Table Of ContentSolution Manual
1
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 ⎞ ⎛ 1 1 ⎞
A= B=
⎜ ⎟ ⎜ ⎟
⎝ 2 −1 ⎠ ⎝ 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.
2
1.3 Given the two matrices
⎛ 4 1 ⎞
⎛ 1 4 −3 ⎞ ⎜ ⎟
A= and B= 2 6
⎝⎜ 2 5 4 ⎠⎟ ⎜ ⎟
⎜ 0 3 ⎟
⎝ ⎠
Find the matrix products AB and BA.
Solution:
⎛ 4 1 ⎞
⎛ 1 4 −3 ⎞⎜ ⎟ ⎛ 12 16 ⎞
AB= 2 6 =
⎝⎜ 2 5 4 ⎠⎟⎜ ⎟ ⎝⎜ 18 44 ⎠⎟
⎜ 0 3 ⎟
⎝ ⎠
⎛ 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:
CTABT →(n×s)T(n×m)(p×m)T →(s×n)(n×m)(m× p)→(s× p)
1.5 Given
⎛ ab b2 ⎞
A=⎜ ⎟
⎝ −a2 −ab ⎠
Determine A2.
Solution: From Eq. (1.13)
3
⎛ a a ⎞⎛ a a ⎞ ⎛ a2 +a a a (a +a ) ⎞
AA=⎜ 11 12 ⎟⎜ 11 12 ⎟ =⎜ 11 12 21 12 11 22 ⎟
⎝ a21 a22 ⎠⎝ a21 a22 ⎠ ⎝⎜ a21(a11+a22) a21a12 +a222 ⎠⎟
⎛ a2b2 −a2b2 b2(ab−ab) ⎞
=⎜ ⎟ =0
⎜ −a2(ab−ab) −a2b2 +a2b2 ⎟
⎝ ⎠
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 ⎞
A2 =⎜ −2 −7 3 ⎟
⎜ ⎟
⎜ −4 −22 10 ⎟
⎝ ⎠
⎛ −14 −1 −7 ⎞
A3 =⎜ 6 −7 7 ⎟
⎜ ⎟
⎜ 12 −30 22 ⎟
⎝ ⎠
Then,
⎛ 1 0 0 ⎞ ⎛ −4 −3 −1 ⎞
4I −4A− A2 + A3 = 4⎜ 0 1 0 ⎟ −4⎜ 2 1 1 ⎟
⎜ ⎟ ⎜ ⎟
⎝⎜ 0 0 1 ⎠⎟ ⎝ 4 −2 4 ⎠
⎛ 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}};
4
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 ⎫ ⎛ 2 a ⎞ ⎛ 6 4 ⎞
x= ⎨ ⎬, A=⎜ ⎟, B=⎜ ⎟
⎩ 2 ⎭ ⎝ 3 4 ⎠ ⎝ 7 5 ⎠
What is the value of a that satisfies the following equation?
xTAx=detB
Solution:
{ }⎛ 2 a ⎞⎧ 1 ⎫ { }⎧ 2+2a ⎫
xTAx= 1 2 ⎜ ⎟⎨ ⎬= 1 2 ⎨ ⎬
⎝ 3 4 ⎠⎩ 2 ⎭ ⎩ 11 ⎭
=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 b 1+b ⎠
Solution:
⎛ 1+a a a ⎞
det⎜ b 1+b b ⎟ =(1+a)⎡⎣(1+b)2 −b2⎤⎦−a⎡⎣b(1+b)−b2⎤⎦+a⎡⎣b2 −b(1+b)⎤⎦
⎜ ⎟
⎝ b b 1+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 x3+1 ⎠⎟
Solution:
⎛ x3+1 1 1 ⎞
det⎜⎜ 1 x3+1 1 ⎟⎟ =(x3+1)⎡⎣(x3+1)2 −1⎤⎦−⎡⎣x3+1−1⎤⎦+⎡⎣1−x3−1⎤⎦
⎝⎜ 1 1 x3+1 ⎠⎟
=(x3+1)⎡(x3+1)2 −1⎤−2x3
⎣ ⎦
=(x3+1)⎡x6 +2x3⎤−2x3
⎣ ⎦
= x3(x6 +2x3)+x6 +2x3−2x3
= x9 +2x6 +x6 = x6(x3+3)
Mathematica verification
6
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 sinx cosx ⎞
⎜ ⎟
det ex cosx sinx
⎜ ⎟
⎜ ⎟
⎝ 1 1−a a ⎠
a function of a?
Solution:
⎛ ex sinx cosx ⎞
det⎜⎜ ex cosx sinx ⎟⎟ =det⎛⎝⎜ c1o−sax sianx ⎞⎠⎟ex −det⎛⎝⎜ e1x sianx ⎞⎠⎟sinx
⎜ ⎟
⎝ 1 1−a a ⎠
⎛ ex cosx ⎞
+det cosx
⎜ ⎟
⎝ 1 1−a ⎠
=ex(acosx−(1−a)sinx)−sinx(aex −sinx)
+cosx(ex(1−a)−cosx)
=ex(acosx−sinx+asinx−asinx+cosx−acosx)
+sin2x−cos2x
=ex(cosx−sinx)+sin2x−cos2x
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
⎛ ⎞
⎜ x2 x 1 ⎟
⎜⎜ 1 1 ⎟⎟⎟
det⎜⎜ x2 x 1 ⎟⎟=(x −x )(x −x )(x −x )
⎝⎜⎜⎜⎜⎜ x322 x23 1 ⎠⎟⎟⎟⎟⎟⎟ 1 2 1 3 2 3
Solution:
7
⎛ ⎞
⎜ x2 x 1 ⎟
⎜⎜ 1 1 ⎟⎟⎟
det⎜⎜ x2 x 1 ⎟⎟=x2(x −x )−x (x2−x2)+(x2x −x2x )
⎜⎜ 2 2 ⎟⎟ 1 2 3 1 2 3 2 3 3 2
⎝⎜⎜⎜ x32 x3 1 ⎠⎟⎟⎟⎟
=x2(x −x )−x (x −x )(x +x )+x x (x −x )
1 2 3 1 2 3 2 3 2 3 2 3
=(x −x )⎡x2−x (x +x )+x x ⎤
2 3 ⎣⎢ 1 1 2 3 2 3⎦⎥
=(x −x )⎡x (x −x )−x (x −x )⎤
2 3 ⎣⎢ 1 1 2 3 1 2 ⎦⎥
=(x −x )(x −x )(x −x )
1 2 1 3 2 3
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)
−1
⎛ a a ⎞ 1 ⎛ a −a ⎞
⎜ 11 12 ⎟ = ⎜ 22 12 ⎟
⎜ a a ⎟ a a −a a ⎜ −a a ⎟
⎝ 21 22 ⎠ 11 22 12 21⎝ 21 11 ⎠
−1
⎛ 17 7 ⎞ 1 ⎛ 9 −7 ⎞
=
⎜ ⎟ ⎜ ⎟
⎝ 19 9 ⎠ 17×9−19×7⎝ −19 17 ⎠
⎛ 9/20 −7/20 ⎞
=
⎜ ⎟
⎝ −19/20 17/20 ⎠
Since, from Eq. (1.13)
⎛ a a ⎞⎛ b b ⎞ ⎛ a b +a b a b +a b ⎞
⎜ 11 12 ⎟⎜ 11 12 ⎟ =⎜ 11 11 12 21 11 12 12 22 ⎟
a a b b a b +a b a b +a b
⎝ 21 22 ⎠⎝ 21 22 ⎠ ⎝ 21 11 22 21 21 12 22 22 ⎠
then
8
⎛ 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 ⎞ ⎛ 2 1 3 ⎞
⎜ ⎟ ⎜ ⎟
A= 2 1 2 and B= 1 2 5
⎜ ⎟ ⎜ ⎟
⎜ 4 2 3 ⎟ ⎜ 0 2 1 ⎟
⎝ ⎠ ⎝ ⎠
Show that (AB)T = BTAT.
Solution:
T T
⎛⎛ 3 1 4 ⎞⎛ 2 1 3 ⎞⎞ ⎛ 7 13 18 ⎞
(AB)T =⎜⎜ 2 1 2 ⎟⎜ 1 2 5 ⎟⎟ =⎜ 5 8 13 ⎟
⎜⎜ ⎟⎜ ⎟⎟ ⎜ ⎟
⎜⎜ 4 2 3 ⎟⎜ 0 2 1 ⎟⎟ ⎜ 10 14 25 ⎟
⎝⎝ ⎠⎝ ⎠⎠ ⎝ ⎠
⎛ 7 5 10 ⎞
⎜ ⎟
= 13 8 14
⎜ ⎟
⎜ 18 13 25 ⎟
⎝ ⎠
and
T T
⎛ 2 1 3 ⎞ ⎛ 3 1 4 ⎞ ⎛ 2 1 0 ⎞⎛ 3 2 4 ⎞
BTAT =⎜ 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
9
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?
100y +420y +486y =17
1 2 3
400y +1050y +972y =18
1 2 3
700y +1680y +1458y =−3
1 2 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.
a +2a +3a =1
1 2 3
4a +5a +6a =0
1 2 3
7a +8a +9a =−7
1 2 3
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 2ùìx ü é1 0ùìx ü ì0ü
ê úí 1ý-=lê úí 1ý í ý
5 1 x 0 1 x 0
ë ûî þ ë ûî þ î þ
2 2
When l = 4, what are the values of x and x ? Justify your answer.
1 2
Solution: When
10
