# Lecture 1 Notes: MIT Linear Algebra (18.06)

### The Geometry of Linear Equations

Ax = b is a big theme.
A = matrix, x = vector of unknowns, and b = solution vector or point

Main questions:
Can I solve Ax = b for every b?
Do the linear combinations of the columns fill 3D space? (for our 3D example below, yes it does)

### 2 equations, 2 unknowns

Find the x and y values that satisfy both equations. $\hphantom{-}2x - y = 0 \\ -x + 2y = 3$

Row Picture

The coefficients of the left side of the first equation become the first row of the first matrix. The coefficients of the the left side of the second question become the second row of the first matrix.

First equation coefficients are -2 and -1.
Second equation coefficients are -1 and 2.
Both are displayed below in their own matrix.

Next is the vector containing the 2 unknowns: x and y.

Lastly the matrix times the vector (also known as taking the dot product)
is equal to the vector containing the numbers on the right side of the given equations. $\Bigg[ \begin{matrix} \hphantom{-}2 & -1 \\ -1 & \hphantom{-}2 \end{matrix} \hphantom{-}\Bigg] \Bigg[ \begin{matrix} \hphantom{-}x \hphantom{-}y \end{matrix} \hphantom{-}\Bigg] \hphantom{-}= \hphantom{-}\Bigg[ \begin{matrix} \hphantom{-}0 \hphantom{-}3 \end{matrix} \hphantom{-}\Bigg]$

With the column picture, you plot the lines, see where they intersect and that is your solution: x = 1, y =2.

Column Picture

Take the first and second columns from the previous matrix and multiply them by their corresponding unknown: x and y. Add these vectors together and have them equal to the same previous vector. $x\Bigg[ \begin{matrix} \hphantom{-}2 -1 \end{matrix} \hphantom{-}\Bigg] \hphantom{-}+ \hphantom{-}y\Bigg[ \begin{matrix} -1 \hphantom{-}2 \end{matrix} \hphantom{-}\Bigg] \hphantom{-}= \hphantom{-}\Bigg[ \begin{matrix} \hphantom{-}0 \hphantom{-}3 \end{matrix} \hphantom{-}\Bigg]$

When we plot these vectors and use the answer x = 1, y = 2 from from before, we start by putting together 1 of the x vectors and 2 of the y vectors. The vector between the beginning point and the ending point is your solution. Again x = 1, y = 2.

The solutions from the column and row pictures are easy to see at this point, no reason to use one over the other when dealing with 2 equations and 2 unknowns. Now on the 3 equations and 3 unknowns.

### 3 equations, 3 unknown $\hphantom{-}2x - y \hphantom{-0z}= 0 \\ -x + 2y - z = -1 \\ \hphantom{-0x}-3y + 4z = 4$

#### Row Picture

Same process as above. However, when it comes to plotting the 3 planes and finding the intersecting point, it gets tricky to visualize.

#### Column Picture

A ton easier when it comes to plotting, only 3 vectors need to be plotted. It’s a lot easier to visualize and to find the solution vector.