Linear Algebra 2

Contents: Vectors Lines & Planes Eigenvalues & Eigenvectors Sample Problems

Vectors

In general, vectors are written with bolded letters: $\mathbf{v}=\vec{AB}$. The vector v starts from the point A, and to the point B. The point A is referred as the Initial Point, and the point B is referred as the Terminal Point.

The attribute of vectors:

• Direction: The direction of the vector's arrowhead.
• Magnitude: The length of the arrow.

Multiple vectors are equivalent, if they share the same Direction and Magnitude (or length).

If a vector's initial and terminal points coincide, then its length will be 0. That vector can be called as a Zero Vector, which is denoted with O.

The component of a vector in Rⁿ, can be written as a set of real numbers. For example: $\mathbf{v}=(v_1,v_2,...,v_n)$

The vector v in R² from point A to B can also be written as:

$\mathbf{v}=(x_B-x_A,y_B-y_A)$

The addition, subtraction, and scalar multiplication of vectors are the same as when operating matrices.

For examples, let $\mathbf{v}=(v_1,v_2,...,v_n)$ and $\mathbf{w}=(w_1,w_2,...,w_n)$.

Vector Addition:
If vector v and w is added, then the resulting vector will be: $\mathbf{v+w}=(v_1+w_1,v_2+w_2,...,v_n+w_n)$.



Vector Subtraction: A negative of a vector's length will be the same as its positive counterpart, but its direction is the opposite.


The resulting vector will be: $\mathbf{v-w}=(v_1-w_1,v_2-w_2,...,v_n-w_n)$.

Vector Scalar Multiplication:If vector v is multiplied by a scalar (or a constant) k, then the resulting vector is: $k\mathbf{ v}=(k v_1,k v_2,...,k v_n)$



The norm, or a length of vector is denoted as $\left \| v \right \|$. Let $\mathbf{v}=(v_1,v_2,...,v_n)$, a vector in Rⁿ. Norm of v is defined as:

$\left \| \mathbf{ v} \right \|=\sqrt{v{_{1}}^{2}+v{_{2}}^{2}+...+v{_{n}}^{2}}$

The properties of norm:

• $\left \| \mathbf{ v} \right \|\geq 0$
• $\left \| \mathbf{ v} \right \|=0$, if and only if v is a Zero Vector.
• $\left \| k\mathbf{ v} \right \|=|k|.\left \| \mathbf{ v} \right \|$

Vectors whose norm is equal to 1, is called Unit Vectors. In order to find an unit vector u, in the same direction as vector v, the following formula: $\mathbf{u}=\frac{\mathbf{v}}{\left \| \mathbf{v} \right\|}$ can be used. Standard Unit Vectors, are Unit Vectors in the positive directions of the coordinate axes.

Dot product (Euclidean inner product) of two vectors v and w (and θ is the angle formed) are defined as:

$\mathbf{v}\cdot \mathbf{w}=\left \| \mathbf{ v} \right \|\left \| \mathbf{ w} \right \|\cos{\theta}$

Dot product can also be calculated based on the components of each vector:

$\mathbf{v}\cdot \mathbf{w}=v_1w_1 + v_2w_2 + ... + v_nw_n$

Based on the dot product formula, the angle formed can be identified:

$\theta\begin{cases} & \mathbf{v}.\mathbf{w} > 0 \text{ if } \theta \text{ is acute} \\ & \mathbf{v}.\mathbf{w} = 0 \text{ if } \theta=90^{\circ} \\ & \mathbf{v}.\mathbf{w} < 0 \text{ if } \theta\text{ is obtuse}\end{cases}$

Let u, v and w as vectors in Rⁿ, and k as a scalar. The properties of dot product:

• Symmetry property: $\mathbf{v}\cdot \mathbf{w}=\mathbf{w}\cdot \mathbf{v}$
• Distributive property: $\mathbf{v}\cdot (\mathbf{u}\pm\mathbf{w})=\mathbf{v}\cdot \mathbf{u}\pm\mathbf{v}\cdot\mathbf{w}$
• Homogeneity property: $k(\mathbf{v} \cdot \mathbf{w})=(k\mathbf{v})\cdot\mathbf{w}$
• $\mathbf{v}\cdot \mathbf{v}=\left \| \mathbf{v} \right \|^2$
• $\mathbf{v}\cdot \mathbf{v}\geq 0$ and $\mathbf{v}\cdot \mathbf{v}= 0$ if and only if v is a Zero Vector.

Cross product of two vectors are defined as:

$\begin{align*} \mathbf{v} \times \mathbf{w}&=\begin{vmatrix} i & j & k\\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix} \\ &= \begin{vmatrix} v_2 & v_3\\ w_2 & w_3 \end{vmatrix}i-\begin{vmatrix} v_1 & v_3\\ w_1 & w_3 \end{vmatrix}j+ \begin{vmatrix} v_1 & v_2\\ w_1 & w_2 \end{vmatrix}k\\ \end{align*}$

Let u, v and w as vectors in Rⁿ, and k as a scalar. The properties of cross product:

• $\mathbf{v} \times \mathbf{w}= -\mathbf{w} \times \mathbf{v}$
• $\mathbf{v} \times (\mathbf{w}\pm\mathbf{u})= \mathbf{v} \times \mathbf{w} \pm \mathbf{v} \times \mathbf{u}$
• $(\mathbf{v}\pm\mathbf{w}) \times \mathbf{u}= \mathbf{v} \times \mathbf{u} \pm \mathbf{w} \times \mathbf{u}$
• $k(\mathbf{v} \times \mathbf{w})=(k\mathbf{v})\times \mathbf{w}=\mathbf{v}\times (k\mathbf{w})$
• $\mathbf{v}\times\mathbf{0}=\mathbf{0}\times\mathbf{v}=\mathbf{0}$
• $\mathbf{v}\times\mathbf{v}=\mathbf{0}$

The dot product and cross product have the following relations:

• $\mathbf{v} \cdot (\mathbf{v} \times \mathbf{w})= 0$: v and v+w are orthogonal to each other.
• Lagrange's Identity: $\left \| \mathbf{v} \times \mathbf{w} \right \|^2 = \left \| \mathbf{v}\right \|^2\left \| \mathbf{w}\right \|^2-(\mathbf{v} \cdot \mathbf{w})^2$
• $\mathbf{u}\times(\mathbf{v} \times \mathbf{w})= (\mathbf{u}\cdot\mathbf{w})\mathbf{v}-(\mathbf{u}\cdot\mathbf{v})\mathbf{w}$
• $(\mathbf{u}\times\mathbf{v}) \times \mathbf{w}= (\mathbf{u}\cdot\mathbf{w})\mathbf{v}-(\mathbf{v}\cdot\mathbf{w})\mathbf{u}$

Geometric interpretation of cross product (derived from Lagrange's Identity):

$\left \| \mathbf{v}\times\mathbf{w}\right\|=\left \| \mathbf{v} \right \| \left \| \mathbf{w} \right \| \sin \theta$

Projection Theorem
Let v and w are vectors in Rⁿ (v is not a Zero Vector). v can be expressed as $\mathbf{v}=\mathbf{u_1}+\mathbf{u_2}$, where u₁ is the scalar multiple of w, and u₂ is orthogonal to w.


In the figure provided, the vector component of v along w:

$\text{proj}_\mathbf{w}\mathbf{v}=\frac{\mathbf{v}\mathbf{w}}{{\left \|\mathbf{w} \right \|}^2}\mathbf{w}$

and, the vector component of v orthogonal to w:

$\mathbf{v}-\text{proj}_\mathbf{w}\mathbf{v}=\mathbf{v}-\frac{\mathbf{v}\mathbf{w}}{{\left \|\mathbf{w} \right \|}^2}\mathbf{w}$

Linear Combination
Let v and w are vectors in Rⁿ, and k_i is a scalar. Vector v can be a linear combination if it fulfuills:

$\mathbf{v}=k_1\mathbf{w_1}+k_2\mathbf{w_2}+...+k_n\mathbf{w_n}$

Let S be a non-empty set of vectors in a vector space V:
$k_1\mathbf{v_1}+k_2\mathbf{v_2}+...+k_n\mathbf{v_n}=0$
has only one solution, which is the Trivial Solution:
$k_1=k_2=...=k_n=0$
So, S is a Linearly Independent Set. Otherwise, if there is another solution, S is a Linearly Dependent Set.

Lines & Planes

Let v be a vector parallel to the line L, and has x₀ as the initial point, and x as the point in the line, and k as a parameter with arbitrary value:

$x=x_0+k\mathbf{v}$

Since vector can be defined by the difference between the terminal points (x₁) and initial points (x₀), the above formula can be derived to:

$x=x_0+k(x_1-x_0)$

In R³, the formula is:

$x=x_0+k_1v_1+k_2v_2$

Let P and P_o be a point on the plane, and r as a vector from P_o to (a,b,c). So, the point-normal equation of the plane can be written as:

$a(x-x_o)+b(y-y_o)+c(z-z_o)=0$

Let P_o be a point with the coordinate x_o, y_o and z_o, and a plane $ax+by+cz+d=0$. The distance between the point and the plane can written as:

$\text{Distance}=\frac{|ax_o+by_o+cz_o+d|}{\sqrt{a^2+b^2+c^2}}$

Eigenvalues & Eigenvectors

If A is a n × n matrix, then a non-zero vector x in Rⁿ is called the eigenvector of A, if: $Ax=\lambda x$, where λ is a scalar, and it is the eigenvalue of A. x is also called the eigenvector corresponding to λ.

λ is the eigenvalue of A, if and if only it satisifies the Characteristic Equation:

$det(\lambda I-A)=0$

The eigenspace of A corresponding to λ is the solution space of the above equation.

Diagonalizable Matrix
The square matrix A (n × n) is a Diagonalizable Matrix if there exists an invertible matrix P so that $P^{-1}AP$ is a diagonal matrix. For a matrix to be diagonalizable, that matrix must have n linearly independent eigenvectors.

With this concept, we can calculate a matrix to the k-th power. Let k be a positive integer, D as a diagonal matrix, and P as the matrix that diagonalizes A into D:

$A^{k}=PD^{k}P^{-1}$

Sample Problems

Problem 1: Vectors
If the terminal point of a vector is (3, 4), and that vector v is (5, 3), what is the initial point of that vector?

Know:

• Terminal Point = (3,4)
• v = (5, 3)

Want:

• initial point

Let A be an initial point of the vector, and B be an terminal point of the vector. So:
$\begin{align*} \mathbf{v} &= \overrightarrow{AB}\\ \begin{bmatrix} 5\\3 \end{bmatrix} &= \begin{bmatrix} 3\\ 4 \end{bmatrix}-\begin{bmatrix} x\\ y \end{bmatrix}\\ \begin{bmatrix} x\\y \end{bmatrix} &= \begin{bmatrix} -2\\ 1 \end{bmatrix}\\ \end{align*}$

The initial point is (-2, 1).

Problem 2: Vectors
Let u = (−3, 2, 1, 0), v = (4, 7,−3, 2), and w = (5,−2, 8, 1). Find the components of $3u-2v+5w$.

Know:

• u = (−3, 2, 1, 0)
• v = (4, 7,−3, 2)
• w = (5,−2, 8, 1)

Want:

• 3u-2v+5w

$\begin{align*} 3u-2v+5w &= 3\begin{bmatrix} -3\\ 2\\ 1\\ 0 \end{bmatrix}-2\begin{bmatrix} 4\\ 7\\ -3\\ 2 \end{bmatrix}+5\begin{bmatrix} 5\\ -2\\ 8\\ 1 \end{bmatrix} \\ &= \begin{bmatrix} -9\\ 6\\ 3\\ 0 \end{bmatrix}-\begin{bmatrix} 8\\ 14\\ -6\\ 4 \end{bmatrix}+\begin{bmatrix} 25\\ -10\\ 40\\ 5 \end{bmatrix} \\ &= \begin{bmatrix} -8\\ -18\\ 49\\ 1 \end{bmatrix} \\ \end{align*}$

The resulting vector is (-8, -18, 49, 1).

Problem 3: Vectors
If there exists point A(1, 3) and B(7, 5), find the middle point of a line that connects AB.

Know:

• A(1, 3)
• B(7, 5)

Want:

• middle point of line AB

Let v be a vector whose initial point is A, and terminal point is B. So:
$\begin{align*} \mathbf{v} &= \overrightarrow{AB}\\ &= \begin{bmatrix} 7\\ 5 \end{bmatrix}-\begin{bmatrix} 1\\ 3 \end{bmatrix}\\ &= \begin{bmatrix} 6\\ 2 \end{bmatrix}\\ \end{align*}$

Now that the vector v is found, in order to find the middle point, the vector can be halved so that its terminal points and the middle point we're looking for are the same. So, let C be the middle point:
$\begin{align*} \frac{1}{2}\mathbf{v} &= \overrightarrow{AC}\\ \begin{bmatrix} 3\\ 1 \end{bmatrix}&= \begin{bmatrix} x\\ y \end{bmatrix}-\begin{bmatrix} 1\\ 3 \end{bmatrix}\\ \begin{bmatrix} x\\ y \end{bmatrix}&= \begin{bmatrix} 4\\ 4 \end{bmatrix}\\ \end{align*}$

So, the middle point of the line that connects A and B is (4, 4).

Problem 4: Vectors
Determine the values for x, y, z so that $x(1, 2, 3) + y(1, 4, 6) + z(2, -3, -5) = (9, 1, 0)$ .

Know:

• x(1, 2, 3) + y(1, 4, 6) + z(2, -3, -5) = (9, 1, 0)

Want:

• x, y, z

This is a system of linear equations. In order to find the x, y and z, many methods can be used. For this solution, the row operation method will be used:
$\begin{align*} &\begin{bmatrix} 1 & 1 & 2 & 9\\ 2 & 4 & -3 & 1\\ 3 & 6 & -5 & 0 \end{bmatrix} \\ &\rightarrow \begin{bmatrix} 1 & 1 & 2 & 9\\ 0 & 2 & -7 & -17\\ 0 & 3 & -11 & -27 \end{bmatrix}\begin{matrix} \text{No Change}\\ {R_2}'=R_2-2R_1\\ {R_3}'=R_3-3R_1 \end{matrix} \\ &\rightarrow \begin{bmatrix} 1 & 1 & 0 & 3\\ 0 & 2 & -7 & -17\\ 0 & 0 & 1 & 3 \end{bmatrix}\begin{matrix} {R_1}'=R_1-2{R_3}'\\ \text{No Change}\\ {R_3}'=3R_2-2R_3 \end{matrix} \\ &\rightarrow \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 3 \end{bmatrix}\begin{matrix} {R_1}'=R_1-{R_2}'\\ {R_2}'=\frac{1}{2}(R_2+7R_3)\\ \text{No Change} \end{matrix} \\ \end{align*}$

So, x = 1, y = 2 and z = 3.

Problem 5: Vectors
Given two vectors v = (3, -2, 1, 3) and w = (-4, 1, -3, 7):

1. Find the euclidean distance between both vectors.
2. Find cos θ, where θ is the angle formed between both vectors.
3. State whether θ is an acute, obtuse or right angle.

Know:

• v = (3, -2, 1, 3)
• w = (-4, 1, -3, 7)

Want:

• distance between both vectors.
• cos θ
• to know if the angle is an acute, obtuse, or right angle.

Sub-Problem A
$\begin{align*} \text{Distance} &= \sqrt{(v_1-w_1)^2+(v_2-w_2)^2+(v_2-w_2)^2+(v_2-w_2)^2}\\ &= \sqrt{7^2+(-3)^2+4^2+(-4)^2}\\ &= \sqrt{49+9+16+16}\\ &= \sqrt{90}\\ &= 3\sqrt{10}\\ \end{align*}$

So, the distance is $3\sqrt{10}$.

Sub-Problem B
Using the relation between dot product and angle formula:
$\begin{align*} \mathbf{v}\cdot \mathbf{w}&= \left \| \mathbf{v}\right \| \left \| \mathbf{w}\right \| \cos \theta \\ 3(-4)-2(1)+1(-3)+3(7)&=\sqrt{9+4+1+9}\sqrt{16+1+9+49}\cos\theta\\ 4&= \sqrt{1725}\cos\theta\\ \cos\theta&= \frac{4\sqrt{69}}{345}\\ \end{align*}$

So, the cosine of the angle is $\frac{4\sqrt{69}}{345}$.

Sub-Problem C
Since the dot product of both vectors are 4, which is greater than 0, it can be concluded the angle formed is an acute angle.

Problem 6: Vectors
Given v = (2, 3, 3) and w = (3, 0, 4):

1. Find the component of vector v along w.
2. Find the component of vector v orthogonal to w

Know:

• v = (2, 3, 3)
• w = (3, 0, 4)

Want:

• proj_wv
• v - proj_wv

Sub-Problem A
$\begin{align*} \text{proj}_{\mathbf{w}}\mathbf{v}&=\frac{\mathbf{v}\cdot\mathbf{w}}{\left \| \mathbf{w}\right \|^2}\mathbf{w}\\ &= \frac{18}{25}\begin{bmatrix} 3\\ 0\\ 4 \end{bmatrix} \\ &= \begin{bmatrix} \frac{54}{25}\\ 0\\ \frac{72}{25} \end{bmatrix} \end{align*}$

So, the projection of the vector v along w is $(\frac{54}{25},0,\frac{72}{25})$.

Sub-Problem B
$\begin{align*} \mathbf{v}-\text{proj}_{\mathbf{w}}\mathbf{v}&=\begin{bmatrix} 2\\ 3\\ 3 \end{bmatrix}-\begin{bmatrix} \frac{54}{25}\\ 0\\ \frac{72}{25} \end{bmatrix}\\ &= \begin{bmatrix} -\frac{4}{25}\\ 0\\ \frac{3}{25} \end{bmatrix} \\ \end{align*}$

So, the projection of the vector v orthogonal to w is $(-\frac{4}{25},0,-\frac{3}{25})$.

Problem 7: Vectors
Given the vertices that form a shape, calculate the area:

1. Parallelogram: A(1, 2), B(4, 4), C(7, 5), D(4, 3)
2. Triangle: X(1, 1), Y(2, 2), Z(3, −3)

Know:

• Parallelogram: A(1, 2), B(4, 4), C(7, 5), D(4, 3)
• Triangle: X(1, 1), Y(2, 2), Z(3, −3)

Want:

• area of ABCD
• area of XYZ

Sub-Problem A
Let v be a vector that connects A to B, and w as a vector that connects B to D. So:

• v = (3, 2)
• w = (0,-1)

Though cross product can only be done with 3D vectors, the third component of each 2D vector (or vertices) will be 0. So:

• v = (3, 2, 0)
• w = (0,-1, 0)

The area of the parallelogram can be calculated:
$\begin{align*} \text{Area} &=\left \| \mathbf{v}\times\mathbf{w} \right \| \\ \mathbf{v}\times\mathbf{w}&= \begin{pmatrix} \begin{vmatrix} 2 & 0\\ -1 & 0 \end{vmatrix}, & -\begin{vmatrix} 3 & 0\\ 0 & 0 \end{vmatrix}, & \begin{vmatrix} 3 & 2\\ 0 & -1 \end{vmatrix} \end{pmatrix}\\ &= \begin{pmatrix} 0,&0, &-3 \end{pmatrix}\\ \left \| \mathbf{v}\times\mathbf{w} \right \| &= 3 \end{align*}$

So, the area of the parallelogram formed is 3.

Sub-Problem B
Let v be a vector that connects X to Y, and w as a vector that connects X to Z. So:

• v = (1, 1)
• w = (2,-4)

Though cross product can only be done with 3D vectors, the third component of each 2D vector (or vertices) will be 0. So:

• v = (1, 1, 0)
• w = (2,-4, 0)

Finding the θ, which is the angle formed between both vectors:
$\begin{align*} \left \|\mathbf{v}\times\mathbf{w} \right \| &=\left \| \mathbf{v}\right \| \left \| \mathbf{w}\right \| \sin \theta \\ \sin \theta &=\frac{\left \|\mathbf{v}\times\mathbf{w} \right \|}{\left \| \mathbf{v}\right \| \left \| \mathbf{w}\right \|} \\ &= \frac{\left \| \begin{pmatrix} \begin{vmatrix} 1 & 0\\ -4 & 0 \end{vmatrix}, & -\begin{vmatrix} 1 & 0\\ 2 & 0 \end{vmatrix}, & \begin{vmatrix} 1 & 1\\ 2 & -4 \end{vmatrix} \end{pmatrix}\right \|}{\sqrt{40}}\\ &= \frac{6}{\sqrt{40}}\\ &=\frac{3\sqrt{40}}{20} \end{align*}$

Now that sin θ is found, the formula for the area of a triangle can be used:
$\begin{align*} \text{Area} &=\frac{1}{2}\left \| \mathbf{v}\right \| \left \| \mathbf{w}\right \| \sin \theta \\ &= \frac{1}{2}\cdot\sqrt{40}\cdot\frac{6}{\sqrt{40}} \\ &= 3 \end{align*}$

So, the area of the triangle formed is 3.

Problem 8: Vectors
Using the same vectors from Problem 6, and another vector $\mathbf{u}=(9,7,5)$: Find the components of the following vectors:

1. (u × v) × w
2. v × (w × u)

Know:

• u = (9, 7, 5)
• v = (2, 3, 3)
• w = (3, 0, 4)

Want:

• (u × v) × w
• v × (w × u)

Sub-Problem A
$\begin{align*} (\mathbf{u} \times \mathbf{v}) \times \mathbf{w}&= (\mathbf{u}\cdot\mathbf{w})\mathbf{v}-(\mathbf{v}\cdot\mathbf{w})\mathbf{u}\\ &=47\mathbf{v}-18\mathbf{u} \\ &=\begin{bmatrix} 94-162\\ 141-126\\ 141-90 \end{bmatrix} \\ &= \begin{bmatrix} -68\\ 15\\ 51 \end{bmatrix} \end{align*}$

So, the resulting vector is (-68, 15, 51).

Sub-Problem B
$\begin{align*} \mathbf{v} \times (\mathbf{w} \times \mathbf{u})&= (\mathbf{v}\cdot\mathbf{u})\mathbf{w}-(\mathbf{v}\cdot\mathbf{w})\mathbf{u}\\ &=54\mathbf{w}-18\mathbf{u} \\ &=\begin{bmatrix} 162-162\\ 0-126\\ 216-90 \end{bmatrix} \\ &= \begin{bmatrix} 0\\ -126\\ 126 \end{bmatrix} \end{align*}$

So, the resulting vector is (0, -126, 126).

Problem 9: Lines & Planes
Find the vector, and parametric equations of the line containing the given point, and also parallel to the given vector(s):

1. A(0,0,0) and $\mathbf{v}=(-3,0,1)$
2. A(-9,3,4) and $\mathbf{v}=(-1,6,0)$
3. A(0,5,-4) and $\mathbf{v}=(0,0,-5)$ and $\mathbf{w}=(1,-3,-2)$

Know:

• A(0,0,0) and $\mathbf{v}=(-3,0,1)$
• A(-9,3,4) and $\mathbf{v}=(-1,6,0)$
• A(0,5,-4) and $\mathbf{v}=(0,0,-5)$ and $\mathbf{w}=(1,-3,-2)$

Want:

• vector
• parametric equation

Sub-Problem A
$\begin{align*} \mathbf{x} &= x_o+k\mathbf{v}\\ &= \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}+k\begin{bmatrix} -3\\ 0\\ 1 \end{bmatrix}\\ &= \begin{bmatrix} -3k\\ 0\\ k \end{bmatrix} \end{align*}$

So, the resulting vector is (-3k, 0, k). The parametric equations are:
$\begin{align*} x &= -3k\\ y &= 0\\ z &= k \end{align*}$

Sub-Problem B
$\begin{align*} \mathbf{x} &= x_o+k\mathbf{v}\\ &= \begin{bmatrix} -9\\ 3\\ 4 \end{bmatrix}+k\begin{bmatrix} -1\\ 6\\ 0 \end{bmatrix}\\ &= \begin{bmatrix} -9-k\\ 3+6k\\ 4 \end{bmatrix} \end{align*}$

So, the resulting vector is (-9-k, 3+6k, 4). The parametric equations are:
$\begin{align*} x &= -9-k\\ y &= 3+6k\\ z &= 4 \end{align*}$

Sub-Problem C
$\begin{align*} \mathbf{x} &= x_o+k_1\mathbf{v}+k_2\mathbf{w}\\ &= \begin{bmatrix} 0\\ 5\\ -4 \end{bmatrix}+k_1\begin{bmatrix} 0\\ 0\\ -5 \end{bmatrix}+k_2\begin{bmatrix} 1\\ -3\\ -2 \end{bmatrix}\\ &= \begin{bmatrix} k_2\\ 5-3k_2\\ -4-5k_1-2k_2 \end{bmatrix} \end{align*}$

So, the resulting vector is $(k_2,5-3k_2,-4-5k_1-2k_2)$. The parametric equations are:
$\begin{align*} x &= k_2\\ y &= 5-3k_2\\ z &= -4-5k_1-2k_2 \end{align*}$

Problem 10: Lines & Planes
Find the general solution to the linear systems provided, and confirm the orthogonality between row vectors of the coefficient matrix and the solution vectors:
$\begin{align*} &x_1+5x_2+x_3+2x_4-x_5=0\\ &x_1-2x_2-x_3+3x_4+2x_5=0 \end{align*}$

Know:

• $\begin{align*} &x_1+5x_2+x_3+2x_4-x_5=0\\ &x_1-2x_2-x_3+3x_4+2x_5=0 \end{align*}$

Want:

• general solution & orthogonality between row vectors of coefficient matrix & solution vectors.

For this solution, the row operation method will be used:
$\begin{align*} &\begin{bmatrix} 1 & 5 & 1 & 2 & -1 & 0\\ 1 & -2 & -1 & 3 & 2 & 0 \end{bmatrix} \\ &\rightarrow \begin{bmatrix} 7 & 35 & 7 & 14 & -7 & 0\\ 0 & -35 & -10 & 5 & 15 & 0 \end{bmatrix}\begin{matrix} {R_1}'=7R_1\\ {R_2}'=5(R_2-R_1) \end{matrix} \\ &\rightarrow \begin{bmatrix} 7 & 0 & -3 & 19 & 8 & 0\\ 0 & -7 & -2 & 1 & 3 & 0 \end{bmatrix}\begin{matrix} {R_1}'=R_1+R_2\\ {R_2}'=\frac{1}{5}R_2 \end{matrix} \\ \end{align*}$

Since there are 5 variables, and there's 3 empty rows (not shown), there will be 3 parameters introduced:
$\begin{align*} 7x_1-3x_3+19x_4+8x_5&=0\\ 7x_1&=3x_3-19x_4-8x_5\\ x_1&=\frac{3}{7}x_3-\frac{19}{7}x_4-\frac{8}{7}x_5\\ &=\frac{3}{7}p-\frac{19}{7}q-\frac{8}{7}r\\ \\ -7x_2-2x_3+x_4+3x_5&=0\\ -7x_2&=2x_3-x_4-3x_5\\ x_2&=-\frac{2}{7}x_3+\frac{1}{7}x_4+\frac{3}{7}x_5\\ &=-\frac{2}{7}p+\frac{1}{7}q+\frac{3}{7}r\\ \\ x_3=p\\ x_4=q\\ x_5=r \end{align*}$

So, the solution matrix is:
$\begin{bmatrix} \frac{3}{7}p-\frac{19}{7}q-\frac{8}{7}r\\ -\frac{2}{7}p+\frac{1}{7}q+\frac{3}{7}r\\ p\\ q\\ r \end{bmatrix}$

Confirming orthogonality:
$\begin{align*} &\begin{bmatrix} \frac{3}{7}p-\frac{19}{7}q-\frac{8}{7}r\\ -\frac{2}{7}p+\frac{1}{7}q+\frac{3}{7}r\\ p\\ q\\ r \end{bmatrix}\cdot \begin{bmatrix} 1\\ 5\\ 1\\ 2\\ -1 \end{bmatrix}\\ &=\frac{3}{7}p-\frac{19}{7}q-\frac{8}{7}r+5 \left ( -\frac{2}{7}p+\frac{1}{7}q+\frac{3}{7}r \right ) +p +2q -r \\ &=\frac{3}{7}p-\frac{19}{7}q-\frac{8}{7}r-\frac{10}{7}p+\frac{5}{7}q+\frac{15}{7}r +\frac{7}{7}p +\frac{14}{7}q -\frac{7}{7}r \\ &= 0 \\ \\ &\begin{bmatrix} \frac{3}{7}p-\frac{19}{7}q-\frac{8}{7}r\\ -\frac{2}{7}p+\frac{1}{7}q+\frac{3}{7}r\\ p\\ q\\ r \end{bmatrix}\cdot \begin{bmatrix} 1\\ -2\\ -1\\ 3\\ 2 \end{bmatrix}\\ &=\frac{3}{7}p-\frac{19}{7}q-\frac{8}{7}r-2 \left ( -\frac{2}{7}p+\frac{1}{7}q+\frac{3}{7}r \right ) -p +3q -2r \\ &=\frac{3}{7}p-\frac{19}{7}q-\frac{8}{7}r+\frac{4}{7}p-\frac{2}{7}q-\frac{6}{7}r -\frac{7}{7}p +\frac{21}{7}q -\frac{14}{7}r \\ &= 0 \end{align*}$

Problem 11: Lines & Planes
Given a vertice that a plane passes through a given vertice, and a normal vector v, find the point-normal equation:

1. A(-1,3,-2) and $\mathbf{v}=(-2,1,-1)$
2. A(-9,3,4) and $\mathbf{v}=(-1,6,0)$

Know:

• A(-1,3,-2) and $\mathbf{v}=(-2,1,-1)$
• A(-9,3,4) and $\mathbf{v}=(-1,6,0)$

Want:

• point-normal equation

Sub-Problem A
$\begin{align*} (-2)(x+1)+(1)(y-3)+(-1)(z+2)&=0\\ -2x-2+y-3-z-2&=0 \\ -2x+y-z-7&=0 \end{align*}$

So, the point-normal equation is:
$-2x+y-z-7=0$

Sub-Problem B
$\begin{align*} (-1)(x+9)+(6)(y-3)+(0)(z-4)&=0\\ -x-9+6y-18&=0 \\ -x+6y-27&=0 \end{align*}$

So, the point-normal equation is:
$-x+6y-27=0$

Problem 12: Lines & Planes
Find the distance between the given planes:

• 2x − y + z = 1
• -2x + y - z = 1

Know:

• 2x − y + z = 1
• -2x + y - z = 1

Want:

• distance between both planes

Let x = 0, y = 0. So, a point on the first plane has been found: (0, 0, 1). Now, calculate the distance between that point and the second plane:
$\begin{align*} \text{Distance}&=\frac{|ax_o+by_o+cz_o+d|}{\sqrt{a^2+b^2+c^2}}\\ &=\frac{2}{\sqrt{6}} \\ &= \frac{1}{3}\sqrt{6} \end{align*}$

So, the distance between both planes are $\frac{1}{3}\sqrt{6}$.

Problem 13: Eigenvalues & Eigenvectors
Given the matrix $A=\begin{bmatrix} 3 & 0 & 0\\ 0 & 2 & 0\\ 0 & 1 & 2 \end{bmatrix}$:

1. Find the eigenvalues of A.
2. Find the bases for the eigenspaces of A.
3. Is the matrix A diagonalizable? Justify.

Know:

• $A=\begin{bmatrix} 3 & 0 & 0\\ 0 & 2 & 0\\ 0 & 1 & 2 \end{bmatrix}$

Want:

• eigenvalues
• base for eigenspace
• to know if A is diagonalizable or not

Sub-Problem A
$\begin{align*} det(\lambda I-A) &=0 \\ \begin{vmatrix} \lambda-3 & 0 & 0\\ 0 & \lambda-2 & 0\\ 0 & -1 & \lambda-2 \end{vmatrix}&=0 \\ (\lambda-3)\begin{vmatrix} \lambda-2 & 0 \\ -1 & \lambda-2 \\ \end{vmatrix} &= 0\\ (\lambda-3)(\lambda-2)^2&=0 \end{align*}$

So, the eigenvalue for the matrix above is 2 and 3.

Sub-Problem B
Now that λ has been found, insert the value for λ = 2:
$\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix}$
From the linear system of the matrix above, it can be seen that these are the solutions:
$\begin{align*} x &= 0\\ y &= 0\\ z&= p \end{align*}$

So, if written in matrix form (which is the eigenvector):
$\mathbf{x}=\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ p \end{bmatrix}=p\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$

So, the basis for the eigenvector for λ = 2 is
$\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$.

Now, do that again for λ = 3:
$\begin{bmatrix} 0 & 0 & 0\\ 0 & -1 & 0\\ 0 & 1 & -1 \end{bmatrix}$
From the linear system of the matrix above, it can be seen that these are the solutions:
$\begin{align*} x &= p\\ y &= 0\\ z&= 0 \end{align*}$

So, if written in matrix form (which is the eigenvector):
$\mathbf{x}=\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} p\\ 0\\ 0 \end{bmatrix}=p\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$

So, the basis for the eigenvector for λ = 3 is
$\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$.

Sub-Problem C
A square n × n matrix is diagonalizable if it has n linearly independent eigenvectors. The given matrix only has 2 eigenvectors. So, it is not diagonalizable.

Problem 14: Eigenvalues & Eigenvectors
Given the matrix $A=\begin{bmatrix} -1 & 7 & -1\\ 0 & 1 & 0\\ 0 & 15 & 2 \end{bmatrix}$, calculate A¹¹!

Know:

• $A=\begin{bmatrix} -1 & 7 & -1\\ 0 & 1 & 0\\ 0 & 15 & 2 \end{bmatrix}$

Want:

• A¹¹

Start by finding eigenvalues:
$\begin{align*} det(\lambda I-A) &=0 \\ \begin{vmatrix} \lambda+1 & -7 & 1\\ 0 & \lambda-1 & 0\\ 0 & -15 & \lambda-2 \end{vmatrix}&=0 \\ (\lambda+1)\begin{vmatrix} \lambda-1 & 0 \\ 15 & \lambda-2 \\ \end{vmatrix} &= 0\\ (\lambda+1)(\lambda-1)(\lambda-2)&=0\\ \end{align*}$

So, the eigenvalue for the matrix above is -1, 1 and 2. Now that λ has been found, the matrix for $\lambda=1$:
$\begin{bmatrix} 2 & -7 & 1\\ 0 & 0 & 0\\ 0 & -15 & -1 \end{bmatrix}$

Converting to augmented matrix, and perform row operation:
$\begin{align*} &\begin{bmatrix} 2 & -7 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & -15 & -1 & 0 \end{bmatrix} \\ &\rightarrow \begin{bmatrix} 1 & -11 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 15 & 1 & 0 \end{bmatrix}\begin{matrix} {R_1}'=\frac{1}{2}(R_1+R_3)\\ \text{No Change}\\ \text{No Change} \end{matrix} \\ \end{align*}$

So, if the solutions are written in matrix form (which is the eigenvector): (1 parameter will be introduced)
$\mathbf{x}=\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} -\frac{11}{15}p\\ -\frac{1}{15}p\\ p \end{bmatrix}=p\begin{bmatrix} -\frac{11}{15}\\ -\frac{1}{15}\\ 1 \end{bmatrix}$

So, the basis for the eigenvector (1) is
$\begin{bmatrix} -\frac{11}{15}\\ -\frac{1}{15}\\ 1 \end{bmatrix}$.

The matrix for $\lambda=-1$:
$\begin{bmatrix} 0 & -7 & -1\\ 0 & -2 & 0\\ 0 & -15 & -3 \end{bmatrix}$

Converting to augmented matrix, and perform row operation:
$\begin{align*} &\begin{bmatrix} 0 & -7 & -1 & 0\\ 0 & -2 & 0 & 0\\ 0 & -15 & -3 & 0 \end{bmatrix} \\ &\rightarrow \begin{bmatrix} 0 & 0 & -1 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -3 & 0 \end{bmatrix}\begin{matrix} {R_1}'=R_1-7R_2\\ {R_2}'=\frac{1}{2}R_2\\ {R_3}'=R_3-15R_2 \end{matrix} \\ \end{align*}$

So, if the solutions are written in matrix form (which is the eigenvector): (1 parameter will be introduced)
$\mathbf{x}=\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} p\\ 0\\ 0 \end{bmatrix}=p\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$

So, the basis for the eigenvector (-1) is
$\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$.

The matrix for $\lambda=2$:
$\begin{bmatrix} 3 & -7 & 1\\ 0 & 1 & 0\\ 0 & -15 & 0 \end{bmatrix}$

Converting to augmented matrix, and perform row operation:
$\begin{align*} &\begin{bmatrix} 3 & -7 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & -15 & 0 & 0 \end{bmatrix} \\ &\rightarrow \begin{bmatrix} 3 & 0 & 1 & 0\\ 0 & 7 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\begin{matrix} {R_1}'=R_1+{R_2}'\\ {R_2}'=7R_2\\ {R_3}'=R_3-15R_2 \end{matrix} \\ \end{align*}$

So, if the solutions are written in matrix form (which is the eigenvector): (1 parameter will be introduced)
$\mathbf{x}=\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} -\frac{1}{3}p\\ 0\\ p \end{bmatrix}=p\begin{bmatrix} -\frac{1}{3}\\ 0\\ 1 \end{bmatrix}$

So, the basis for the eigenvector (2) is
$\begin{bmatrix} -\frac{1}{3}\\ 0\\ 1 \end{bmatrix}$.

Now that all three eigenvectors are found, the matrix P that diagonalizes A can be arranged from these eigenvectors (in any order). So, let's take:
$P=\begin{bmatrix} -\frac{11}{15} & 1 & -\frac{1}{3}\\ -\frac{1}{15} & 0 & 0\\ 1 & 0 & 1 \end{bmatrix}$

Now, using the P^-1AP formula in order to find D. But, we must find P^-1 first:

$\begin{align*} &\begin{bmatrix} -\frac{11}{15} & 1 & -\frac{1}{3} & 1 & 0 &0\\ -\frac{1}{15} & 0 & 0 & 0 & 1 & 0\\ 1 & 0 & 1 & 0 & 0 & 1 \end{bmatrix} \\ &\rightarrow \begin{bmatrix} 1 & -\frac{15}{11} & 0 & -\frac{15}{11} & 0 &0\\ -1 & 0 & 0 & 0 & 15 & 0\\ 0 & 0 & 1 & 0 & 15 & 1 \end{bmatrix}\begin{matrix} {R_1}'=-\frac{15}{11}R_1\\ {R_2}'=15R_2\\ {R_3}'=R_3+{R_2}' \end{matrix} \\ &\rightarrow \begin{bmatrix} 0 & -\frac{15}{11} & 0 & -\frac{15}{11} & \frac{90}{11} &-\frac{5}{11}\\ -1 & 0 & 0 & 0 & 15 & 0\\ 0 & 0 & 1 & 0 & 15 & 1 \end{bmatrix}\begin{matrix} {R_1}'=R_1+R_2-\frac{5}{11}R_3\\ {R_2}'=15R_2\\ {R_3}'=R_3+{R_2}' \end{matrix} \\ &\rightarrow \begin{bmatrix} 0 & 1 & 0 & 1 & -6 &\frac{1}{3}\\ 1 & 0 & 0 & 0 & -15 & 0\\ 0 & 0 & 1 & 0 & 15 & 1 \end{bmatrix}\begin{matrix} {R_1}'=-\frac{11}{15}R_1\\ {R_2}'=15R_2\\ {R_3}'=R_3+{R_2}' \end{matrix} \\ &\rightarrow \begin{bmatrix} 1 & 0 & 0 & 0 & -15 & 0\\ 0 & 1 & 0 & 1 & -6 &\frac{1}{3}\\ 0 & 0 & 1 & 0 & 15 & 1 \end{bmatrix}\\ \end{align*}$

So, the inverse is $\begin{bmatrix} 0 & -15 &0\\ 1 & -6 & \frac{1}{3}\\ 0 & 15 & 1 \end{bmatrix}$

Now, the previously mentioned formula can be used:
$\begin{align*} D &=P^{-1}AP \\ &= \begin{bmatrix} 0 & -15 &0\\ 1 & -6 & \frac{1}{3}\\ 0 & 15 & 1 \end{bmatrix} \begin{bmatrix} -1 & 7 & -1\\ 0 & 1 & 0\\ 0 & 15 & 2 \end{bmatrix} \begin{bmatrix} -\frac{11}{15} & 1 & -\frac{1}{3}\\ -\frac{1}{15} & 0 & 0\\ 1 & 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 0 & -15 & 0 \\ -1 & 6 & -\frac{1}{3}\\ 0 & 30 & 2 \end{bmatrix} \begin{bmatrix} -\frac{11}{15} & 1 & -\frac{1}{3}\\ -\frac{1}{15} & 0 & 0\\ 1 & 0 & 1 \end{bmatrix}\\ &= \begin{bmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 2 \end{bmatrix} \\ \end{align*}$

Finally:
$\begin{align*} A^{11}&=PD^{11}P^{-1}\\ &=\begin{bmatrix} -\frac{11}{15} & 1 & -\frac{1}{3}\\ -\frac{1}{15} & 0 & 0\\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 2 \end{bmatrix}^{11} \begin{bmatrix} 0 & -15 &0\\ 1 & -6 & \frac{1}{3}\\ 0 & 15 & 1 \end{bmatrix} \\ &=\begin{bmatrix} -1 & -10223 & -683\\ 0 & 1 & 0\\ 0 & 30705 & 2048 \end{bmatrix} \\ \end{align*}$

So, $A^{11}=\begin{bmatrix} -1 & -10223 & -683\\ 0 & 1 & 0\\ 0 & 30705 & 2048 \end{bmatrix} \\ $.