1.3 Parameterized Solutions¶

Example 1¶

Basic and Free Variables¶

Consider the following example that might occur after row-reducing an augemented matrix:

\[\begin{split}A = \left[\begin{array}{rrr|r}-2&2&0&6\\0&0&-1&2\\\end{array}\right]\end{split}\]

We have two pivots, \(b_{11} = -2\) and \(b_{23} = -1\). The pivot columns correspond to basic variables while the other columns correspond to free variables. Let’s solve as far we as we can.

Tip

Usually, we back-substitute, starting from bottom-right and working our way back up toward top-left.

Cleary,

\[x_3 = - 2\]

and \(x_2\) is a free variable, but the best we can do for \(x_1\) is:

\[-2x_1 + 2x_2 = 6\]

We solve for the basic variables in terms of any free variables.

\[x_1 = -3 + x_2\]

Parameterized Solutions¶

Now we know the solutions have a very specific form. The Margalit and Robanoff textbook uses the following notation to express it:

\[(x_1,x_2,x_3) = (-3 +x_2, x_2, -2)\]

In the column-vector notation we have been using:

\[\begin{split}\vec x = \left[\begin{array}{rrr} -3 & + & x_2 \\ && x_2 \\ -2 && \end{array}\right]\end{split}\]

or

\[\begin{split}\vec x = \left[\begin{array}{r} -3\\ 0 \\ -2\end{array}\right] + \left[\begin{array}{rrr} x_2 \\ x_2 \\ 0 \end{array}\right]\end{split}\]

which, after factoring out \(x_2\), is equivalent to the vector equations

\[\begin{split}\vec x = \left[\begin{array}{r} -3\\ 0 \\ -2\end{array}\right] + \left[\begin{array}{rrr} 1 \\ 1 \\ 0 \end{array}\right]x_2\end{split}\]

where the \(x_2\) can be any real number. Since \(x_2\) is a free variable, then we can plug any value for it that we like, and the result will still be a solution to our linear system.

Example 2¶

\[\begin{split}B = \left[\begin{array}{rrr|r}15&9&-4&-17\\-5&-3&4&11\\0&0&0&1\\\end{array}\right]\end{split}\]

The steps in the code block will row-reduce the augmented matrix. Only the result is shown below.

Tip

Placing a semicolon at the end of a line tells MATLAB to execute that line but suppress the result.

B= [ 15     9    -4   -17 ; -5    -3     4    11 ; 0     0     0     1 ];
B([1 2],:) = B([2 1],:);
B(2,:) = B(2,:) + 3 * B(1,:)

B =

    -5    -3     4    11
     0     0     8    16
     0     0     0     1

The bottom row of the matrix

\[\begin{split}A = \left[\begin{array}{rrrr}-5&-3&4&11\\0&0&8&16\\0&0&0&1\\\end{array}\right]\end{split}\]

claims that \(0 = 1\) which is not possible, which means there are no solutions to this linear system.

Example 3¶

Suppose we have the following REF of the augmented matrix

\[\begin{split}C = \left[\begin{array}{rrrrr|r}3&-3&0&9&0&6\\0&0&1&5&0&1\\0&0&0&0&5&5\\\end{array}\right]\end{split}\]

where we see basic variables \(x_1, x_3, x_5\) and free variables \(x_2, x_4\). Then \(x_5 = 1\),

\[x_3 = 1 - 5x_4\]

and

\[3x_1 = 6 + 3x_2 - 9x_4\]

which simplifies to

\[x_1 = 2 + x_2 - 3x_4\]

Using the Margilit and Robanoff notation: \((x_1,x_2,x_3,x_4, x_5) = (2+x_2 -3x_4,x_2,1-5x_4,x_4,5)\), but we will use the expanded form shown below in this supplement.

This means our solutions will have the form:

\[\begin{split}\vec x = \left[ \begin{array}{ccccc}2 & +x_2 & -3x_4\\ &x_2&\\1 && -5x_4 \\ &&x_4 \\ 5&&\end{array}\right]= \left[\begin{array}{r} 2 \\ 0 \\ 1 \\ 0 \\ 5\end{array}\right] + \left[\begin{array}{c} x_2 \\ x_2 \\ 0 \\ 0 \\ 0\end{array}\right] + \left[\begin{array}{c} -3x_4 \\ 0 \\ -5x_4 \\ x_4 \\ 0\end{array}\right] \end{split}\]

which simplifies, after factoring, to

\[\begin{split}\vec x = \left[\begin{array}{r} 2 \\ 0 \\ 1 \\ 0 \\ 5\end{array}\right] + \left[\begin{array}{r} 1 \\ 1 \\ 0 \\ 0 \\ 0\end{array}\right]x_2 + \left[\begin{array}{r} -3 \\ 0 \\ -5 \\ 1 \\ 0\end{array}\right]x_4 \end{split}\]

Linear Algebra with MATLAB