Exam (elaborations) TEST BANK FOR Advanced Engineering Mathematics with Mathematica By Edward B. Magrab (Solution manual)

Exam (elaborations) TEST BANK FOR Advanced Engineering Mathematics with Mathematica By Edward B. Magrab (Solution manual) Solutions to Exercises in Chapter 1 Section 1.2 1.1 A matrix is an orthogonal matrix if Is the following matrix an orthogonal matrix? Solution: x={{-1.,-1},{1,-1},{-1,1},{1,1}}/2; Transpose[x].x//MatrixForm yields Therefore, X is an orthogonal matrix. 1.2 If does (A + B)2 = A 2 + B 2? Solution: a={{1,-1},{2,-1}}; b={{1,1},{4,-1}}; ((a+b).(a+b)-a.a-b.b)//MatrixForm yields Therefore, the expressions are equal. XT X = I X = 1 2 −1 −1 1 −1 −1 1 1 1 ⎛ ⎝ ⎜⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟⎟ 1 0 0 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ A = 1 −1 2 −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ B = 1 1 4 −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 0 0 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 1.3 Given the two matrices Find the matrix products AB and BA. Solution: Aa={{1,4,-3},{2,5,4}}; Bb={{4,1},{2,6},{0,3}}; Aa.Bb//MatrixForm Bb.Aa//MatrixForm 1.4 Given the following matrices and their respective orders: A (n´m), B (p´m), and C (n´s). Show one way in which these three matrices can be multiplied. What is the order of the resulting matrix? Solution: 1.5 Given Determine A2. Solution: From Eq. (1.13) A = 1 4 −3 2 5 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ and B = 4 1 2 6 0 3 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ AB = 1 4 −3 2 5 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 1 2 6 0 3 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ = 12 16 18 44 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ BA = 4 1 2 6 0 3 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ 1 4 −3 2 5 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 6 21 −8 14 38 18 6 15 12 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ CT ABT →(n × s)T (n × m)(p × m)T →(s × n)(n × m)(m × p)→(s × p) A = ab b2 −a2 −ab ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 Aa={{a b, b^2},{-a^2,-a b}}; Aa.Aa//MatrixForm 1.6 Given the matrix Determine the value of 4I - 4A - A2 + A3. Solution: Then, Mathematica verification Aa={{-4,-3,-1},{2,1,1},{4,-2,4}}; AA = a11 a12 a21 a22 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ a11 a12 a21 a22 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = a11 2 + a12a21 a12 a11 + a22 ( ) a21 a11 + a22 ( ) a21a12 + a22 2 ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ = a2b2 − a2b2 b2 (ab − ab) −a2 (ab − ab) −a2b2 + a2b2 ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ = 0 A = −4 −3 −1 2 1 1 4 −2 4 ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ A2 = 6 11 −3 −2 −7 3 −4 −22 10 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ A3 = −14 −1 −7 6 −7 7 12 −30 22 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ 4I − 4A− A2 + A3 = 4 1 0 0 0 1 0 0 0 1 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ − 4 −4 −3 −1 2 1 1 4 −2 4 ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ − 6 11 −3 −2 −7 3 −4 −22 10 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ + −14 −1 −7 6 −7 7 12 −30 22 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ = 0 0 0 0 0 0 0 0 0 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ 5 A2=Aa.Aa; MatrixForm[A2] A3=A2.Aa; MatrixForm[A3] (4 IdentityMatrix[3]-4 Aa-A2+A3)//MatrixForm Section 1.3 1.7 Given the following matrices: What is the value of a that satisfies the following equation? Solution: Therefore, Mathematica verification Solve[{1,2}.{{2,a},{3,4}}.{{1},{2}}==Det[{{6,4},{7,5}}],a] 1.8 Show that Solution: x = 1 2 ⎧⎨⎩ ⎫⎬⎭ , A = 2 a 3 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ , B = 6 4 7 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xT Ax = det B xT Ax = { 1 2 } 2 a 3 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 ⎧⎨⎩ ⎫⎬⎭ = { 1 2 } 2 + 2a 11 ⎧⎨⎩ ⎫⎬⎭ = 24 + 2a detB = det 6 4 7 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 6 × 5 − 4 × 7 = 2 24 + 2a = 2 a = −11 det a b + c 1 b a + c 1 c a + b 1 ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ = 0 det a b + c 1 b a + c 1 c a + b 1 ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ = a a + c 1 a + b 1 − (b + c) b 1 c 1 + b a + c c a + b = a(c − b)− (b + c)(b − c)+ b(a + b)− c(a + c) = 0 6 Mathematica verification Det[{{a,b+c,1},{b,a+c,1},{c,a+b,1}}] 1.9 Expand the following determinants and reduce them to their simplest terms. a) Solution: Mathematica verification Det[{{1+a,a,a},{b,1+b,b},{b,b,1+b}}] b) Solution: Mathematica verification det 1+ a a a b 1+ b b b b 1+ b ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ det 1+ a a a b 1+ b b b b 1+ b ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ = (1+ a) (1+ b)2 − b2 ⎡⎣ ⎤⎦ − a b(1+ b)− b2 ⎡⎣ ⎤⎦ + a b2 − b(1+ b) ⎡⎣ ⎤⎦ = (1+ a)[1+ 2b]− ab − ab = 1+ a + 2b + 2ab − 2ab = 1+ a + 2b det x3 +1 1 1 1 x3 +1 1 1 1 x3 +1 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ det x3 +1 1 1 1 x3 +1 1 1 1 x3 +1 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ = x( 3 +1) x( 3 +1)2 −1 ⎡⎣ ⎤⎦ − x3 +1−1 ⎡⎣ ⎤⎦ + 1− x3 −1 ⎡⎣ ⎤⎦ = x( 3 +1) x( 3 +1)2 −1 ⎡⎣ ⎤⎦ − 2x3 = x( 3 +1) x6 + 2x3 ⎡⎣ ⎤⎦ − 2x3 = x3 x6 + 2x( 3 )+ x6 + 2x3 − 2x3 = x9 + 2x6 + x6 = x6 x( 3 + 3)