By J J Connor
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This follows from the fact that the coefficient matrix is singular. (Oii — %1)(a22 — — a12a21 = 0 Since only one equation isindcpendcnt and there are two unknowns, the soluAssuming* tiOn is not unique. We define as the solution for A = 0, the solution of the first equation is that a12 xi') = 1) C1 = 012 where c1 is an arbitrary constant. Continuing, we let and take c1 such that = 1. This operation is called normalization, and the resulting column matrix, denoted by Q1, is referred to as the characteristic vector for = = —_2112 + L By definition, if a12 (2—10) a12 Q1Q1 = 1 0, we work with the second equation.
Determine where °1 [ L0 01 [A11 Iqj [A21 01 cj [A21 lqj [0 A — 1—36. AIA Aizi AW — A22J — [o Iq \1 Suppose we want to rearrange the columns of a in the following way: 1 2 3 a= 2 1 3 3 4 5 col2—+coll col3—*col2 INTRODUCTION TO MATRIX ALGEBRA 42 (a) CHAP. 1 Show that postmultiplication byIl(which is called a permutation matrix) results in the desired column rearrangement: o 0 11 1 0 01 o i oj H= Note that we just rearrange the corresponding columns of 13. (b) rearranges the rows of a in the Show that pre;nultiplication by same way.
Is . , an even permutation (1—43) . , a,,) is an odd permutation Using (1—43), the definition equation for an ,ith-order determinant can be written as a11 a12 a1,, a21 a22 a2,, = (1—44) 1 where the summation is taken over all possible permutations of (1, 2, Factorial n = = n(n — 1)(n — 2) . • (2)(1). . , n). INTRODUCTION TO MATRIX ALGEBRA 18 CHAP. 1 Example 1—8 The permutations for n = 3 are cxi—1 x23 a33 a32 =2 1 =3 a1=1 z1=2 a3=1 a32 a3=1 e123=+1 e132=—1 e231=+1 e312=+1 e321—-—1 Using (1—44), we obtain a11a22a33 — a11a23a32 a11 a12 a13 a21 a22 a23 = —a12a21a33 + a12a23a31 a32 a33 +a13a21a32 — a13a22a31 This result coincides with (1—42).
Analysis of Structural Member Systems by J J Connor