Write the set of equations as an augmented matrix:

1 0 5 | 2

-2 1 -6 | -1

0 2 8 | 6

Reduce to echelon form by gaussian elimination

This will be of the form

1 a b | c or 1 a b | c or 1 a b | c

0 1 d | e 0 1 d | e 0 0 1 | d

0 0 1 | f 0 0 0 | 0 0 0 0 | 0

Add 2*row 1 to row 2:

1 0 5 | 2

0 1 4 | 3

0 2 8 | 6

Take 2*row 2 from row 3:

1 0 5 | 2

0 1 4 | 3

0 0 0 | 0

This is now in row echelon form

` ` The coefficient associated with a(3) which we will call `z` can take any value. It is a free parameter. Call the coefficients associated with a(1) and a(2) `x`and `y`respectively.

Then `x + 5z = 2` and `y + 4z = 3`

`implies x = 2 -5z` and `y = 3 -4z`

`therefore` `b = (2-5z)a(1) + (3-4z)a(2) + za(3)` for any ` ``z`

So `b` can be written as a linear combination of `a(1), a(2)` and `a(3)` in an infinite number of ways

**Yes, b is a linear combination of a(1), a(2) and a(3)**

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