# Lecture 1: The Geometry of Linear Equations. Video

Notes

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.

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.

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.

So again we turn to the column picture.

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

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.

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.

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

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