home
Δβα\Delta _{\beta \alpha}
The Δelta βeta αrchivesThe \ \Delta elta \ \beta eta \ \alpha rchives
Algebra
MAA10

Sequences and Series

MAA11

Exponents and Logarithms

MAA12

Functions

Functions
MAA20

Polynomial Functions

MAA21

Rational Functions

MAA22

Exponential and Log Functions

MAA23

Trigonometric Functions

Geometry and Trigonometry
MAA30

Trigonometric Identities

MAA31

Solving Triangles

MAA32

Vectors

Calculus
MAA40

Limits and Continuity

MAA41

Derivatives

MAA42

Integrals

Statistics and Probability
MAA50

Descriptive Statistics

MAA51

Probability Concepts

MAA52

Probability Distributions

Advanced Topics
MAA60

Complex Numbers

MAA61

Further Calculus

MAA62

Matrices

Applications and Modeling
MAA70

Applications and Modeling

IA preparation
MAA80

IA Preparation (Math)

MAA81

IA Preparation (Physics)

Chapter MAA62

Matrices

Advanced Topics

Introduction to Matrices

A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. Matrices provide a concise way to represent and solve systems of linear equations. The size of a matrix is described by its dimensions: the number of rows and columns.
A=(a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮am1am2⋯amn)A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}
A matrix with
mm
rows and
nn
columns is called an
m×nm \times n
matrix. When
m=nm = n
, the matrix is a square matrix. The element in the
ii
-th row and
jj
-th column is denoted by
aija_{ij}
.

Systems of Linear Equations

One of the most important applications of matrices is representing systems of linear equations. Consider a system of
mm
linear equations with
nn
unknowns:
a11x1+a12x2+⋯+a1nxn=b1a21x1+a22x2+⋯+a2nxn=b2⋮⋮⋱⋮⋮am1x1+am2x2+⋯+amnxn=bm\begin{array}{rcrcrcrl} a_{11}x_1 & + & a_{12}x_2 & + & \cdots & + & a_{1n}x_n & = b_1 \\ a_{21}x_1 & + & a_{22}x_2 & + & \cdots & + & a_{2n}x_n & = b_2 \\ \vdots & & \vdots & & \ddots & & \vdots & \vdots \\ a_{m1}x_1 & + & a_{m2}x_2 & + & \cdots & + & a_{mn}x_n & = b_m \end{array}
This system can be represented using an augmented matrix, which combines the coefficient matrix and the constant vector:
(a11a12⋯a1n∣b1a21a22⋯a2n∣b2⋮⋮⋱⋮∣⋮am1am2⋯amn∣bm)\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} & | & b_1 \\ a_{21} & a_{22} & \cdots & a_{2n} & | & b_2 \\ \vdots & \vdots & \ddots & \vdots & | & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} & | & b_m \end{pmatrix}

Elementary Row Operations

To solve systems of linear equations using matrices, we use elementary row operations that transform the matrix while maintaining the equivalence of the represented system. These operations are:
Elementary Row Operations:
1. Row Swap
Interchange two rows:
Ri↔RjR_i \leftrightarrow R_j
2. Row Scaling
Multiply a row by a non-zero scalar:
Ri→c⋅RiR_i \rightarrow c \cdot R_i
where
c≠0c \neq 0
3. Row Addition
Add a multiple of one row to another:
Ri→Ri+c⋅RjR_i \rightarrow R_i + c \cdot R_j
where
i≠ji \neq j
These operations do not change the solution set of the system, making them powerful tools for solving linear equations.

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations. The goal is to transform the augmented matrix into row echelon form (REF) or reduced row echelon form (RREF), from which the solution can be easily determined.

Row Echelon Form (REF) Properties

1. All rows consisting entirely of zeros are at the bottom of the matrix.
2. The leading entry (first non-zero element) of each non-zero row appears to the right of the leading entry of the row above it.
3. The leading entry in each non-zero row is 1.

Reduced Row Echelon Form (RREF) Properties

1. The matrix is in row echelon form.
2. Each leading 1 is the only non-zero entry in its column.
To perform Gaussian elimination, we follow these steps:
Gaussian Elimination Algorithm:
Step 1: Identify the leftmost column that doesn't consist entirely of zeros.
Step 2: Find a non-zero element in this column (pivot element). If necessary, swap rows to move this element to the top row being considered.
Step 3: Scale the pivot row to make the pivot element equal to 1.
Step 4: Use row operations to create zeros in all other entries in the pivot column.
Step 5: Ignore the row containing the pivot and repeat steps 1-4 on the submatrix consisting of the remaining rows.
Step 6: After achieving row echelon form, work backward from the bottom row to create zeros above each leading 1 to obtain reduced row echelon form (Gauss-Jordan elimination).

Solving Systems with Gaussian Elimination

Problem: Solve the following system of linear equations using Gaussian elimination:
2x+y−z=8−3x−y+2z=−11−2x+y+2z=−3\begin{array}{rcrcr} 2x & + & y & - & z & = 8 \\ -3x & - & y & + & 2z & = -11 \\ -2x & + & y & + & 2z & = -3 \end{array}
1
First, we write the augmented matrix for the system:
(21−1∣8−3−12∣−11−212∣−3)\begin{pmatrix} 2 & 1 & -1 & | & 8 \\ -3 & -1 & 2 & | & -11 \\ -2 & 1 & 2 & | & -3 \end{pmatrix}
2
We start by creating a pivot in the first position. Since there's already a 2 in the top-left, we scale the first row by
12\frac{1}{2}
to get a leading 1:
(112−12∣4−3−12∣−11−212∣−3)\begin{pmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 4 \\ -3 & -1 & 2 & | & -11 \\ -2 & 1 & 2 & | & -3 \end{pmatrix}
3
Eliminate the entries below the first pivot by adding multiples of the first row:
R2+3R1R_2+3R_1
:
(112−12∣401212∣1−212∣−3)\begin{pmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & | & 1 \\ -2 & 1 & 2 & | & -3 \end{pmatrix}

R3+2R1R_3+2R_1
:
(112−12∣401212∣1021∣5)\begin{pmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & | & 1 \\ 0 & 2 & 1 & | & 5 \end{pmatrix}
4
Now we create a pivot in the second position. Multiply the second row by 2 to get a leading 1:
(112−12∣4011∣2021∣5)\begin{pmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 4 \\ 0 & 1 & 1 & | & 2 \\ 0 & 2 & 1 & | & 5 \end{pmatrix}
5
Eliminate the entry below the second pivot:
R3−2R2R_3-2R_2
:
(112−12∣4011∣200−1∣1)\begin{pmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 4 \\ 0 & 1 & 1 & | & 2 \\ 0 & 0 & -1 & | & 1 \end{pmatrix}
6
Scale the third row to get a leading 1:
−R3-R_3
:
(112−12∣4011∣2001∣−1)\begin{pmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 4 \\ 0 & 1 & 1 & | & 2 \\ 0 & 0 & 1 & | & -1 \end{pmatrix}
7
Now we work backwards (Gauss-Jordan elimination) to create zeros above each pivot:
R2−R3R_2-R_3
:
(112−12∣4010∣3001∣−1)\begin{pmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 4 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & -1 \end{pmatrix}

R1+12R3R_1+\frac{1}{2}R_3
:
(1120∣72010∣3001∣−1)\begin{pmatrix} 1 & \frac{1}{2} & 0 & | & \frac{7}{2} \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & -1 \end{pmatrix}
8
Finally, eliminate the remaining non-zero element above the second pivot:
R1−12R2R_1-\frac{1}{2}R_2
:
(100∣2010∣3001∣−1)\begin{pmatrix} 1 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & -1 \end{pmatrix}
9
From the final matrix, we can read the solution directly:
x=2,y=3,z=−1x = 2, \quad y = 3, \quad z = -1

Types of Solutions

When solving systems of linear equations, we may encounter three possible scenarios:
Solution Possibilities:
1. Unique Solution
The system has exactly one solution. This occurs when no zero rows appear in the reduced row echelon form.
2. Infinitely Many Solutions
The system has infinitely many solutions. This happens when at least one bottom row is entirely zeros (like
0=00 = 0
). Each such row creates a free variable, resulting in parameterized solutions.
3. No Solution
The system is inconsistent and has no solution. This occurs when any row has all zeros on the left side but a non-zero constant on the right (like
0=50 = 5
), creating a contradiction.
Problem: Solve the following system, which has infinitely many solutions:
x+2y−z=32x+4y−2z=6x+2y+z=7\begin{array}{rcrcr} x & + & 2y & - & z & = 3 \\ 2x & + & 4y & - & 2z & = 6 \\ x & + & 2y & + & z & = 7 \end{array}
1
Write the augmented matrix:
(12−1∣324−2∣6121∣7)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 2 & 4 & -2 & | & 6 \\ 1 & 2 & 1 & | & 7 \end{pmatrix}
2
Eliminate entries below the first pivot:
R2−2R1R_2-2R_1
:
(12−1∣3000∣0121∣7)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 0 & 0 & | & 0 \\ 1 & 2 & 1 & | & 7 \end{pmatrix}

R3−R1R_3-R_1
:
(12−1∣3000∣0002∣4)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 2 & | & 4 \end{pmatrix}
3
Scale the third row to get a pivot of 1:
12R3\frac{1}{2}R_3
:
(12−1∣3000∣0001∣2)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 1 & | & 2 \end{pmatrix}
4
Eliminate the entry above the third pivot:
R1+R3R_1+R_3
:
(120∣5000∣0001∣2)\begin{pmatrix} 1 & 2 & 0 & | & 5 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 1 & | & 2 \end{pmatrix}
5
The system is now in row echelon form. Notice that there are only 2 pivots for 3 variables, which means there will be infinitely many solutions with one free variable.
6
We can parameterize the solution. Let
y=ty = t
where
tt
is a free parameter. From the first row, we get
x+2t=5x + 2t = 5
, so
x=5−2tx = 5 - 2t
. From the third row, we have
z=2z = 2
.
7
The complete solution is:
x=5−2t,y=t,z=2for any real value tx = 5 - 2t, \quad y = t, \quad z = 2 \quad \text{for any real value } t