Linear Algebra for Team-Based Inquiry Learning: Fall 2024 Math 0520 Edition

(a)

1. One algebraic approach is to first consider the corresponding augmented matrix:

   \begin{equation*} \left[\begin{array}{cc|c}1 & -1 & 1\\ 3& -3& k\end{array}\right]. \end{equation*}

   Using row operations, the corresponding RREF is equal to:

   \begin{equation*} \left[\begin{array}{cc|c}1 & -1 & 1\\ 0 & 0& k-3\end{array}\right]. \end{equation*}

   Then, by Fact 1.3.7, the linear system is inconsistent whenever \(k\neq 3\) and is consistent with infinitely many solutions when \(k=3\text{.}\)
2. Geometrically, each of the given linear equations describe lines in the \(x_1x_2\)-plane. Re-writing the equations as follows:

   \begin{alignat*}{2} x_1 &\,=\,& 1+x_2\\ x_1 &\,=\,&\frac{k}{3}+x_2, \end{alignat*}

   we see that the lines have the same slopes, i.e., they are parallel. It follows that the lines must be identical (in which case the corresponding) solution is infinite, or else they do not intersect at all, in which case there are no solutions. This corresponds to the case where \(k=3\) and \(k\neq 3\) respectively.

(b)

1. One algebraic approach is to first consider the corresponding augmented matrix:

   \begin{equation*} \left[\begin{array}{cc|c}1 & -1 & 1\\ 3& -k& 3\end{array}\right]. \end{equation*}

   Using row operations, the corresponding equivalent to:

   \begin{equation*} \left[\begin{array}{cc|c}1 & -1 & 1\\ 0 & k-3& 0\end{array}\right]. \end{equation*}

   If \(k=3\text{,}\) then this augmented matrix is in RREF and by Fact 1.3.7, the system has infinitely many solutions. Otherwise, when \(k\neq 3\text{,}\) then the RREF is given by:

   \begin{equation*} \left[\begin{array}{cc|c}1 & 0 & 1\\ 0 & 1& 0\end{array}\right] \end{equation*}

   and the linear system is consistent with a unique solution.
2. In the \(x_1x_2\)-plane, the first equation describes the line passing through \((1,0)\) with slope equal to \(1\text{.}\) The second line corresponds to the line passing through \((1,0)\) with slope equal to \(\frac{k}{3}\text{.}\) Two lines passing through the same point are either equal or intersect in a single point (the point of common intersection). This happens when \(k=3\) and \(k\neq 3\) respectively, which correspond to the system having infinitely many solutions or a single solution, as desired.

(a)

(b)

(c)

(a)

• The following system has no free variables:

  \begin{alignat*}{4} x_1 & &\,=\,& 1\\ &x_2 &\,=\,& 2\\ &&x_3\,=\,&3. \end{alignat*}

  Its solution set is \(\left\{\left[\begin{array}{c}1\\2\\3\end{array}\right]\right\}\text{.}\)

• The following system has 1 free variable (namely, \(x_3\)):

  \begin{alignat*}{4} x_1 & &\,=\,& 1\\ &x_2 &\,=\,& 2\\ &&0\,=\,&0. \end{alignat*}

  Its solution set is \(\left\{\left[\begin{array}{c}1\\2\\b\end{array}\right]\,\middle|\, b \in\mathbb R\right\}\text{.}\)

• The following system has 2 free variable (namely, \(x_2,x_3\)):

  \begin{alignat*}{4} x_1 & &\,=\,& 1\\ & &0\,=\,& 0\\ &&0\,=\,&0. \end{alignat*}

  Its solution set is \(\left\{\left[\begin{array}{c}1\\a\\b\end{array}\right]\,\middle|\, a,b \in\mathbb R\right\}\text{.}\)

• In the following system, all three of \(x_1,x_2,x_3\) are free:

  \begin{alignat*}{4} & &0\,=\,& 0\\ & &0\,=\,& 0\\ &&0\,=\,&0. \end{alignat*}

  Its solution set is \(\IR^3\) because these equations are valid no matter what.

(b)