By P. M. Cohn (auth.)

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**Example text**

If the third equation of our system had been 4x 2 - 5X3 = 3, then the same reduction would lead to 1 o o 5 2 9 -4 5 0 0 -3 0 The row of zeros corresponds to the equation 0 = 0, which may be omitted without affecting the solution, so we are left with 2 equations in 3 unknowns. This system cannot have a unique solution, because we can assign X 3 arbitrarily and then solve for Xl' X 2 as before; therefore the system cannot be regular. R2 -3 I R4~R4+R3 I -2 -4 From the triangular form found we obtain 4x 4 = - 4, hence X4 = - 1, x 3 = - 2 - 2x 4 = 0, X2 = -1- 2X3 + 4x 4 = - 41, Xl = 5 + 2x 2 + 7X3 - 4x 4 = 3, so the unique solution is x = (3, - 41,0, - 1)T.

Xnen) = A(xle l ) + ... + A(xne n) = xlAe l + ... 4). Thus we have expressed Ax as a linear combination ofthe n vectors Ae l , ... , Aen, which of course do not depend on the choice ofx. These vectors Aej are nothing but the columns of A: Aej = (alj' a2j, ... , amj)T. 3 EQUALITY OF MATRICES Two matrices A and Bare said to be equal, A = B, if they have the same number of rows and columns - say, they are both m x n matrices - and if corresponding entries are equal: aij = bij (i = 1, ... , m, j = 1, ...

E. b;'ib;'j. Suppose first that i # j and consider the Ath entry in this sum, corresponding to the Ath edge. It is 1 if the edge joins Pi and Pj' and 0 otherwise, so in all the total is just 1 if there is an edge joining Pi and Pj' and 0 otherwise, and this is just aij . b~i' which is the number of edges ending at Pi and this is just the order di at Pi' while aii = 0, by hypo thesis. 27). To illustrate the result, we have for the incidence matrices of the graphs in Figs. ~} (: ~ l) and for the products BTB we obtain The diagonal terms give the order matrices, while the non-diagonal terms give the adjacency matrices.