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Linear Algebra Decoded
 

Linear Algebra Decoded Example

Linear Algebra Decoded
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This example has been taken directly from the solution given by Linear Algebra Decoded to the formulated problem.
Express the vector u as a linear combination of the vectors of V.

V = {(1  -4  1  2); (4  -8  4  4); (-4  4  -5  -3); (2  -2  3  2)}
u = (-4  2  -5  -2)

How to solve this problem?
A vector u is a linear combinations of vectors v1, v2, ..., vn if there exist n scalars so u = a1•v1+a2•v2+...+an•vn. In practice this problem is reduced to solving a system of linear equations, where the unknowns are the scalars to determine, and the columns of the coefficient matrix of the system are the vectors of the set V, adding the vector u to compose the augmented matrix. If this system has solution, then a vector u can be expressed as a linear combination using the scalars provided for any particular solution (if there is an infinite number of solutions), or for the unique solution, if it is the case, according to the classification of the system.

Step 1: Write the augmented matrix of the system of linear equations associated to this problem. The columns of the coefficient matrix are the vectors of the set V, and the last column to compose the augmented matrix is the vector u.

┌                      ┐
│  1   4  -4   2 |  -4 │
│ -4  -8   4  -2 |   2 │
│  1   4  -5   3 |  -5 │
│  2   4  -3   2 |  -2 │
└                      ┘

Step 2: Transform the augmented matrix to row echelon form.

r2 <---> r2 + 4•r1
r3 <---> r3 - r1
r4 <---> r4 - 2•r1
┌                       ┐
│ 1   4   -4   2 |   -4 │
│ 0   8  -12   6 |  -14 │
│ 0   0   -1   1 |   -1 │
│ 0  -4    5  -2 |    6 │
└                       ┘
r2 <---> r4
┌                       ┐
│ 1   4   -4   2 |   -4 │
│ 0  -4    5  -2 |    6 │
│ 0   0   -1   1 |   -1 │
│ 0   8  -12   6 |  -14 │
└                       ┘
r4 <---> r4 + 2•r2
┌                     ┐
│ 1   4  -4   2 |  -4 │
│ 0  -4   5  -2 |   6 │
│ 0   0  -1   1 |  -1 │
│ 0   0  -2   2 |  -2 │
└                     ┘
r4 <---> r4 - 2•r3
┌                     ┐
│ 1   4  -4   2 |  -4 │
│ 0  -4   5  -2 |   6 │
│ 0   0  -1   1 |  -1 │
│ 0   0   0   0 |   0 │
└                     ┘

Step 3: Classify the system.

The rank of the matrices are equal, but less than the number of variables. There is an infinite number of solutions (The system is consistent dependent).

Step 4: Translate the row echelon form matrix to the associated system of linear equations, eliminating the null equations.

x+4y-4z+2w = -4
-4y+5z-2w = 6
-z+w = -1

Step 5: Select the free variables, which are the ones associated to those columns in the coefficient matrix that do not contain leading entries.

{w }

Step 6: Back substitution. Starting by the last equation, isolate the dependent variable expressing it in terms of the free variables, and replaces it in the equation above. Do this process until the first equation.


-z+w = -1
z = 1+w

-4y+5z-2w = 6
y = (-6+5z-2w)/4
y = (-6+5•(1+w)-2w)/4
y = (-1+3w)/4

x+4y-4z+2w = -4
x = -4-4y+4z-2w
x = -4-4•((-1+3w)/4)+4•(1+w)-2w
x = 1-w

Step 7: Solution.

coef =  {(1-w ; (-1+3w)/4 ; 1+w ; w) | w ∈ R }

Step 8: Particular solution.

coef =  { (1 , -1/4 , 1 , 0) }

coef =  { (1 , -0.25 , 1 , 0) }
  
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