Quizes and solutions

Quiz 1

Use Gauss-Jordan reduction to solve the following system of linear equations:

\[\begin{eqnarray*} x_1 + 2x_2 + x_3 &=& 0, \\ 2x_1 + 3x_2 &=& -1, \\ 3x_1 + 4x_2 + x_3 &=& 0, \end{eqnarray*}\]

Solution

  1. Write the problem in augmented matrix form:

    \[\begin{equation} \left( \begin{array}{ccc|c} 1 & 2 & 1 & 0\\ 2 & 3 & 0 & -1\\ 3 & 4 & 1 & 0 \end{array} \right) \end{equation}\]
  2. Add -2 and -3 times the first equation to the second and third equation.

    \[\begin{equation} \left( \begin{array}{ccc|c} 1 & 2 & 1 & 0\\ 0 & -1 & -2 & -1\\ 0 & -2 & -2 & 0 \end{array} \right) \end{equation}\]
  3. Change signs of the second and third equation.

    \[\begin{equation} \left( \begin{array}{ccc|c} 1 & 2 & 1 & 0\\ 0 & 1 & 2 & 1\\ 0 & 2 & 2 & 0 \end{array} \right) \end{equation}\]
  4. Add -2 times the second equation to the third.

    \[\begin{equation} \left( \begin{array}{ccc|c} 1 & 2 & 1 & 0\\ 0 & 1 & 2 & 1\\ 0 & 0 & -2 & -2 \end{array} \right) \end{equation}\]
  5. Multiply the last equation by -1/2.

    \[\begin{equation} \left( \begin{array}{ccc|c} 1 & 2 & 1 & 0\\ 0 & 1 & 2 & 1\\ 0 & 0 & 1 & 1 \end{array} \right) \end{equation}\]
  6. Add -2 and -1 times the last equation to the second and first equations.

    \[\begin{equation} \left( \begin{array}{ccc|c} 1 & 2 & 0 & -1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 1 \end{array} \right) \end{equation}\]
  7. Add -1 times the second equation to the first equation.

    \[\begin{equation} \left( \begin{array}{ccc|c} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 1 \end{array} \right) \end{equation}\]
  8. The solution is \(x_1 = 1, \, x_2 = -1, \, x_3 = 1\).

Quiz 1

Use Gauss-Jordan reduction compute the inverse of:

\[\begin{equation*} T = \left[ \begin{array}{rrr} 3 & 2 & -3 \\ -2 & -1 & 2 \\ 2 & 0 & -1 \end{array} \right] \end{equation*}\]

Solution

  1. Write the problem in augmented matrix form:

    \[\begin{equation*} \left( \begin{array}{rrr | rrr} 3 & 2 & -3 & 1 & 0 & 0\\ -2 & -1 & 2 & 0 & 1 & 0\\ 2 & 0 & -1 & 0 & 0 & 1\\ \end{array} \right) \end{equation*}\]
  2. Using the standard manipulations we find

    \[\begin{equation*} \left( \begin{array}{rrr | rrr} 3 & 2 & -3 & 1 & 0 & 0\\ 0 & -1 & 1 & 0 & 1 & 1\\ 2 & 0 & -1 & 0 & 0 & 1\\ \end{array} \right) \end{equation*}\]
    \[\begin{equation*} \left( \begin{array}{rrr | rrr} 3 & 0 & -1 & 1 & 2 & 2\\ 0 & -1 & 1 & 0 & 1 & 1\\ 2 & 0 & -1 & 0 & 0 & 1\\ \end{array} \right) \end{equation*}\]
    \[\begin{equation*} \left( \begin{array}{rrr | rrr} 1 & 0 & 0 & 1 & 2 & 1\\ 0 & -1 & 1 & 0 & 1 & 1\\ 2 & 0 & -1 & 0 & 0 & 1\\ \end{array} \right) \end{equation*}\]
    \[\begin{equation*} \left( \begin{array}{rrr | rrr} 1 & 0 & 0 & 1 & 2 & 1\\ 0 & -1 & 1 & 0 & 1 & 1\\ 0 & 0 & -1 & -2 & -4 & -1\\ \end{array} \right) \end{equation*}\]
    \[\begin{equation*} \left( \begin{array}{rrr | rrr} 1 & 0 & 0 & 1 & 2 & 1\\ 0 & -1 & 1 & 0 & 1 & 1\\ 0 & 0 & 1 & 2 & 4 & 1\\ \end{array} \right) \end{equation*}\]
    \[\begin{equation*} \left( \begin{array}{rrr | rrr} 1 & 0 & 0 & 1 & 2 & 1\\ 0 & -1 & 0 & -2 & -3 & 0\\ 0 & 0 & 1 & 2 & 4 & 1\\ \end{array} \right) \end{equation*}\]
    \[\begin{equation*} \left( \begin{array}{rrr | rrr} 1 & 0 & 0 & 1 & 2 & 1\\ 0 & 1 & 0 & 2 & 3 & 0\\ 0 & 0 & 1 & 2 & 4 & 1\\ \end{array} \right) \end{equation*}\]
  3. Thus \(T^{-1}\) is

    \[\begin{equation*} T^{-1} = \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 3 & 0 \\ 2 & 4 & 1 \end{array} \right] \end{equation*}\]