斯特朗线性代数第五版习题答案详解

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"《线性代数入门》第五版是Gilbert Strang所著的一本经典教材，本书深入浅出地介绍了线性代数的基本概念和理论，旨在帮助读者掌握这一核心数学领域的基础知识。该书不仅包含理论阐述，还配以了大量的习题和解答，以便于学生通过实践巩固理论理解。在第五版的习题答案部分，我们看到了以下知识点： 1. 第一章的习题1.1涉及向量空间的表示：题目要求分析不同组合如何对应到三维空间中的线、平面以及整个空间。这有助于学生理解向量如何构成几何图形，如线段（直线）、平面区域以及整个空间。 2. 习题涉及到平行四边形的对角线和边的关系，通过给定的向量v和w，求解它们的线性组合表示的图形特征。这涉及向量加法与减法在几何图形中的应用。 3. 另一个习题要求根据已知的对角线v+w和v-w，找出平行四边形的两个边。这展示了如何逆向思考问题，即已知几何结构如何导出向量关系。 4. 习题探讨了线性方程组的解，例如3v+w和cv+dw的表达式，展示了线性变换如何影响向量的组合结果，并且强调了系数c和d对于解的存在性的影响。 5. 对于一组向量u、v和w，通过加法和线性组合，习题展示了向量共面的概念。当线性组合给出零向量时，意味着这些向量在同一平面上。 6. 最后，习题强调了向量加法的性质，即向量的分量和等于零的必要条件，这对于理解线性独立性和零向量的重要性具有重要意义。通过实例说明，只有当向量v和w的分量和相等时，才能找到满足特定线性关系的系数c和d。这些习题解答不仅涵盖了线性代数的基础运算，还包括了向量空间、线性变换、几何意义以及线性方程组的解决策略，对学生理解和应用线性代数有着直接的帮助。通过深入学习这些问题，读者可以提升对线性代数的理解，为后续的数学学习和实际应用打下坚实基础。"

Solutions to Exercises

5 6x + 4y is 2 times 3x + 2y. There is no solution unless the right side is 2 · 10 = 20.

Then all the points on the line 3x + 2y = 10 are solutions, including (0, 5) and (4, −1).

The t wo lines in the row picture are th e same line, contain ing all solutions.

6 Singular system i f b = 4, because 4x + 8y is 2 times 2x + 4y. Then g = 32 makes the

lines 2x + 4y = 16 an d 4x + 8y = 32 beco me the same: inﬁnitely many solutions like

(8, 0) an d (0, 4).

7 If a = 2 el imination must fail (two parallel lines in the row picture). The equations

have no solution. With a = 0, elimination will stop for a row exchange. Then 3y = −3

gives y = −1 and 4x + 6y = 6 gives x = 3.

8 If k = 3 elimination must fail: n o solution. If k = −3, elimination gives 0 = 0 in

equation 2: inﬁnitely many solutions. If k = 0 a row exchang e is needed: one solution.

9 On the left side, 6x −4y is 2 times (3x −2y). Therefore we need b

= 2b

on the right

side. Then there will be inﬁnitely many solutions (two parallel l ines become o ne single

line in the row picture). The column picture has both columns along the same line.

10 The equation y = 1 comes from elimination (subtract x + y = 5 from x + 2y = 6).

Then x = 4 and 5x − 4y = 20 − 4 = c = 1 6.

11 (a) Another solution is

(x + X, y + Y, z + Z). (b) If 25 pl anes meet at two points,

they meet along the whole line through t hose two points.

12 Elimination leads to this upper triangular system; then comes back substitution.

2x + 3y + z = 8

y + 3z = 4

8z = 8

gives

x = 2

y = 1 If a zero is at the start of row 2 or row 3,

z = 1 that avoids a row operation.

2x − 3y = 3

4x − 5y + z = 7

2x − y − 3z = 5

gives

2x − 3y = 3

y + z = 1

2y + 3z = 2

and

2x − 3y = 3

y + z = 1

− 5z = 0

and

x = 3

y = 1

z = 0

Here are steps 1, 2, 3 : Subtract 2 × row 1 from row 2, subtract 1 × row 1 from row 3,

subtract 2 × row 2 from row 3

Solutions to Exercises

14 Subtract 2 times row 1 from row 2 t o reach (d−10)y−z = 2. Equation (3) is y−z = 3.

If d = 10 exchange rows 2 and 3. If d = 11 the system becomes singular.

15 The second pivot position will contain −2 − b. If b = −2 we exchange with row 3.

If b = −1 (singular case) the second equation is −y − z = 0. But equati on (3) is the

same so there i s a line of solution s (x, y, z) = (1, 1, −1).

16 (a)

Example of

2 exchanges

0x + 0y + 2z = 4

x + 2y + 2z = 5

0x + 3y + 4z = 6

(exchange 1 and 2, then 2 and 3)

(b)

Exchange

but then

breakdown

0x + 3y + 4z = 4

x + 2y + 2z = 5

0x + 3y + 4z = 6

(rows 1 and 3 are not consistent)

17 If row 1 = row 2, then row 2 is zero after the ﬁrst step; exchange the zero row with row

3 and row 3 has no pivot. If column 2 = column 1, then column 2 has no pivot.

18 Example x + 2y + 3z = 0, 4x + 8y + 12z = 0, 5x + 10y + 15z = 0 has 9 different

coefﬁcien ts but rows 2 and 3 become 0 = 0: inﬁnitely many solutions to Ax = 0 but

almost surely no solution to Ax = b for a random b.

19 Row 2 becomes 3y − 4z = 5, then row 3 becomes (q + 4)z = t − 5. If q = −4 the

system is singular—no thi rd pivot. Then if t = 5 the third equat ion is 0 = 0 which

allows inﬁnitely many solutions. Choosin g z = 1 the equ ation 3y −4z = 5 gives y = 3

and equation 1 gives x = −9 .

20 Singular if row 3 is a combinati on of rows 1 and 2. From the end view, the three planes

form a triangle. This happens if rows 1+2 =row 3 on t he left side but not the right side:

x+y +z = 0, x−2y −z = 1, 2x−y = 4 . No parallel pl anes but still no solut ion. The

three p lanes i n th e row picture form a triangular tunnel.

21 (a) Pivots 2,

in the equations 2x + y = 0,

y + z = 0,

z + t = 0,

t = 5

after elimination. Back substitution gives t = 4, z = −3, y = 2, x = −1. (b) If

the off-diagonal entries change from +1 to −1, the pivots are the same. The solution is

(1, 2, 3, 4) instead of (−1, 2, −3, 4).

22 The ﬁfth pivot is

for both matrices (1’s or −1’s off the diagonal). The nth pivot i s

n+1

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