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Systems of Linear Equations.

It is forbidden to kill; therefore all murderers are punished unless they kill in large numbers and to the sound of trumpets, Voltaire

Recall

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 $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.

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Solving Systems of Linear Equations

In linear algebra, solving a system of linear equations can be efficiently done using matrices.

Consider a system of linear equations:

$\begin{cases} a_{11}x_1 +a_{12}x_2 + ···a_{1n}x_n = b_1 \\ a_{21}x_1 +a_{22}x_2 + ···a_{2n}x_n = b_2 \\ . \\ . \\ . \\ a_{n1}x_1 +a_{n2}x_2 + ···a_{nn}x_n = b_n \end{cases}$

It can be written or represented in matrix form as: AX = B, where:

  1. A is a n×n matrix known as the coefficient matrix.
  2. X is an n×1 column matrix (vector) of variables.
  3. B is an n×1 column matrix (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.

To solve AX = B, follow these steps:

  1. Ensure A is non-singular. This means that the determinant of A is not zero (det(A) ≠ 0), then we can multiply both sides of the previous equation form A-1.
  2. Calculate the inverse of A.
  3. Multiply both sides of the equation by A-2: A-1(AX) = A-1B⇒[Matrix multiplication is associative, so we can regroup the terms:] (A-1A)X =[The product of a matrix and its inverse is the identity matrix I] A-1B ⇒ IX = A-1B ⇒X = A-1B.

Examples

$\begin{cases} x + y + z = 6 \\ 2y + 5z = -4 \\ 2x + 5y -z = 27 \end{cases}$

In matrix form, this is, $\Bigl(\begin{smallmatrix}1 & 1 & 1\\ 0 & 2 & 5\\ 2 & 5 & -1\end{smallmatrix}\Bigr)\Bigl(\begin{smallmatrix}x\\ y\\ x\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}6\\ -4\\ 27\end{smallmatrix}\Bigr)$.

To solve for X, we need to calculate A-1.

  1. Calculate the determinant of A. det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) = 1·(-27)-1(-10)+1(-4) = -27 +10 -4 = -21.
  2. Find the Cofactor matrix C. $C_{ij} = (-1)^{i+j} \times \text{minor}(A_{ij})$ and C = $\Bigl(\begin{smallmatrix}-27 & 10 & -4\\ 6 & -3 & -3\\ 3 & -5 & 2\end{smallmatrix}\Bigr)$
  3. Transpose the cofactor matrix to obtain the adjugate matrix adj(A): adj(A) = CT = $\Bigl(\begin{smallmatrix}-27 & 6 & 3\\ 10 & -3 & -5\\ -4 & -3 & 2\end{smallmatrix}\Bigr)$
  4. Calculate the inverse A-1: $A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A)$.
  5. Multiply A-1 by B to find X: X = A-1B = $\frac{-1}{21}\Bigl(\begin{smallmatrix}-27 & 6 & 3\\ 10 & -3 & -5\\ -4 & -3 & 2\end{smallmatrix}\Bigr)\Bigl(\begin{smallmatrix}6\\ -4\\ 27\end{smallmatrix}\Bigr) = \frac{-1}{21}\Bigl(\begin{smallmatrix}−27⋅6+6⋅(−4)+3⋅27\\ 10⋅6+(−3)⋅(−4)+(−5)⋅27\\ −4⋅6+(−3)⋅(−4)+2⋅27\end{smallmatrix}\Bigr) = \frac{-1}{21}\Bigl(\begin{smallmatrix}−162−24+81\\ 60+12−135\\ −24+12+54\end{smallmatrix}\Bigr) = \frac{-1}{21}\Bigl(\begin{smallmatrix}-105\\ -63\\ 42\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}5\\ 3\\ -2\end{smallmatrix}\Bigr)$.

Therefore, The solution to our original system of equation is X = ⟨5, 3, -2⟩. This method provides a systematic way to solve systems of linear equations using matrix operations, making it a powerful tool in linear algebra.

$\begin{cases} x + y + z = 7 \\ 3x - 2y -z = 4 \\ x + 6y +5z = 24 \end{cases}$

In matrix form, this is, $\Bigl(\begin{smallmatrix}1 & 1 & 1\\ 3 & -2 & -1\\ 1 & 6 & 5\end{smallmatrix}\Bigr)\Bigl(\begin{smallmatrix}x\\ y\\ x\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}7\\ 4\\ 24\end{smallmatrix}\Bigr)$.

To solve for X, we need to calculate A-1.

  1. Calculate the determinant of A. det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) = 1⋅((−2⋅5)−(−1⋅6)) −1⋅(3⋅5−(−1⋅1)) +1⋅(3⋅6−(−2⋅1)) = 1⋅(−10+6) −1⋅(15+1) +1⋅(18+2) = 1⋅(−4) −1⋅16 +1⋅20 = −4 −16 +20 = 0. The matrix A is singular, so the system could either be inconsistent or have infinitely many solutions.

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.

  1. 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: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 7\\ 3 & -2 & -1 &\bigm| & 4\\ 1 & 6 & 5 &\bigm| & 24\end{smallmatrix}\Biggr)$

  1. Eliminate the x terms from the second and third rows, R2 = R2 - 3R1 and R3 = R3 -R1: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 7\\ 0 & -5 & -4 &\bigm| & -17\\ 0 & 5 & 4 &\bigm| & 17\end{smallmatrix}\Biggr)$
  2. Eliminate the y term from the third row, R3 = R3 + R2: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 7\\ 0 & -5 & -4 &\bigm| & -17\\ 0 & 0 & 0 &\bigm| & 0\end{smallmatrix}\Biggr)$.

The final augmented matrix shows that we have a row of all zeros, indicating that the system is consistent (0x+0y+0z = 0 ↭ 0 = 0) and has free variables.

$\begin{cases} x + y + z = 7 (i) \\ -5y -4z = -17 (ii) \end{cases}$

-5y -4z = -17 (ii)⇒$ y = \frac{-17+4z}{-5}$⇒[Substituting y into the first equation (i)] $x + \frac{-17+4z}{-5} +z = 7$ ⇒[Multiply through by 5 to clear the fraction] $5x +17 -4z + 5z = 35 ⇒ 5x + z = 35-17 ⇒ z = 18-5x$

$ y = \frac{-17+4z}{-5} = \frac{-17+4(18-5x)}{-5} = \frac{-17+72-20x}{-5} = \frac{55-20x}{-5} = -11+4x$

The system has infinitely many solutions of the form ⟨x, 4x -11, -5x +18⟩.

$\begin{cases} x -3y + z = 4 \\ -x + 2y -5z = -4 \\ 5x - 13y +13z = 8 \end{cases}$

In matrix form, this is, $\Bigl(\begin{smallmatrix}1 & -3 & 1\\ -1 & 2 & -5\\ 5 & -13 & 13\end{smallmatrix}\Bigr)\Bigl(\begin{smallmatrix}x\\ y\\ x\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}4\\ -4\\ 8\end{smallmatrix}\Bigr)$.

To solve for X, we need to calculate A-1.

  1. 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)+3·12+1·3 = -39 + 36 + 3 = 0. The matrix A is singular, so the system could either be inconsistent or have infinitely many solutions.
  2. 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:

$\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ -1 & 2 & -5 &\bigm| & -4\\ 5 & -13 & 13 &\bigm| & 8 \end{smallmatrix}\Biggr)$

  1. Eliminate the x term from the second and third rows: R2 = R2 + R1 (Add row 1 to row 2): $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ 0 & -1 & -4 &\bigm| & 0\\ 5 & -13 & 13 &\bigm| & 8 \end{smallmatrix}\Biggr)$
  2. R3 = R3 -5R1 (Subtract 5 times row 1 from row 3): $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ 0 & -1 & -4 &\bigm| & 0\\ 0 & 2 & 8 &\bigm| & -12 \end{smallmatrix}\Biggr)$
  3. Eliminate the y term from the third row: R3 = R3 + 2R2 (Multiply row 2 by 2 and add to row 3): $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ 0 & -1 & -4 &\bigm| & 0\\ 0 & 0 & 0 &\bigm| & -12 \end{smallmatrix}\Biggr)$

Identify Inconsistency. The third rows indicates 0x +0y + 0z = -12 ⇒ 0 = -12. This is a contradiction, which means the system of equations is inconsistent. Therefore, there are no solutions to this system.

Equations of planes

A plane in three-dimensional space can be thought of as a flat, two-dimensional surface that extends infinitely in all directions.

To define a plane mathematically, we need two things:

  1. A point on the plane, often denoted as $\vec{P_0}$ = (x0, y0, z0).
  2. A normal vector to the plane, denoted as $\vec{N} = ⟨a, b, c⟩$. This vector is perpendicular (orthogonal) to every vector that lies on the plane.

Given a point P0 and a normal vector $\vec{N}$, we can find the equation of the plane. Here’s how:

  1. Vectors in the plane. Any point P = (x, y, z) on the plane creates a vector $\vec{P_0P}$ with the point P0. This vector is given by $\vec{P_0P} = ⟨x -x_0, y - y_0, z - z_0⟩$.
  2. Perpendicularity Condition. The vector $\vec{N}$ is perpendicular to the plane, it’s also perpendicular to any vector that lies in the plane. Mathematically, this means that their dot product is zero: $\vec{P_0P}·\vec{N}=0$.
  3. Dot Product Calculation. The dot product is calculated: $⟨x -x_0, y -y_0, z-z_0⟩·⟨a, b, c⟩ = 0 ↭ a(x-x_0)+b(y-y_0)+c(z-z_0) =0$, and this is called the scalar equation of the plane.

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General Form of the Plane Equation

The scalar equation of the plane can be rearranged into a more familiar form: ax + by + cz = d where d = ax0 + by0 + cz0, ⟨a, b, c⟩ = $\vec{N}$.

Solved exercises

  1. Understanding the Normal Vector. A normal vector to a plane is a vector that is perpendicular to the surface of the plane. $\vec{N} = ⟨1, 5, 10⟩$
  2. Point on the plane. Since the plane passes through the origin, the point we use is $P_0 = ⟨0, 0, 0⟩$.
  3. Equation of the plane. We use the fact that any point P on the plane must satisfy the condition that the vector from the origin (P0) to P is perpendicular to the normal vector the plane: $\vec{OP}·\vec{N} = 0 ↭ ⟨x,y,z⟩⋅⟨1,5,10⟩.$
  4. Calculating the dot product: x + 5y +10z = 0.

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  1. Identify the normal vector of the plane. The normal vector $\vec{N}$ of the plane x +4y -2z = 5 can be directly obtained from the coefficients of x, y, and z in the plane equation, $\vec{N} = ⟨1, 4, -2⟩$. Recall that the normal vector is perpendicular to the plane.

  2. The line passing through the point P(2, -4, 3) and in the direction of $\vec{N}$ can be written using the point-direction form of the line equation. If $\vec{r_0}$ is the position vector of the point P and $\vec{d}$ is the direction vector (which, in our particular case, is the normal vector $\vec{N}$), the vector equation of the line is (illustration ii) $\vec{r(t)}=\vec{r_0}+t\vec{d} = \vec{r_0}+t\vec{N}$.

  3. Substitute the known values: $\vec{r(t)} = ⟨2, -4, 3⟩ + t⟨1, 4, -2⟩ = ⟨2+t, -4+4t, 3-2t⟩.$

    Rotation Matrices

In conclusion, the vector equation of the line passing through the point P(2, -4, 3) and perpendicular to the plane x +4y -2z = 5 is $\vec{r(t)}=⟨2+t, -4+4t, 3-2t⟩.$

  1. Understanding the Normal Vector. A normal vector to a plane is a vector that is perpendicular to the surface of the plane. $\vec{N} = ⟨1, 5, 10⟩$
  2. Point on the plane: $P_0 = ⟨2, 1, -1⟩$.
  3. Equation of the plane. We use the fact that any point P on the plane must satisfy the condition that the vector from P0 to P, $\vec{P_0P}$, is perpendicular to the normal vector the plane: $\vec{P_0P}·\vec{N}=0↭ ⟨x -x_0, y -y_0, z-z_0⟩·⟨a, b, c⟩$ ↭[Calculating the dot product] $a(x-x_0)+b(y-y_0)+c(z-z_0) = 0$
  4. Substitute the given values: 1(x-2) +5(y-1) +10(z+1)=0 ↭[Expand and simplify] x −2 +5y −5 +10z +10 = 0 ↭[Combine like terms] x +5y +10z +3 = 0 ↭[Rearrange the equation to standard form:] x + 5y + 10z = -3.

Notice that if we are given the equation of a plane in this form, we can quickly get our normal vector, namely $\vec{N} = ⟨1, 5, 10⟩$ or alternatively, if know our normal vector, we can write down the left-hand side of the plane’s equation, and we plug in P0 in the left-hand side to get the value of the right-hand side of the plane’s equation.

  1. Extract the Normal Vector from the Plane Equation. The equation of the plane is given as x +y +3z = 5. This is the scalar equation of the plane. From the plane equation x +y +3z = 5, we can extract the normal vector $\vec{N}$ that is perpendicular to the plane. Its components are given by the coefficients of x, y, and z in the plane equation: $\vec{N} = ⟨1, 1, 3⟩$
  2. Calculate the dot product of $\vec{v}$ and $\vec{N}$: $\vec{v}·\vec{N}$ =[Recall, $\vec{a}·\vec{b}=a_1b_1+a_2b_2+a_3b_3$] 1·1 + 2·1 + (-1)3 = 1 + 2 -3 = 0.
  3. To determine if $\vec{v}$ is parallel, perpendicular, or neither to the plane, we will interpret the dot product result of $\vec{v}$ and $\vec{N}$. If the dot product is zero, it indicates that $\vec{v}$ and $\vec{N}$ are perpendicular to each other. Since $\vec{v}$ is perpendicular to $\vec{N}$, and $\vec{N}$ is the normal vector to the plane, then $\vec{v}$ is parallel to the plane.

Rotation Matrices

  1. Determine two vectors in the plane. Since the plane passes through points Q, R, and S, then we can find two vectors $\vec{QR}= R−Q = ⟨−4−(−1),2−1,2−2⟩ = ⟨-3, 1, 0⟩$ and $\vec{QS} = S−Q = ⟨−2−(−1),1−1,5−2⟩ = ⟨-1, 0, 3⟩$ that lie in the plane.

  2. Calculate the Normal Vector to the plane. The normal vector $\vec{N}$ to the plane can be found by taking the cross product of $\vec{QR}$ and $\vec{QS}$

    $\vec{N} = |\begin{smallmatrix} \vec{i} & \vec{j} & \vec{k}\\ -3 & 1 & 0 \\ -1 & 0 & 3 \end{smallmatrix}| = 3\vec{i} +9\vec{j} + \vec{k} = ⟨3, 9, 1⟩$

  3. Determine the plane equation. Let P(x, y, z) be an arbitrary point P in the plane, the vector $\vec{QP} = ⟨x +1, y -1, z -2⟩$ lie in the plane, and therefore is perpendicular to $\vec{N}$. Therefore, their dot product must be zero: $\vec{N}·\vec{QP} = 0 ↭ ⟨3, 9, 1⟩·⟨x +1, y -1, z -2⟩ = 0$ ↭[Calculate the dot product:] 3(x+1)+9(y-1)+(z-2) = 0 ↭[Simplify the equation:] 3x +3 +9y −9 +z −2 = 0 ↭ 3x + 9y + z = 8. Hence, 3x + 9y + z = 8 is the plane that passes through Q, R, and S.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, Andrew Misseldine, blackpenredpen, and MathMajor, YouTube’s channels.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
  8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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