J J Connor's Analysis of Structural Member Systems PDF

By J J Connor

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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). The following properties of determinants can be established* from (1—44): 1. 2. 3. 4. 5. 6. 7. If all elements of any row (or column) are zero, the determinant is zero.

Where a is a square matrix. (a + b)(a + b) Show that the product of two symmetrical matrices is symmetrical only when they are commutative. 1—8. Show that the following products are symmetrical: (a) aTa (b) aTba where b is symmetrical bTaTcab where c is symmetrical (c) INTRODUCTION TO MATRtX ALGEBRA 36 CHAP. 1 1-9. Evaluate the following matrix product, using the indicated submatrices: 1 3 3 1 51 4 1—10. Let c = ab. Show that the horizontal partitions of c correspond to those of a and the vertical partitions of c correspond to those of b.

A11a22 — a12a21 Properties 1 and 2 are obvious. It follows from property 2 that laTl * See Probs. 1—17, 1—18, 1—19. a!. We COFACTOR EXPANSION FORMULA SEC. 1—8. illustrate the third by interchanging the rows of a: [a21 a' = a22 a12 = a21a12 — a11a22 = —Ia! a'! Property 4 is also obvious from (b). To demonstrate the fifth, we take a21 = a22 = ka12 ka11 Then a! ii and a12 in (b). Finally, to illustrate property 7, we take This b12 = a12 + ka22 b21 = a21 b22 = a7, Then, ibi 1-8. = (a11 + ka21)a22 — (a12 + ka22)a21 = a!

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Analysis of Structural Member Systems by J J Connor


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