Back to Gauss’s Method

Exercise 3

Avg difficulty 3.0

Use Gauss’s Method to solve each system or conclude ‘many solutions’ or ‘no solutions’.

  • (a)  2x+2y=5x−4y=0\begin{array}{rcrcr}2x&+&2y&=&5\\ x&-&4y&=&0\end{array}     (b)  −x+y=1x+y=2\begin{array}{rcrcr}-x&+&y&=&1\\ x&+&y&=&2\end{array}     (c)  x−3y+z=1x+y+2z=14\begin{array}{rcrcrcr}x&-&3y&+&z&=&1\\ x&+&y&+&2z&=&14\end{array}     (d)  −x−y=1−3x−3y=2\begin{array}{rcrcr}-x&-&y&=&1\\ -3x&-&3y&=&2\end{array}     (e)  4y+z=202x−2y+z=0x+z=5x+y−z=10\begin{array}{rcrcrcr}&&4y&+&z&=&20\\ 2x&-&2y&+&z&=&0\\ x&&&+&z&=&5\\ x&+&y&-&z&=&10\end{array}     (f)  2x+z+w=5y−w=−13x−z−w=04x+y+2z+w=9\begin{array}{rcrcrcrcr}2x&&&+&z&+&w&=&5\\ &&y&&&-&w&=&-1\\ 3x&&&-&z&-&w&=&0\\ 4x&+&y&+&2z&+&w&=&9\end{array}

Hefferon, Linear Algebra, Linear Systems, Gauss’s Method, Exercise 3

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Adapted from Linear Algebra © 2016 Jim Hefferon. Licensed under CC BY-SA 3.0 US. Source: Linear Algebra. Imported at revision df2262e.