Problem of the week - Does the vector belong to the span of the set of vectors?

Like last week, this week's problem is also related to spanning sets of vectors, but this time to determine if a vector belongs to a subspace spanned by a set a vectors. Once again, it will be revealed the importance of mastering the basic concepts of rank of a matrix, row operations to convert a matrix to row echelon form and systems of linear equations, to solve problems related to vector spaces.

Find λ and μ, real numbers, so that the vector (-4  -5  λ  μ) belongs to the subspace F spanned by the vectors (-1  -1  -4  4), (-1  -1  -2  1) and (1  2  2  -2).

For (-4  -5  λ  μ) to belong to the subspace F, it is a necessary and sufficient condition that it can be expressed as a linear combination of the spanning set of F, that is, we can find a, b and c, real numbers, such that:

(-4  -5  λ  μ) = a(-1  -1  -4  4) + b(-1  -1  -2  1) + c(1  2  2  -2)

The previous linear combination leads to the following system of linear equations, where the unknowns are a, b and c.

 -a -  b +  c = -4
 -a -  b + 2c = -5
-4a - 2b + 2c =  λ 
 4a +  b - 2c =  μ

If the system of linear equations is consistent, then, the vector can been expressed as a linear combination of the spanning set of vectors, and therefore, it belongs to F.

Let's represent the system in its matrix form, using the associated augmented matrix.

┌                 ┐
│ -1  -1   1 | -4 │
│ -1  -1   2 | -5 │
│ -4  -2   2 |  λ │
│  4   1  -2 |  μ │
└                 ┘

Perform row operations to convert the augmented matrix to row echelon form.

r2 <———> r2 - r1
r3 <———> r3 - 4•r1
r4 <———> r4 + 4•r1
┌                    ┐
│ -1  -1   1 |    -4 │
│  0   0   1 |    -1 │
│  0   2  -2 |  λ+16 │
│  0  -3   2 |  μ-16 │
└                    ┘
r2 <———> r4
r2 <———> r3
┌                    ┐
│ -1  -1   1 |    -4 │
│  0   2  -2 |  λ+16 │
│  0  -3   2 |  μ-16 │
│  0   0   1 |    -1 │
└                    ┘
r3 <———> 2•r3
┌                     ┐
│ -1  -1   1 |     -4 │
│  0   2  -2 |   λ+16 │
│  0  -6   4 |  2μ-32 │
│  0   0   1 |     -1 │
└                     ┘
r3 <———> r3 + 3•r2
┌                       ┐
│ -1  -1   1 |       -4 │
│  0   2  -2 |     λ+16 │
│  0   0  -2 | 3λ+2μ+16 │
│  0   0   1 |       -1 │
└                       ┘
r3 <———> r4
┌                       ┐
│ -1  -1   1 |       -4 │
│  0   2  -2 |     λ+16 │
│  0   0   1 |       -1 │
│  0   0  -2 | 3λ+2μ+16 │
└                       ┘
r4 <———> r4 + 2•r3
┌                       ┐
│ -1  -1   1 |       -4 │
│  0   2  -2 |     λ+16 │
│  0   0   1 |       -1 │
│  0   0   0 | 3λ+2μ+14 │
└                       ┘

The system of linear equations is consistent if and only if the rank of the coefficient matrix is equal to that of the augmented matrix.

The rank of the coefficient matrix is equal to 3, and the rank of the augmented matrix will be also equal to 3 when the expression 3λ + 2μ + 14 is equal to zero.

Isolating one of the variables, this expression is equal to zero when μ = - 7 - 3λ/2, and in such a case the system will be consistent, and consequently the vector will belong to the span of the set of vectors, so, the solution to the problem is:

λ, μ ∈ R: μ = - 7 - 3λ/2

Where a particular solution is: λ = 2, μ = -10.