=29\text{,}\) \(\det(AB)=-348\)

6.5 Applications of the Determinant
6.5.3 Exercises

6.5.3.13.

Answer.
\(A^{-1} = \frac{1}{8}\bbm -11 \amp 10 \amp 13\\6\amp -4\amp -2\\9\amp -6\amp -7\ebm\)

6.5.3.17.

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

7.1.13.
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

7.2.5.
Answer.
  1. \(\lambda _1 = -4\) 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 \(\vx[3] = \bbm 9\\1\\22\ebm\)
  2. \(\lambda _1 = -4\) 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 \(\vx[3] = \bbm-1\\1\\0\ebm\)
  3. \(\lambda _1 = -1/4\) 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 \(\vx[3] = \bbm 9\\1\\22\ebm\)
  4. \(\displaystyle -48\)

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.