Lecture 1: The Geometry of Linear Equations. Video




In this session simultaneous equations are looked at three ways.

  • Row picture
  • *Column picture
  • Matrix form

Special attention should be given to the Column picture.

The first two equations are:

{$$ \left\{ \array{ 2x - y &= 0 \cr -x + 2y &= 3} \right. $$}

As a look ahead these are immediately rendered as the product of a coefficient matrix and a column vector equal to a column vector.

{$$ \begin{align} \left[ \matrix{ 2 & -1 \cr -1 & 2 }\right] \left[ \matrix{ x \cr y} \right] &= \left[ \matrix{ 0 \cr 3} \right] \cr \mathbf A \mathbf x &= \mathbf b \end{align} $$}

{$ \mathbf A ,\ \mathbf x , \ \& \ \mathbf b $} will be the standard names for the coefficient matrix, the vector of unknowns, and the vector of the right side of the equations.

Reading the equations accross is called the row picture. Two points are quickly identified in the first equation: (0,0) and (1,2), and the line is graphed. With only a little more trouble, two points are found of the second equation: (-3, 0) and (-1, 1). The line is drawn and it is seen the lines cross at (1,2) which are the values of x and y which solve the system of equations.

The "Row Picture" is the familiar intersecting lines of grade-school algebra. The solution at the intersection of the two lines is x=1 and y=2 which is easily verified algebraically.

This is contrasted with the column picture, which is greatly recommended.

The column picture comes from expressing the equations as a linear combination of column vectors:

{$$ x \left[ \array{2 \cr -1}\right] + y \left[ \array{-1 \cr 2}\right] = \left[ \array{0 \cr 3}\right] $$}

The answer as to how an equation like this could be solved analytically is left in abeyance, but we learned the solution in composing the row picture. So the column picture is drawn.

Two of the (2,-1) vectors add to one of the (-1,2) vector to reach the (0,3) vector.

Mention is made that with these two vectors any solution could be produced as a linear combination, and such combinations of vectors could cover the plane.

However, we move on to the case of three equations in three unknowns.


{$$ \left\{ \array{ 2x& - y& \ &= 0& \cr -x& + 2y& -z &= -1& \cr & -3y& +4z &= 4&} \right. $$}

{$$ \mathbf A = \left[ \matrix{ 2 & -1 & 0 \cr -1 & 2 &-2 \cr 0 & -3 & 5 }\right] ;\ \mathbf b = \left[ \matrix{ 0 \cr -1 \cr 4} \right] $$}

The attempt to produce the "row picture" on the blackboard is fairly doomed, but these points are made: the graph of one of the equations is a plane, when there are two planes, they intersect on a line, and if all three planes could be sketched correctly the three planes would intersect at a point. Clearly the idea of graphic solutions in higher dimensions is unappealing. Mention is made that we should be sure each equation produces a plane and that the planes should be unique is left tacit.

The point of intersection is (0,0,1), but you might have to know that already to see it in the graph. Equations are solved for z in order to graph them, and we see the bottom of one of the planes which is why its color does not agree with the legend.


So again we turn to the column picture.

{$$ x \left[ \array{2 \cr -1 \cr 0}\right] + y \left[ \array{-1 \cr 2 \cr -3 }\right] + z \left[ \array{ 0 \cr -1 \cr 4 }\right] = \left[ \array{0 \cr -1 \cr 4 }\right] $$}

The three vectors are drawn. but how to find the solution?

Once again there is no attempt to solve the vector equations analytically, but the fix is in. There was a sly set-up when the equations were choose with the appearance of casual randomness. The z vector is exactly the same as the solution vector, so it is obvious that x=0, y=0, and z=`. (This is lavender vector in the graph.)


The fix is admitted to, and a new fix is proposed. {$ \mathbf b $} is changes so it uses one of x and one of y with nothing of z.

{$$ x \left[ \array{2 \cr -1 \cr 0}\right] + y \left[ \array{-1 \cr 2 \cr -3 }\right] + z \left[ \array{ 0 \cr -1 \cr 4 }\right] = \left[ \array{1 \cr 1 \cr -3 }\right] $$}

It is promised that next lecture will introduce methods for solving such equations algebraically. In the mean time, a column picture is offered.

The three vectors are drawn, and because we are in on the fix, we know we can get the result by adding one of vector x and one of vector y.

The lecturer does not attempt a row picture, but it might look like this.


Now the question occurs, "Can I solve every {$ \mathbf A \mathbf x = \mathbf b \text{ for every } \mathbf b $}?" This is claimed to be equivalent to the question "Do the linear combinations of the columns fill 3D space?"

So far the answer has been yes for the matrices we have considered. But what could go wrong?

  • If the column vectors are coplanar, "singular," the linear combinations could only be solved for result vectors in the same plane. (if that)
  • If two of the columns are equal

There is brief consideration given of 9D space. If the 9th column were the same as the 8th column, we would have an 8D plane in 9D space, and we could not fill 9D space with linear combinations of those 9 columns.


Now the way of multiply a matrix by a vector is considered. Again the column view is favored:

{$$ \left[ \matrix{ 2 & 5 \cr 1 & 3 }\right] \left[ \matrix{ 1 \cr 2} \right] = 1 \left[ \array{2 \cr 1 }\right] + 2 \left[ \array{5 \cr 3 }\right] = \left[ \matrix{ 12 \cr 7} \right] $$}

Well this is exactly what we have been doing when the vector contained unknowns. Again this is emphasized to be a linear combination of the columns of the matrix.

With moving fingers it is demonstrated that this produces the same result as row multiplication, which is called "taking the dot product," but it is clear we should concentrate on the column method for the time being.

{$$ \tag*{$\blacksquare$} $$}

Sources: Lecture 1: The Geometry of Linear Equations.

View the complete course at: License: Creative Commons BY-NC-SA More information at More courses at

File:Linear subspaces with shading.svg by Alksentrs Wikimedia. Creative Commons Attribution-Share Alike 3.0 Unported license. The three-dimensional Euclidean space R3 is a vector space, and lines and planes passing through the origin are vector subspaces in R3.

Strang, Gilbert. Introduction to Linear Algebra. 4th ed. Wellesley, MA: Wellesley-Cambridge Press, February 2009. [SG]

Venkatraman, Dheera. Fooplot Source code available under GNU Lesser General Public License (LGPL) v3. Produced plots are dedicated to the public domain, CCD 1.0. Many machine generated graphs have been manually enhanced by Laurence Eighner Hexamer.

Williams, Thomas. Colin Kelley. Gnuplot Copyright 1986 - 1993, 1998, 2004 Thomas Williams, Colin Kelley. Many machine generated graphs have been manually enhanced by Laurence Eighner Hexamer.





This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.


This page is LinearAlgebraAllVideos1