MAT223 Tutorial 02

• 1
  • Let $x=\left[\begin{array}{c}1\\ -1\\ 3\\ 5\end{array}\right]$
  • $y=\left[\begin{array}{c}-2\\ 0\\ 1\\ -2\end{array}\right]$
  • $z=\left[\begin{array}{c}-1\\ 2\\ 0\\ 1\end{array}\right]$
  • Compute $3x+y-2z$
  • $3x=\left[\begin{array}{c}3\\ -3\\ 9\\ 15\end{array}\right]$
  • $2z=\left[\begin{array}{c}-2\\ 4\\ 0\\ 2\end{array}\right]$
  • $3x+y-2z=\left[\begin{array}{c}3-2-(-2)\\ -3+0-4\\ 9+1-0\\ 15-2-2\end{array}\right]$
  • $3x+y-2z=\left[\begin{array}{c}3\\ -7\\ 10\\ 11\end{array}\right]$
• 2
  • $\left[\begin{array}{ccccc}1& 1& -2& -1& 3\\ 2& 1& -1& -3& 4\\ 3& 2& -3& -4& 7\end{array}\right]$
  • $R_2=R_2-2R_1$
  • $\left[\begin{array}{ccccc}1& 1& -2& -1& 3\\ 0& -1& 3& -1& -2\\ 3& 2& -3& -4& 7\end{array}\right]$
  • $R_2=\frac{R_2}{-1}$
  • $\left[\begin{array}{ccccc}1& 1& -2& -1& 3\\ 0& 1& -3& 1& 2\\ 3& 2& -3& -4& 7\end{array}\right]$
  • $R_1=R_1-R_2$
  • $\left[\begin{array}{ccccc}1& 0& 1& -2& 1\\ 0& 1& -3& 1& 2\\ 3& 2& -3& -4& 7\end{array}\right]$
  • $R_3=R_3-3R_1$
  • $\left[\begin{array}{ccccc}1& 0& 1& -2& 1\\ 0& 1& -3& 1& 2\\ 0& 2& -6& 2& 4\end{array}\right]$
  • $R_3=\frac{R_3}{2}$
  • $\left[\begin{array}{ccccc}1& 0& 1& -2& 1\\ 0& 1& -3& 1& 2\\ 0& 1& -3& 1& 2\end{array}\right]$
  • $R_3=R_3-R_2$
  • $\left[\begin{array}{ccccc}1& 0& 1& -2& 1\\ 0& 1& -3& 1& 2\\ 0& 0& 0& 0& 0\end{array}\right]$
  • $x_1=-x_3+2x_4+1$
  • $x_2=3x_3-x_4+2$
  • $x_3=s$
  • $x_4=t$
  • $x_1=-s+2t+1$
  • $x_2=3s-t+2$
  • Solutions:
    • $\left[\begin{array}{c}1\\ 2\\ 0\\ 0\end{array}\right]+s\left[\begin{array}{c}-2\\ 3\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}2\\ -1\\ 0\\ 1\end{array}\right]$
• 3
  • Let $x=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$
  • $y=\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]$
  • $z=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$
  • If $w=\left[\begin{array}{c}5\\ 3\\ 5\end{array}\right]$ can be a linear combination of $x,y,z$
  • $5=a+c$
  • $3=b$
  • $5=a+b$
  • $\left[\begin{array}{cccc}1& 0& 1& 5\\ 0& 1& 0& 3\\ 1& 1& 0& 5\end{array}\right]$
  • $R_3=R_3-R_1$
  • $\left[\begin{array}{cccc}1& 0& 1& 5\\ 0& 1& 0& 3\\ 0& 1& -1& 0\end{array}\right]$
  • $R_3=R_3-R_2$
  • $\left[\begin{array}{cccc}1& 0& 1& 5\\ 0& 1& 0& 3\\ 0& 0& -1& -3\end{array}\right]$
  • $R_3=\frac{R_3}{-1}$
  • $\left[\begin{array}{cccc}1& 0& 1& 5\\ 0& 1& 0& 3\\ 0& 0& 1& 3\end{array}\right]$
  • $c=3$
  • $b=3$
  • $a+c=5$
  • $a+3=5$
  • $a=2$
  • $w=2x+3y+3c$
  • $\overrightarrow{w}=2\overrightarrow{x}+3\overrightarrow{y}+3\overrightarrow{c}$
  • Because the rank is the same as the number of leading ones, we don't have a free variable. So it's one solution.
• 4
  • 1
    • If a consistent system has 6 variables and 3 equations then the solution must have 3 parameters.
    • $3\times 6$
    • $\left[\begin{array}{ccccccc}1& 0& 1& 0& 1& 0& 1\\ 0& 1& 0& 1& 0& 1& 0\\ 0& 0& 1& 0& 1& 0& 1\end{array}\right]$
    • $x_1,x_2,x_3$ are leading
    • $x_4,x_5,x_6$ are free
    • False
    • Doesn't tell us about the rank.
    • Counterexample:
      • $\left[\begin{array}{ccccccc}1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]$
  • 2
    • If a consistent $4\times 5$ system has rank $3$ then the rref of the augmented coefficient matrix has a row of $0$s.
    • $\left[\begin{array}{cccccc}1& 0& 1& 0& 1& 0\\ 0& 1& 0& 1& 0& 1\\ 0& 0& 1& 0& 1& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right]$
    • $3$ leading variables by rank. $4$ equations. So one must be the $0$ row.
    • True
    • If it's inconsistent we can have something like this in the last row.
    • $\left[\begin{array}{cccccc}1& 0& 1& 0& 1& 0\\ 0& 1& 0& 1& 0& 1\\ 0& 0& 1& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]$
    • Our condition means we can't have this problem.