You will be able to
- Write the elimination matrix that subtracts a multiple of one row from another
- Apply elimination matrices to a matrix and a right-hand side to reach upper triangular form
- Multiply matrices row by row and column by column using the associative law
- Compute the inverse of a matrix with Gauss-Jordan elimination
- Recognize when a matrix is singular and explain why elimination fails for it
Good to know first
- Solving a small system of linear equations by substitution or elimination
- Multiplying a matrix by a column vector
- Basic arithmetic with fractions and negative numbers
Syllabus
7 lessons and a final project
- 1
Recap of Elimination on a 3x3 System Free
This lesson recaps how elimination solves a 3 by 3 linear system by turning it into a matrix and removing the entries below each pivot. It shows how the same steps are captured by writing the system as A x = b.
2 code5 build steps2 exercises3 questions - 2
Elimination as a matrix: the E21 multiplier Beyond Pass
Elimination is a row operation that can be written as multiplying by a matrix, starting with the matrix E21 that removes the 3 below the first pivot. You will see how the multiplier 3 comes from dividing the entry you want to kill by the pivot.
4 build steps3 exercises3 questions - 3
Applying elimination to the right-hand side b Beyond Pass
Elimination changes the matrix A and the right-hand side b in exactly the same way, so we can carry b along as an extra column while we eliminate. We then solve the resulting triangular system by back-substitution.
1 exercise3 questions - 4
Matrix multiplication rules and the associative law Beyond Pass
This lesson shows how to multiply a matrix by a column vector, row by row and column by column, and why a row times a matrix gives a row. It also explains that matrix multiplication is associative, so (AB)C equals A(BC), which lets us group products any way we like.
1 code4 build steps2 exercises3 questions - 5
Completing elimination to get the upper triangular U Beyond Pass
Elimination is finished by a second row operation that clears the entry below the diagonal in column 2, leaving the upper triangular matrix U. Each step is a multiplication by an elimination matrix E, and the whole process is written as E32(E21 A) = U.
1 code3 build steps2 exercises3 questions - 6
Finding inverse matrices with Gauss-Jordan Beyond Pass
This lesson shows how elimination on a matrix can be used to find its inverse, by applying the same row operations to the identity matrix. It also introduces permutation matrices, which swap rows.
1 code4 build steps2 exercises3 questions - 7
Singular matrices: when elimination breaks down Beyond Pass
Elimination can only turn a square matrix into a full set of pivots when the matrix is invertible; a singular matrix runs out of pivots. This lesson shows how the elimination matrices undo each other, and how the inverse of a lower-triangular elimination matrix is found by flipping the sign of its entry.
2 formulas3 exercises3 questions
Final project
Build a Mini Elimination Solver with Matrices
Take a 3 by 3 system Ax = b and solve it entirely with elimination matrices, then find A inverse with Gauss-Jordan. Finish by showing what happens when you change A so that it becomes singular. This ties together multipliers, E matrices, U, inverses and pivots.
Its 8 tasks open with Beyond Pass.
Free preview · Lesson 1 in full
Recap of Elimination on a 3x3 System
This lesson recaps how elimination solves a 3 by 3 linear system by turning it into a matrix and removing the entries below each pivot. It shows how the same steps are captured by writing the system as A x = b.
Draw and write next to the video.
The idea
Elimination is the systematic way to solve a system of linear equations. Take three equations in three unknowns: x + 2y + z = 2, 3x + 8y + z = 12, and 4y + z = 2. The goal is to remove variables one at a time until the system is simple enough to solve from the bottom up.
The first step is to write the system as a matrix. The left-hand side numbers form the coefficient matrix A, with rows 1 2 1, 3 8 1, and 0 4 1. Elimination works on the rows of A. The first pivot is the 1 in the top left corner. Multiply the first row by 3 and subtract it from the second row to make the entry below the pivot zero. The second row becomes 0 2 -2 on the left, with right-hand side 12 - 6 = 6.
The second pivot is the 2 in the middle of the new second row. The entry below it is already 4 in the third row. Subtract 2 times the second row from the third row to remove the 4. The third row becomes 0 0 5 on the left, with right-hand side adjusted to match. Now the matrix is upper triangular, with pivots 1, 2 and 5 on the diagonal.
Back-substitution then solves from the bottom. The last equation gives z directly, then the middle equation gives y, and the top equation gives x. This is why elimination ends with a triangular system: each equation has fewer unknowns than the one above it.
Elimination can fail if a pivot becomes zero. In that case you must swap rows to bring a nonzero entry into the pivot position. If no swap works, the system has no unique solution. In the video the lecturer shows both the success and the failure case side by side.
The lecturer points out that the whole process can be written as a matrix equation A x = b, where A holds the coefficients, x holds the unknowns, and b holds the right-hand sides. The elimination steps are row operations on A, and they must be applied to b as well, so the two stay in step.
Key points
- Elimination turns the system into an upper triangular one by removing entries below each pivot. 2:00
- The coefficient matrix A is written with rows 1 2 1, 3 8 1, 0 4 1 for this system. 2:50
- Each row operation applied to A must also be applied to the right-hand side b. 2:51
- Back-substitution solves the triangular system from the bottom equation upward. 4:11
- If a pivot is zero, swap rows; if no swap works, elimination fails. 4:20
Code from the screen
x + 2y + z = 2
3x + 8y + z = 12
4y + z = 21 2 1
3 8 1
0 4 1Build along in the practice terminal
- Write the three equations on paper, lining up the x, y, z terms in columns.
- Copy the coefficients into a grid: row 1 is 1 2 1, row 2 is 3 8 1, row 3 is 0 4 1. This grid is A.
- Use row 1 as the pivot row. Subtract 3 times row 1 from row 2 to make the first entry of row 2 zero.
- Look at the new second pivot and subtract a multiple of row 2 from row 3 so the entry below it becomes zero.
- Check that the matrix is now triangular, then solve from the bottom row up.
Practice
For the system with rows 1 2 1 and 3 8 1, subtract 3 times row 1 from row 2. What is the new entry in row 2, column 2?
Hint
Multiply the top row's middle entry by 3, then subtract from the second row's middle entry.
Solution
- Row 2 column 2 is 8.
- Row 1 column 2 is 2, so 3 times it is 6.
- New entry = 8 - 6 = 2.
2
In your own words, why does elimination need to change the right-hand side b as well as the matrix A?
Solution
- Each row of the system is an equation that must stay true.
- If you subtract a multiple of one equation from another, you must do the same to both sides.
- Changing only A would give a different system with different answers.
Model answer: Each row is an equation, so subtracting a multiple of one equation from another must be done on both sides, both in A and in b. Otherwise the equations no longer describe the same solution.
On your scratchpad
Draw a 3 by 3 grid of the coefficients, circle the pivot in each column, and put an X under each entry you make zero.
Quiz
What is the first pivot in the elimination of this system?
- 1, the top left entry
- 3, the entry below it
- 4, the entry in row 3 column 2
- 0, the entry in row 3 column 1
Show the answer
A. 1, the top left entry · The first pivot is the top left entry, 1. It is used to eliminate the 3 below it.
What multiple of row 1 is subtracted from row 2 to remove the 3?
- 2 times row 1
- 3 times row 1
- 4 times row 1
- 1 times row 1
Show the answer
B. 3 times row 1 · Row 2 starts with 3, and row 1 starts with 1, so subtracting 3 times row 1 makes the entry 0.
Which step ends elimination and begins back-substitution?
- When the matrix is upper triangular
- When the matrix is the identity
- When the right-hand side is zero
- When all entries are positive
Show the answer
A. When the matrix is upper triangular · Once zeros fill the entries below the pivots, the matrix is upper triangular and you solve from the bottom up.
Lessons 2–7 and the final project
They open with Beyond Pass: every lesson of every course, in full. One payment: $14.99 for 90 days or $29 for a year. No subscription.
Made by Beyond from the video “2. Elimination with Matrices.” by MIT OpenCourseWare. The video plays in YouTube's own player. Watch it on YouTube · All lessons