Author Jonathan David

Title: Construct a Matrix with Given Column Space and Nullspace (Step-by-Step Proof)

In this walkthrough, we show how to construct a matrix whose column space contains (1,1,1) and whose nullspace is the line of multiples of (1,1,1,1).

Step 1: Understand the Problem

We want:

• Column space to contain the vector (1,1,1)

• Nullspace to be the set of all scalar multiples of (1,1,1,1)

This means we are looking for a matrix A such that:

• A * x = 0 has solution space spanned by (1,1,1,1)

• The matrix A has 3 rows (since the column vector has 3 entries)

• The matrix A has 4 columns (since the nullspace vector has 4 components)

Step 2: Definitions

Column Space: All linear combinations of the columns
Nullspace: All solutions to A * x = 0
So x = t(1,1,1,1) is the general solution of the homogeneous system.

Step 3: Deconstruct and Solve

We assume:

x = (1,1,1,1)t
So, x₁ = x₂ = x₃ = x₄

This gives:

• x₁ - x₂ = 0

• x₂ - x₃ = 0

• x₃ - x₄ = 0

Which gives us a system of equations:

x₁ - x₂ = 0  
x₂ - x₃ = 0  
x₃ - x₄ = 0

Written in matrix form:

[ 1 -1  0  0 ]  
[ 0  1 -1  0 ]  
[ 0  0  1 -1 ]

Thus, a correct matrix A is:

A = [ 1  0  0  -1  
      0  1  0  -1  
      0  0  1  -1 ]

This has:

• Column space containing (1,1,1)

• Nullspace spanned by (1,1,1,1)

Step 4: Validation

Any vector in the nullspace must satisfy A * x = 0.
Let x = (1,1,1,1), then:

A * x = [1 -1 0 0]·x = 1 - 1 = 0  
        [0 1 -1 0]·x = 1 - 1 = 0  
        [0 0 1 -1]·x = 1 - 1 = 0  

✅ Valid

And since (1,1,1) can be expressed as a linear combination of the columns, it's in the column space.

Conclusion

One valid matrix is:

A = [ 1  0  0  -1  
      0  1  0  -1  
      0  0  1  -1 ]

This matches the conditions exactly. There are infinitely many matrices that can satisfy these conditions, but this form is simplest and easily validated.

—

This lesson aligns with Chapter 3.2 in Strang’s Linear Algebra (4th ed.). Always cite sources and avoid shortcut tools like AI or uncredited platforms during exams unless explicitly allowed.

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Title: How to Use Gauss-Jordan Elimination Step-by-Step – RREF Made Easy

This lesson shows how to perform Gauss-Jordan Elimination to reduce a system of equations into Reduced Row-Echelon Form (RREF). This is a continuation from the prior Gaussian Elimination example but takes it further to fully solve for each variable.

Matrix Setup (from previous work)

Starting augmented matrix:

[ 1  -2   3 | -2 ]
[ 0   3  -5 |  6 ]
[ 0   0   0 |  1 ]

Since the third row is [ 0 0 0 | 1 ], we already concluded this is inconsistent (no solution).

But we continue Gauss-Jordan to learn the process.

Gauss-Jordan Elimination Strategy:

Use counterclockwise elimination:

1. Eliminate below pivot a11

2. Then below a22

3. Then below a33

4. Then eliminate above each pivot

5. Finally, divide rows to get leading 1s

Step-by-Step:

1. Multiply row 2 by 3/5
   → 3/5 * R2 + R1 → R1

2. Result:

[ 1   9/5   2   8/5 ]
[ 0     3  -5     6 ]
[ 0     0   0     1 ]

1. Now eliminate back into R1 using R3:
   → -6*R3 + R2, etc.

Eventually, you reach:

[ 1  0  0 | -1/3 ]
[ 0  1  0 |  5/3 ]
[ 0  0  1 |    1 ]

Final Result:

This is Reduced Row-Echelon Form (RREF). But since the matrix had [ 0 0 0 | 1 ] earlier, it’s still inconsistent.

Key Takeaway:

Gauss-Jordan Elimination gives you full control over the system and reveals inconsistencies. If you’re asked to fully reduce a matrix into RREF, this is the path to follow.

This tutorial is part of Chapter 9: Gauss-Jordan Elimination from the Ultimate Crash Course for STEM Majors. It’s designed to help you understand how matrix algebra is used to prove solutions exist—or not.

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Search tags: gauss-jordan elimination example, how to find rref, rref inconsistent matrix, full matrix row reduction steps, how to check if a matrix has a solution, back elimination matrix, step-by-step rref method.

Title: Using Gaussian Elimination to Determine If a System Has a Solution

This post explains how to use Gaussian Elimination to test whether a system of linear equations is consistent, inconsistent, or has infinite solutions. It walks through the full process from the augmented matrix to the final conclusion, showing how even one row can prove there's no solution.

Given System:

x - 2y + 3z = -2  
2x - y + z = 2  
2x - 7y + 11z = -9

Augmented matrix:

[ 1  -2   3  | -2 ]  
[ 2  -1   1  |  2 ]  
[ 2  -7  11  | -9 ]

Step 1: Use Row Operations to Reach Row-Echelon Form

Apply Gaussian Elimination to zero out lower values:

• Multiply row 1 and subtract from rows 2 and 3

• Resulting matrix:

[ 1  -2   3  | -2 ]  
[ 0   3  -5  |  6 ]  
[ 0  -3   5  | -5 ]

Add row 2 and 3 to eliminate the leading term in row 3:

[ 1  -2   3  | -2 ]  
[ 0   3  -5  |  6 ]  
[ 0   0   0  |  1 ]

Step 2: Interpret the Final Matrix

The third row says:

0x + 0y + 0z = 1

That is a contradiction → no solution

This system is inconsistent.

Visual Summary of Matrix Solution Types:

• If last row is [ 0 0 0 | 1 ] → no solution (inconsistent)

• If last row is [ 0 0 0 | 0 ] → infinite solutions (if others are consistent)

• If matrix reduces cleanly → unique solution

Final Answer:

No solution.
The system is inconsistent based on the Gaussian elimination result.

This walkthrough is from Chapter 8 of the Ultimate Crash Course for STEM Majors and is designed to help you recognize how and why systems fail before wasting time solving.

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Search tags: Gaussian Elimination step by step, inconsistent matrix system, how to use row operations, no solution matrix, false row in matrix, linear system contradiction, matrix solution cases.