SOLUTIONS MANUAL
Solution Manual
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Solutions to Exercises in Chapter 1 Section 1.2 1.1 A matrix is an orthogonal matrix if
XTX = I Is the following matrix an orthogonal matrix?
⎛ −1 −1 ⎞ 1 ⎜ 1 −1 ⎟ ⎟ X= ⎜ 2 ⎜ −1 1 ⎟ ⎜⎝ 1 1 ⎟⎠ Solution: x={{-1.,-1},{1,-1},{-1,1},{1,1}}/2; Transpose[x].x//MatrixForm yields
⎛ 1 0 ⎞ ⎜ ⎟ ⎝ 0 1 ⎠ Therefore, X is an orthogonal matrix. 1.2 If
⎛ 1 −1 ⎞ A=⎜ ⎟ ⎝ 2 −1 ⎠
⎛ 1 1 ⎞ B=⎜ ⎟ ⎝ 4 −1 ⎠
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
⎛ 0 0 ⎞ ⎜ ⎟ ⎝ 0 0 ⎠ Therefore, the expressions are equal.
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1.3 Given the two matrices ⎛ 1 4 −3 ⎞ A=⎜ ⎝ 2 5 4 ⎟⎠
and
⎛ 4 1 ⎞ B=⎜ 2 6 ⎟ ⎜ ⎟ ⎜⎝ 0 3 ⎟⎠
Find the matrix products AB and BA. Solution:
⎛ 4 1 ⎞ ⎛ 1 4 −3 ⎞ ⎜ ⎛ 12 16 ⎞ AB = ⎜ 2 6 ⎟ =⎜ ⎟ ⎝ 2 5 4 ⎠ ⎜⎜ 0 3 ⎟⎟ ⎝ 18 44 ⎟⎠ ⎝ ⎠ ⎛ 4 1 ⎞ ⎛ 6 21 −8 ⎞ ⎛ 1 4 −3 ⎞ ⎜ ⎜ ⎟ BA = 2 6 ⎜ = 14 38 18 ⎟ ⎜ ⎟ ⎝ 2 5 4 ⎟⎠ ⎜ ⎟ ⎜⎝ 0 3 ⎟⎠ ⎜⎝ 6 15 12 ⎟⎠ 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:
C T ABT → (n × s)T (n × m)( p × m)T → (s × n)(n × m)(m × p) → (s × p) 1.5 Given
⎛ ab A=⎜ ⎝ −a 2 Determine A2. Solution: From Eq. (1.13)
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b2 ⎞ ⎟ −ab ⎠
⎛ a11 a12 ⎞ ⎛ a11 a12 ⎞ ⎛ a112 + a12 a21 a12 ( a11 + a22 ) ⎞ ⎟ AA = ⎜ ⎟⎜ ⎟ =⎜ 2 a a a a ⎜ a a + a a a + a ⎟⎠ ( ) ⎝ 21 22 ⎠ ⎝ 21 22 ⎠ ⎝ 21 11 22 21 12 22 ⎛ a 2b 2 − a 2b 2 b 2 ( ab − ab ) ⎞ ⎟ =0 =⎜ ⎜⎝ −a 2 ( ab − ab ) −a 2b 2 + a 2b 2 ⎟⎠ Aa={{a b, b^2},{-a^2,-a b}}; Aa.Aa//MatrixForm 1.6 Given the matrix ⎛ −4 −3 −1 ⎞ A=⎜ 2 1 1 ⎟ ⎜ ⎟ ⎝ 4 −2 4 ⎠
Determine the value of 4I - 4A - A2 + A3. Solution:
⎛ 6 11 −3 ⎞ ⎜ A = −2 −7 3 ⎟ ⎜ ⎟ ⎜⎝ −4 −22 10 ⎟⎠ 2
⎛ −14 −1 −7 ⎞ A3 = ⎜ 6 −7 7 ⎟ ⎜ ⎟ ⎜⎝ 12 −30 22 ⎟⎠ Then,
⎛ 1 0 0 ⎞ ⎛ −4 −3 −1 ⎞ ⎜ ⎟ 4I − 4 A − A + A = 4 0 1 0 − 4 ⎜ 2 1 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ 0 0 1 ⎟⎠ ⎝ 4 −2 4 ⎠ 2
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⎛ 6 11 −3 ⎞ ⎛ −14 −1 −7 ⎞ − ⎜ −2 −7 3 ⎟ + ⎜ 6 −7 7 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ −4 −22 10 ⎟⎠ ⎜⎝ 12 −30 22 ⎟⎠ ⎛ 0 0 0 ⎞ =⎜ 0 0 0 ⎟ ⎜ ⎟ ⎜⎝ 0 0 0 ⎟⎠ Mathematica verification Aa={{-4,-3,-1},{2,1,1},{4,-2,4}};
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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:
⎧ 1 ⎫ x=⎨ ⎬, 2 ⎩ ⎭
⎛ 2 a ⎞ A=⎜ , ⎝ 3 4 ⎟⎠
⎛ 6 4 ⎞ B=⎜ ⎝ 7 5 ⎟⎠
What is the value of a that satisfies the following equation?
x T Ax = det B Solution:
x T Ax =
{ 1 2 }⎛⎜⎝ 23 a4 ⎞⎟⎠ ⎧⎨⎩ 12 ⎫⎬⎭ = { 1 2 } ⎧⎨⎩ 2 +112a ⎫⎬⎭
= 24 + 2a ⎛ 6 4 ⎞ detB = det ⎜ = 6×5−4×7= 2 ⎝ 7 5 ⎟⎠ Therefore,
24 + 2a = 2 a = −11 Mathematica verification Solve[{1,2}.{{2,a},{3,4}}.{{1},{2}}==Det[{{6,4},{7,5}}],a]
1.8 Show that ⎛ a b+c 1 ⎞ det ⎜ b a + c 1 ⎟ = 0 ⎜ ⎟ ⎝ c a+b 1 ⎠
Solution:
⎛ a b+c 1 ⎞ a+c 1 b 1 b a+c det ⎜ b a + c 1 ⎟ = a − (b + c) + ⎜ ⎟ a+b 1 c 1 c a+b ⎝ c a+b 1 ⎠ = a(c − b) − (b + c)(b − c) + b(a + b) − c(a + c) =0 5
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) ⎛ 1+ a a a ⎞ ⎜ det b 1+ b b ⎟ ⎜ ⎟ b 1+ b ⎠ ⎝ b
Solution:
⎛ 1+ a a a ⎞ 2 2 2 2 det ⎜ b 1+ b b ⎟ = (1+ a ) ⎡⎣(1+ b) − b ⎤⎦ − a ⎡⎣b(1+ b) − b ⎤⎦ + a ⎡⎣b − b(1+ b) ⎤⎦ ⎜ ⎟ b 1+ b ⎠ ⎝ b = (1+ a )[1+ 2b ] − ab − ab = 1+ a + 2b + 2ab − 2ab = 1+ a + 2b Mathematica verification Det[{{1+a,a,a},{b,1+b,b},{b,b,1+b}}]
b) ⎛ x3 + 1 1 1 ⎞ ⎜ ⎟ det ⎜ 1 x3 + 1 1 ⎟ ⎜⎝ 1 1 x 3 + 1 ⎟⎠
Solution: ⎛ x3 + 1 1 1 ⎞ 2 ⎜ ⎟ 3 3 3 3 det ⎜ 1 x3 + 1 1 ⎟ = ( x + 1) ⎡⎣( x + 1) − 1⎤⎦ − ⎡⎣ x + 1− 1⎤⎦ + ⎡⎣1− x − 1⎤⎦ ⎜⎝ 1 1 x 3 + 1 ⎟⎠ 2 = ( x 3 + 1) ⎡( x 3 + 1) − 1⎤ − 2x 3 ⎣ ⎦
= ( x 3 + 1) ⎡⎣ x 6 + 2x 3 ⎤⎦ − 2x 3
= x 3 ( x 6 + 2x 3 ) + x 6 + 2x 3 − 2x 3 = x 9 + 2x 6 + x 6 = x 6 ( x 3 + 3)
Mathematica verification
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Det[{{x^3+1,1,1},{1,1+x^3,1},{1,1,1+x^3}}]
1.10 Determine if the following determinant a function of a ⎛ ex det ⎜ e x ⎜ ⎜⎝ 1
sin x cos x ⎞ ⎟ cos x sin x ⎟ 1− a a ⎟⎠
a function of a? Solution: ⎛ ex det ⎜ e x ⎜ ⎜⎝ 1
sin x cos x ⎞ ⎛ x ⎛ cos x sin x ⎞ x ⎟ e − det ⎜ e cos x sin x ⎟ = det ⎜ ⎟ a ⎠ ⎝ 1− a ⎝ 1 1− a a ⎟⎠ ⎛ x + det ⎜ e ⎝ 1
sin x ⎞ sin x ⎟ a ⎠
cos x ⎞ cos x ⎟ 1− a ⎠
= e x ( a cos x − (1− a)sin x ) − sin x ( ae x − sin x ) + cos x ( e x (1− a) − cos x )
= e x ( a cos x − sin x + asin x − asin x + cos x − a cos x ) + sin 2 x − cos 2 x = e x ( cos x − sin x ) + sin 2 x − cos 2 x
which is not a function of a. Mathematica verification Det[{{Exp[x],Sin[x],Cos[x]},{Exp[x],Cos[x],Sin[x]},{1,1-a,a}}] 1.11 Show that
⎛ 2 ⎜⎜ x1 ⎜ det ⎜⎜⎜ x22 ⎜⎜ ⎜⎜ x32 ⎝
⎞ x1 1 ⎟⎟ ⎟⎟ x2 1 ⎟⎟⎟ = ( x1 − x2 )( x1 − x3 )( x2 − x3 ) ⎟⎟ x3 1 ⎟⎟⎟⎠
Solution:
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⎛ 2 ⎜⎜ x1 ⎜ det ⎜⎜⎜ x22 ⎜⎜ ⎜⎜ x32 ⎝
⎞ x1 1 ⎟⎟ ⎟⎟ 2 2 2 2 2 x2 1 ⎟⎟⎟ = x1 ( x2 − x3 ) − x1 ( x2 − x3 ) + ( x2 x3 − x3 x2 ) ⎟⎟ x3 1 ⎟⎟⎟⎠ = x12 ( x2 − x3 ) − x1 ( x2 − x3 )( x2 + x3 ) + x2 x3 ( x2 − x3 ) = ( x2 − x3 ) ⎡⎢ x12 − x1 ( x2 + x3 ) + x2 x3 ⎤⎥ ⎣ ⎦ = ( x2 − x3 ) ⎡⎢ x1 ( x1 − x2 ) − x3 ( x1 − x2 )⎤⎥ ⎣ ⎦ = ( x1 − x2 )( x1 − x3 )( x2 − x3 )
Mathematica verification Simplify[Det[{{x1^2,x1,1},{x2^2,x2,1},{x3^2,x3,1}}]]
Section 1.4 1.12 Given
⎛ 17 7 ⎞ A=⎜ ⎟ ⎝ 19 9 ⎠ Determine A-1 and verify your result. Solution: From Eq. (1.27)
⎛ a 11 ⎜ ⎜⎝ a21
−1
⎛ a a12 ⎞ −a12 ⎞ 1 22 ⎟ = ⎜ ⎟ a11a22 − a12 a21 ⎜⎝ −a21 a11 ⎟⎠ a22 ⎟⎠ −1
⎛ 17 7 ⎞ ⎛ 9 −7 ⎞ 1 ⎜ 19 9 ⎟ = 17 × 9 − 19 × 7 ⎜ −19 17 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 9 / 20 −7 / 20 ⎞ =⎜ ⎟ ⎝ −19 / 20 17 / 20 ⎠ Since, from Eq. (1.13)
⎛ a11 a12 ⎞ ⎛ b11 b12 ⎞ ⎛ a11b11 + a12b21 a11b12 + a12b22 ⎞ ⎜ ⎟⎜ ⎟ =⎜ ⎟ ⎝ a21 a22 ⎠ ⎝ b21 b22 ⎠ ⎝ a21b11 + a22b21 a21b12 + a22b22 ⎠ then
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⎛ 17 7 ⎞ ⎛ 9 / 20 −7 / 20 ⎞ ⎛ 17 × 9 / 20 − 7 × 19 / 20 −17 × 7 / 20 + 7 × 17 / 20 ⎞ ⎜ 19 9 ⎟ ⎜ −19 / 20 17 / 20 ⎟ = ⎜ 19 × 9 / 20 − 9 × 19 / 20 −19 × 7 / 20 + 9 × 17 / 20 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ 1 0 ⎞ =⎜ ⎟ ⎝ 0 1 ⎠ Mathematica verification Aa={{17,7},{19,9}}; Inverse[Aa]//MatrixForm Inverse[Aa].Aa//MatrixForm
Section 1.5 1.13 Given the two matrices ⎛ 3 1 4 ⎞ A=⎜ 2 1 2 ⎟ ⎜ ⎟ ⎜⎝ 4 2 3 ⎟⎠
⎛ 2 1 3 ⎞ B=⎜ 1 2 5 ⎟ ⎜ ⎟ ⎜⎝ 0 2 1 ⎟⎠
and
Show that (AB)T = BTAT. Solution: T
⎛⎛ 3 1 4 ⎞ ⎛ 2 1 3 ⎞⎞ ⎛ 7 13 18 ⎞ ⎜ ⎟ T ⎜ ⎟ ⎜ ⎟ ( AB) = ⎜ 2 1 2 1 2 5 ⎟ = ⎜ 5 8 13 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜⎝ 10 14 25 ⎟⎠ ⎜⎝ ⎜⎝ 4 2 3 ⎟⎠ ⎜⎝ 0 2 1 ⎟⎠ ⎟⎠
T
⎛ 7 5 10 ⎞ = ⎜ 13 8 14 ⎟ ⎜ ⎟ ⎜⎝ 18 13 25 ⎟⎠ and T
T
⎛ 2 1 3 ⎞ ⎛ 3 1 4 ⎞ ⎛ 2 1 0 ⎞⎛ 3 2 4 ⎞ T T ⎜ ⎟ ⎜ ⎟ B A = 1 2 5 2 1 2 =⎜ 1 2 2 ⎟⎜ 1 1 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜⎝ 0 2 1 ⎟⎠ ⎜⎝ 4 2 3 ⎟⎠ ⎜⎝ 3 5 1 ⎟⎠ ⎜⎝ 4 2 3 ⎟⎠ ⎛ 7 5 10 ⎞ = ⎜ 13 8 14 ⎟ ⎜ ⎟ ⎜⎝ 18 13 25 ⎟⎠ Mathematica verification
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Aa={{3,1,4},{2,1,2},{4,2,3}}; Bb={{2,1,3},{1,2,5},{0,2,1}}; Aa.Bb//MatrixForm Transpose[Aa.Bb]//MatrixForm Transpose[Bb]//MatrixForm Transpose[Aa]//MatrixForm Transpose[Bb].Transpose[Aa]//MatrixForm Section 1.7 1.14 Does the following system of equations have a solution? 100 y1 + 420 y2 + 486 y3 = 17 400 y1 +1050 y2 + 972 y3 = 18 700 y1 +1680 y2 +1458y3 = −3
Solution: It is seen from Eq. (1.30) that a = 300, b = 630, c = 486, and k = 2. Therefore, the determinant equals zero. Mathematica verification Det[{{100,420,486},{400,1050,972},{700,1680,1458}}] 1.15 Without solving, determine whether the following system of equations has a solution. a1 + 2a2 + 3a3 = 1 4a1 + 5a2 + 6a3 = 0 7a1 + 8a2 + 9a3 = −7
Solution: It is seen from Eq. (1.28) that c = 3 and d = 6. Therefore, the determinant equals zero. Mathematica verification Det[{{1,2,3},{4,5,6},{7,8,9}}] 1.16 Given the following system of equations
é7 2ù ì x1 ü é1 0ù ì x1 ü ì0ü l í ý ê5 1 ú x ê0 1 ú í x ý = í0ý ë ûî 2þ ë ûî 2þ î þ When l = 4, what are the values of x1 and x2? Justify your answer. Solution: When
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2 ù ì x1 ü ì0ü é3 2 ù ì x1 ü ì0ü é7 - l é3 2 ù í ý=í ý®ê í ý = í ý ® det ê ê 5 ú ú ú = -19 1 - l û î x2 þ î0þ ë5 -3û î x2 þ î0þ ë ë5 -3û Since determinant is not equal to zero, x1 and x2 must equal zero. 1.17 Under what conditions does the following system of equations have a solution? (1− λ ) y1 + 2 y2 + 3y3 = 0 4 y1 + (6− λ ) y2 + 5y3 = 0 7 y1 + 8y2 + (9− λ ) y3 = 0
Solution:
⎛ 1− λ 2 3 ⎞⎟⎟ ⎜⎜ ⎛ 4 5 ⎞⎟⎟ 5 ⎞⎟⎟ ⎟⎟ = (1− λ )det ⎛⎜⎜ 6− λ ⎜⎜ det ⎜⎜ 4 − 2det 6− λ 5 ⎟ ⎟ ⎜⎜ 8 ⎜⎜ 7 9− λ ⎟⎟ ⎟⎠ ⎜⎜ ⎟⎟ 9− λ ⎝ ⎠ ⎝ ⎜⎝ 7 8 9− λ ⎟⎠ ⎛ ⎞ + 3det ⎜⎜⎜ 4 6− λ ⎟⎟⎟ ⎜⎝ 7 8 ⎟⎠ = (1− λ )((6− λ )(9− λ )− 40) − 2(4(9− λ )−35) + 3(32−7(6− λ )) = −λ 3 +16λ 2 − 29λ +14 + 8λ − 2 + 21λ −30 = −λ 3 +16λ 2 −18 Thus, this system of equation has a solution for those values of l that satisfy
−λ 3 +16λ 2 −18 = 0 That is, when l = -1.02814, l = 1.09908, and l = 15.9291. Mathematica verification N[Eigenvalues[{{1,2,3},{4,6,5},{7,8,9}}]] 1.18 Given the following system of equations
(5 − λ )x1 + x2 = 2 −2x1 + (2 − λ )x2 = 1 Under what specific conditions can one solve for x1 and x2. Solution The determinant of the coefficients must not equal zero; that is, we can solve for x1 and x2 provided that
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