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Life is a math equation. In order to gain the most, you have to know how to convert negatives into positives, Anonymous.
Pure mathematics is, in its way, the poetry of logical ideas, Albert Einstein.
Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.
Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.
The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a1+b1), (a2+b2), (a3+b3) >.
The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a1-b1), (a2-b2), (a3-b3) >.
Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction. It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = < ca_1, ca_2, ca_3>$.
The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$
It is the product of their magnitudes multiplied by the cosine of the angle between them, $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$
The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both of the input vectors and has a magnitude equal to the area of the parallelogram formed by the two input vectors.
The cross product $\vec{A}x\vec{B}$ can be computed using the following formula, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}\bigl) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$
Matrices offers an efficient way for solving systems of linear equations. Utilizing matrices, we can represent these equations more concisely and conveniently.
For a system like,
$\begin{cases} 2x_1 +3x_2 + 3x_3 = u_1 \\ 2x_1 +4x_2 + 4x_3 = u_2 \\ x_1 + x_2 + 2x_3 = u_3 \end{cases}$
$(\begin{smallmatrix}2 & 3 & 3\\ 2 & 4 & 4\\ 1 & 1 & 2\end{smallmatrix}\bigl) (\begin{smallmatrix}x_1\\ x_2\\x_3\end{smallmatrix}\bigl) = (\begin{smallmatrix}u_1\\ u_2\\u_3\end{smallmatrix}\bigl)$, A · X = U provides a more convenient and concise notation and an efficient way to solve system of linear equations where A is a 3 x 3 matrix representing coefficients, X is a 3 x 1 column vector representing the results where we basically do dot products between the rows of A (a 3 x 3 matrix) and the column vector of X (a 3 x 1 matrix).
A square matrix is a matrix with an equal number of rows and columns. For example, a 3x3 matrix has 3 rows and 3 columns.
Matrices offers an efficient way for solving systems of linear equations. For instance, consider the system of equations:
$\begin{cases} x + y + z = 7 \\ 3x -2y -z = 4 \\ x + 6y + 5z = 24 \end{cases}$
This can be written in matrix form as: A · X = U
where:
$A = \bigr(\begin{smallmatrix}1 & 1 & 1\\ 3 & -2 & -1\\ 1 & 6 & 5\end{smallmatrix}\bigl), X = \bigr(\begin{smallmatrix}x_1\\ x_2\\x_3\end{smallmatrix}\bigl), B = \bigr(\begin{smallmatrix}7\\ 4\\ 24\end{smallmatrix}\bigl)$.
It provides a more convenient and concise notation and an efficient way to solve system of linear equations.
Given a square matrix A, an inverse matrix A-1 exists if and only if A is non-singular, meaning its determinant is non-zero (det(A)≠ 0).
The inverse matrix of A, denoted as A-1, has the property that $A \times A^{-1} = A^{-1} \times A = I$. Essentially, multiplying a matrix by its inverse reverses (“undoes”) the effect of the original matrix. If det(A) ≠ 0, the inverse can be found using the formula $A^{-1}=\frac{1}{det(A)}adj(A)$ where adj(A) is the adjugate of A, which is the transpose of the cofactor matrix. The cofactor matrix is found by computing the minors of A, adjusting signs according to their position, and then transposing. adj(A) = CT where $C_{ij} = (-1)^{i+j} \times \text{minor}(A_{ij})$ and det(A) =[A is a 3x3 matrix] a(ei - fh) - b(di - fg) + c(dh - eg).
Consider a system of linear equations represented in matrix form as: AX = B, where A is a n×n matrix (coefficient matrix), X is an n×1 matrix (column vector of variables), and B is an n×1 matrix (column vector of constants).
Solving a system of linear equations expressed as AX = B involves finding or isolating the matrix X when we are given both matrices A and B. AX = B ⇒[A should be non-singular (invertible), meaning its determinant is non-zero, det(A)≠0, then we can multiply both side by A-1]⇒ X = A-1B. where $A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A)$.
However, if A is singular (det(A) = 0), we cannot find A-1. Instead, we need to use other methods like row reduction to understand the nature of the solutions.
A system of three equations with three unknowns represents three planes in a three dimensional space. When solving the system, you are determining how the planes intersect.
The solution to AX = B is given by X = A-1B, but this is not always the case. A system of three linear equations in three variables can have one of the following outcomes.
Let R be a commutative ring with unity. Let A ∈ Rn×n be a square matrix of order n. A is invertible if and only if its determinant is invertible in R. If R = ℚ, ℝ or ℂ, A is invertible if and only if its determinant is non-zero.
$\begin{cases} x + z = 1 \\ x + y = 2 \\ x + 2y +3z = 3 \end{cases}$
In matrix form, the system can be written as: $\Bigl(\begin{smallmatrix}1 & 0 & 1\\ 1 & 1 & 0\\ 1 & 2 & 3\end{smallmatrix}\Bigr)\Bigl(\begin{smallmatrix}x\\ y\\ x\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}1\\ 2\\ 3\end{smallmatrix}\Bigr)$.
$\begin{cases} x -3y + z = 4 (i) \\ -x + 2y -5z = 3 (ii) \\ 5x -13y +13z = 8 (iii) \end{cases}$
Add equations (i) and (ii), (i)+(ii): -y - 4z = 7 (iv).
Multiply equation (i) by -5, -5(i): -5x +15y -5z = -20 (vi)
Add equations (vi) and (iii), (vi)+(iii): 2y + 8z = 12 (viii)
Add 2 times equation (iv) to equation (viii), 2(iv)+(viii): -2y -8z + 2y + 8z = 14 -12 ↭ 0 = 2 ⊥ This contradiction (0 ≠ 14) indicates that the system of equations is inconsistent, meaning it has no solutions.
Calculate the determinant of A. det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) = 1⋅(2⋅13−(−5)⋅(−13)) −(−3)⋅(−1⋅13−(−5)⋅5) +1⋅(−1⋅(−13)−2⋅5) = 1⋅(−39) −(−3)⋅(−38) +1⋅3 = -39 + 36 + 3 = 0.
If A is singular (det(A) = 0), we cannot find A-1. Instead, we need to use other methods like row reduction to understand the nature of the solutions.
Convert to Row Echelon Form. A matrix is said to be in row echelon form when all its non-zero rows have a pivot, that is, a non-zero entry such that all the entries to its left and below it are equal to zero.
Start with the augmented matrix: $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ -1 & 2 & -5 &\bigm| & 3\\ 5 & -13 & 13 &\bigm| & 8\end{smallmatrix}\Biggr)$
Make the first element of the first row a pivot (which it already is).
Eliminate the first element (x terms) of the second and third rows by adding suitable multiples of the first row. Add the first row to the second row to eliminate the x term in the second row, R2 = R1 + R2, $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ 0 & -1 & -4 &\bigm| & 3\\ 5 & -13 & 13 &\bigm| & 8\end{smallmatrix}\Biggr)$.
Subtract 5 times the first row from the third row to eliminate the x term in the third row, R3 = R3 −5R1: $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ 0 & -1 & -4 &\bigm| & 3\\ 0 & 2 & 8 &\bigm| & -12\end{smallmatrix}\Biggr)$.
Eliminate the second element (y term) of the third row by subtracting 2 times the second row from the third row, R3 = R3 -2R2: $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ 0 & -1 & -4 &\bigm| & 3\\ 0 & 0 & 0 &\bigm| & 2\end{smallmatrix}\Biggr)$.
The third row indicates a contradiction 0x + 0y + 0z = 2↭ 0 = 2. This confirms that the system of equations is inconsistent and has no solutions.
$\begin{cases} x + y + z = 2 (i) \\ 2x + 2y +2z = 4 (ii) \\ -3x -3y -3z = -6 (iii) \end{cases}$
(-2)*(i): -2x -2y -2z = -4 (iv)
(ii)+(iv): 0=0. The firs pairs of equations is indeed the same equation, 2(i) = (ii). The values of x, y, and z that will make the first equation work will also work for the second and vice-versa.
3*(i)+(iii): 3x + 3y + 3z -3x -3y -3z = 6 - 6 ↭ 0 = 0. Again, the first and third equations are the same. This three equations have infinitely many solution. The solution is a plane defined by any of the equations, e.g., x + y + z = 2.
Start with the augmented matrix: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 2\\ 2 & 2 & 2 &\bigm| & 4\\ -3 & -3 & -3 &\bigm| & -6\end{smallmatrix}\Biggr)$
Convert to Row Echelon Form. A matrix is said to be in row echelon form when all its non-zero rows have a pivot, that is, a non-zero entry such that all the entries to its left and below it are equal to zero.
Make the first element of the first row a pivot (which it already is).
Eliminate the first element (x terms) of the second and third rows. Subtract 2 times the first row from the second row, R2 = R2 -2R1: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 2\\ 0 & 0 & 0 &\bigm| & 0\\ -3 & -3 & -3 &\bigm| & -6\end{smallmatrix}\Biggr)$
Add 3 times the first row to the third row, R3 = R3 + 3R1: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 2\\ 0 & 0 & 0 &\bigm| & 0\\ 0 & 0 & 0 &\bigm| & 0\end{smallmatrix}\Biggr)$
The row echelon form of the matrix indicates that there are two rows of zeros, which means that there are free variables. This implies that the system of equations has infinitely many solutions. The original system reduces to a single equation: x +y +z = 2.
The general solution to the system is: (x, y, z) = (2 -y -z, y, z) where y and z are free variables, so it represents a plane in a three-dimensional space.
Definition: A homogeneous system of linear equations is a set of linear equations where all the constant terms are zero. In other words, the system looks like this:
$\begin{cases} a_{11}x_1 +a_{12}x_2 + ···a_{1n}x_n = 0 \\ a_{21}x_1 +a_{22}x_2 + ···a_{2n}x_n = 0 \\ . \\ . \\ . \\ a_{n1}x_1 +a_{n2}x_2 + ···a_{nn}x_n = 0 \end{cases}$
Such a system always has at least one solution: the zero vector (0, 0, ···, 0), also know as the trivial solution. This is because substituting zeros for all variables satisfies every equation in the plane. Geometrically, all planes pass through the origin.
An n×n homogeneous system of linear equations has a unique solution (the trivial solution) if and only if the determinant of the coefficient matrix is non-zero ⇒ AX = 0 ↭ X = A-1·0 = 0. If this determinant is zero, then the system has infinitely many solutions.
$\begin{cases} x + z = 0 \\ x + y = 0 \\ x + 2y +3z = 0 \end{cases}$
(0, 0, 0) is a trivial solution, det$(\begin{smallmatrix}1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 2 & 3\end{smallmatrix}\bigl) = a(ei−fh)−b(di−fg)+c(dh−eg) = 1∗(1∗3−0∗2)−0∗(1∗3−0∗1)+0∗(1∗2−1∗1)=3 ≠ 0 ⇒$ (0, 0, 0) is the only solution.
In the General case, if Ax = b and det(A) ≠ 0, then A-1 exists, so x = A-1b is the unique solution of the system. If det(A) = 0, then the system of equation either has infinitely many solutions, or no solutions.
We don’t need to calculate the Minors completely, Minors = $(\begin{smallmatrix}x & x & x\\ -2 & x & x\\ -3 & x & x\end{smallmatrix}\bigl)$ ⇒ Cofactors = $(\begin{smallmatrix}x & x & x\\ 2 & x & x\\ -3 & x & x\end{smallmatrix}\bigl)$ ⇒ Transpose = $(\begin{smallmatrix}x & 2 & -3\\ x & x & x\\ x & x & x\end{smallmatrix}\bigl)$ ⇒[Dividing by det(A)=2] A-1 = $\frac{1}{2}(\begin{smallmatrix}1 & 2 & -3\\ -1 & -2 & 5\\ 2 & 2 & -6\end{smallmatrix}\bigl)$.
$\begin{cases} x + 3y + 2z = 1 \\ 2x -z = -2 \\ x +y = 1 \end{cases}$
To solve the system of equations using matrices, we can represent the system in matrix form AX = B, where: A = $(\begin{smallmatrix}1 & 3 & 2\\ 2 & 0 & -1\\ 1 & 1 & 0\end{smallmatrix}\bigl)$ and B = $(\begin{smallmatrix}1\\ -2\\ 1\end{smallmatrix}\bigl)$ ⇒[det(A)≠0] X = A-1·B =[Previous exercise] $\frac{1}{2}(\begin{smallmatrix}1 & 2 & -3\\ -1 & -2 & 5\\ 2 & 2 & -6\end{smallmatrix}\bigl)·(\begin{smallmatrix}1\\ -2\\ 1\end{smallmatrix}\bigl) = \frac{1}{2}(\begin{smallmatrix}-6\\ 8\\ -8\end{smallmatrix}\bigl)=(\begin{smallmatrix}-3\\ 4\\ -4\end{smallmatrix}\bigl)$. Therefore, the solution to the system of equations is x = −3, y = 4, and z = −4.
To determine where a matrix M is not invertible, we need to check when its determinant is zero. A matrix is invertible if and only if its determinant is non-zero. If the determinant is zero, the matrix is singular (not invertible), indicating that its columns are not linearly independent (meaning one of the columns can be expressed as a linear combination of the others).
$\left| M \right| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$ where aij represents the element in the ith row and jth column of matrix M.
det(M) = 1(0⋅0−1⋅1)−3(2⋅0−1⋅1)+c(2⋅1−1⋅0) = 1(−1)−3(−1)+c(2) = 2 + 2c =[For the matrix M to be not invertible, the determinant must be zero:] 0 ⇒[Solving for c] c = 1. o, the matrix M is not invertible when c = −2.
$\begin{cases} x + 3y + z = 0 (i)\\ 2x -z = 0 (ii)\\ x +y = 0 (iii)\end{cases}$
The trivial solution (x, y, z) = (0, 0, 0) always satisfies a homogenous system of linear equations. Now let’s explore if there are other solutions.
If we multiply the third equation by 3 (3iii), 3x + 3y = 0 and subtract it to the second equation, we get x + 3y +z = 0 (3(iii)-(ii) = (i)), that is, the first equation ⇒ The first equation can be derived as a linear combination of the other two equations, indicating that the system is dependent ⇒ the system of equations have infinitely many solutions.
From (iii): x = -y. Substitute x = -y into (ii): 2(-y)-z = 0 ⇒ -2y -z = 0 ⇒ z = -2y. Solution: x = -y, y = y, z = -2y.
Another way of putting this is as follows, our system of equations can be expressed as follows, ⟨x, y, z⟩·⟨1, 3, 1⟩ = 0, ⟨x, y, z⟩·⟨2, 0, -1⟩ = 0, ⟨x, y, z⟩·⟨1, 1, 0⟩ = 0,
if the dot product between a pair of vectors is zero, then the two vectors are orthogonal. Thus, the vector X must be orthogonal to every row vector of M ↭ the vector X is in the null space of M (The null space of a matrix M is the set of all solutions to the homogeneous equation M·X = 0).
As we already know, the cross product of two non-collinear vectors gives a vector perpendicular to the plane determined by the vectors, and therefore, to find the solution ⟨x, y, z⟩ -taking into consideration the (iii) equation is l.d of (i) and (ii), hence redundant-, we take the cross product of ⟨1, 3, 1⟩ and ⟨2, 0, -1⟩.
v = $(\begin{smallmatrix}\vec{i} & \vec{j} & \vec{k}\\ 1 & 3 & 1\\ 2 & 0 & 1\end{smallmatrix}\bigl)$ = ⟨-3, 3, -6⟩. So, any scalar multiple of this vector is also a solution to the system. The line of solution (which passes through O(0, 0, 0) and contains the vector $\vec{N}= ⟨-3, 3, -6⟩$) has parametric form x = -3t, y = 3t, z = -6t.
Formed the Augmented Matrix: $(\begin{smallmatrix}1 & 3 & 1 &\bigm| &0\\ 2 & 0 & -1 &\bigm| & 0\\ 1 & 1 & 0&\bigm| &0\end{smallmatrix}\bigl)$
Row Reduction to Row Echelon Form We will use Gaussian elimination to simplify this matrix.
Eliminate x from rows 2 and 3 using row 1, R2 = R2-2R1: $(\begin{smallmatrix}1 & 3 & 1 &\bigm| &0\\ 0 & -6 & -3 &\bigm| & 0\\ 1 & 1 & 0&\bigm| &0\end{smallmatrix}\bigl)$
R3 = R3 -R1: $(\begin{smallmatrix}1 & 3 & 1 &\bigm| &0\\ 0 & -6 & -3 &\bigm| & 0\\ 0 & -2 & -1&\bigm| &0\end{smallmatrix}\bigl)$
Use row 2 to eliminate y from row 3, R3 = R3 - 3R2: $(\begin{smallmatrix}1 & 3 & 1 &\bigm| &0\\ 0 & -6 & -3 &\bigm| & 0\\ 0 & 0 & 0&\bigm| &0\end{smallmatrix}\bigl)$. Hence, the last equation is l.d. of the other two, there are infinitely many solutions and the system is x + 3y + z = 0 (i), -6y -3z = 0 (ii). Solve for z in (ii), z = $\frac{6y}{-3}=-2y$. Replacing this result in (i) and solving for x, x + 3y -2y = 0↭ x -y = 0 ↭ x = -y.
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