Linearly dependent and independent rows
α_{1}s_{1} + α_{2}s_{2} + ... + α_{l}s_{l}
Solution. Form a linear combination of these rows
α_{1}{2 5} + α_{2}{4 10}
Let us find for what values of α_{1}, α_{2} the linear combination is equal to the zero row
α_{1}{2 5} + α_{2}{4 10} = {0 0}
This equation is equivalent to the following system of equations:
2α_{1} + 4α_{2} = 0 | |
5α_{1} + 10α_{2} = 0 |
Divide the first equation by 2, and the second equation by 5:
α_{1} + 2α_{2} = 0 | |
α_{1} + 2α_{2} = 0 |
The solution of this system may be any number α_{1} and α_{2} such that: α_{1} = -2α_{2}, for example, α_{2} = 1, α_{1} = -2, and this means that the rows s_{1} and s_{2} are linearly dependent.
Solution. Form a linear combination of these rows
α_{1}{2 5 1} + α_{2}{4 10 0}
Let us find for what values of α_{1}, α_{2} the linear combination is equal to the zero row
α_{1}{2 5 0} + α_{2}{4 10 0} = {0 0 0}
This equation is equivalent to the following system of equations:
2α_{1} + 4α_{2} = 0 | |
5α_{1} + 10α_{2} = 0 | |
α_{1} + 0α_{2} = 0 |
From the third equation gives α_{1} = 0. Substituting this value in the first and second equation:
2·0 + 4α_{2} = 0 | => | 4α_{2} = 0 | => | α_{2} = 0 | |||
5·0 + 10α_{2} = 0 | 10α_{2} = 0 | α_{2} = 0 | |||||
α_{1} = 0 | α_{1} = 0 | α_{1} = 0 |
So as a linear combination of rows is equal to zero only when α_{1} = 0 and α_{2} = 0, the rows are linearly independent.
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