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I was watching this lecture on linear algebra, how to a linear system of equations where AX=0, where A is some matrix multiplying the vector X. In the lecture it was stated that if upon doing row eliminations a non pivot column is found, we can freely assign any variable for it. Why is this so?

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This is a consequence of the definition of reduced row echelon form.

By definition, each nonzero row of an augmented matrix in reduced row echelon form corresponds to an equation where exactly one of the variables in a pivot column has coefficient equal to one. This means that the equations defined by a reduced row echelon form system allow one to solve for each "pivot" variable in terms of the "nonpivot" variables.

For example, consider the reduced row echelon form system $$ \left[\begin{array}{rrrrr|r} 1 & 0 & -5 & 0 & -2 & 6 \\ 0 & 1 & 2 & 0 & 11 & 7 \\ 0 & 0 & 0 & 1 & -16 & -9 \end{array}\right] $$ The "pivot" variables are $\{x_1, x_2, x_4\}$ and the "nonpivot" variables are $\{x_3, x_5\}$. The first row is the equation $$ x_1-5\,x_3-2\,x_5=6 $$ which allows us to solve for $x_1$ in terms of $\{x_3, x_5\}$ giving $x_1=6+5\,x_3+2\,x_5$.

The second two equations give $x_2=7-2\,x_3-11\,x_5$ and $x_4=-9+16\,x_5$.

Putting these relations together gives the general solution to the system as $$ \left[\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right] = \left[\begin{array}{r} 5 \, x_{3} + 2 \, x_{5} + 6 \\ -2 \, x_{3} - 11 \, x_{5} + 7 \\ x_{3} \\ 16 \, x_{5} - 9 \\ x_{5} \end{array}\right] =\left[\begin{array}{r} 6 \\ 7 \\ 0 \\ -9 \\ 0 \end{array}\right]+x_3\left[\begin{array}{r} 11 \\ 5 \\ 1 \\ -9 \\ 0 \end{array}\right]+x_5\left[\begin{array}{r} 8 \\ -4 \\ 0 \\ 7 \\ 1 \end{array}\right] $$ So, we may "freely" choose values of $x_3$ and $x_5$. Once these values are set, our formula automatically determines values of $x_1$, $x_2$, and $x_4$.

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