The fundamental problem of linear algebra is to solve n linear equations in n unknowns; for example: 2x − y = 0 −x + 2y = 3. In this first lecture on linear algebra we view this problem in three ways. The system above is two dimensional (n = 2). By adding a third variable z we could expand it to three dimensions. Row Picture Plot the points that satisfy each equation. The intersection of the plots (if they do intersect) represents the solution to the system of equations. Looking at Figure 1 we see that the solution to this system of equations is x = 1, y = 2. −10 1 2 x −2 −1 0 1 2 3 4 y 2x − y = 0 −x + 2y = 3 (1, 2) Figure 1: The lines 2x − y = 0 and −x + 2y = 3 intersect at the point (1, 2). We plug this solution in to the original system of equations to check our work: 2 · 1 − 2 = 0 −1 + 2 · 2 = 3. The solution to a three dimensional system of equations is the common point of intersection of three planes (if there is one). 1 Column Picture In the column picture we rewrite the system of linear equations as a single equation by turning the coefficients in the columns of the system into vectors: � 2 � � −1 � � 0 � x + y = . −1 2 3 Given two vectors c and d and scalars x and y, the sum xc + yd is called a linear combination of c and d. Linear combinations are important throughout this course. Geometrically, we want to find numbers x and y so that x copies of vector � − 2 1 � added to y copies of vector � −1 2 � equals the vector � 0 3 � . As we see from Figure 2, x = 1 and y = 2, agreeing with the row picture in Figure 2. −10 1 2 −1 0 1 2 3 4 2 −1 −1 2 −1 2 0 3 Figure 2: A linear combination of the column vectors equals the vector b. In three dimensions, the column picture requires us to find a linear combination of three 3-dimensional vectors that equals the vector b. Matrix Picture We write the system of equations 2x − y = 0 −x + 2y = 3 2 as a single equation by using matrices and vectors: � 2 −1 � � x � � 0 � −1 2 y = 3 .