Chapter 1
Systems of Linear Equations and Matrices
Section 1.1
Exercise Set 1.1
1. (a), (c), and (f) are linear equations in x1 , x2 , and x3 .
(b) is not linear because of the term x1 x3 .
(d) is not linear because of the term x −2 .
(e) is not linear because of the term x13/ 5 .
3. (a) and (d) are linear systems.
(b) is not a linear system because the first and second equations are not linear.
(c) is not a linear system because the second equation is not linear.
5. By inspection, (a) and (d) are both consistent; x1 = 3, x2 = 2, x3 = −2, x4 = 1 is a solution of (a) and
x1 = 1, x2 = 3, x3 = 2, x4 = 2 is a solution of (d). Note that both systems have infinitely many solutions.
7. (a), (d), and (e) are solutions.
(b) and (c) do not satisfy any of the equations.
9. (a) 7 x − 5 y = 3
5 3
x= y+
7 7
Let y = t. The solution is
5 3
x = y+
7 7
y=t
(b) −8 x1 + 2 x2 − 5 x3 + 6 x4 = 1
1 5 3 1
x1 = x2 − x3 + x4 −
4 8 4 8
Let x2 = r , x3 = s, and x4 = t . The solution is
1 5 3 1
x1 = r− s+ t−
4 8 4 8
x2 =r
x3 =s
x4 =t
⎡2 0 0⎤
11. (a) ⎢ 3 −4 0 ⎥ corresponds to
⎢ ⎥
⎢⎣ 0 1 1⎥⎦
2 x1 =0
3x1 − 4 x2 = 0.
x2 = 1
1
,Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
⎡ 3 0 −2 5⎤ 15. If (a, b, c) is a solution of the system, then
(b) ⎢7 1 4 −3⎥⎥ corresponds to ax12 + bx1 + c = y1 , ax22 + bx2 + c = y2 , and
⎢
⎢⎣ 0 −2 1 7 ⎥⎦ ax32 + bx3 + c = y3 which simply means that the
3x1 − 2 x3 = 5 points are on the curve.
7 x1 + x2 + 4 x3 = −3.
−2 x2 + x3 = 7 17. The solutions of x1 + kx2 = c are x1 = c − kt ,
x2 = t where t is any real number. If these
⎡ ⎤ satisfy x1 + lx2 = d , then c − kt + lt = d or
(c) ⎢7 2 1 −3 5⎥ corresponds to
⎣ 1 2 4 0 1⎦ c − d = (k − l) t for all real numbers t. In
7 x1 + 2 x2 + x3 − 3x4 = 5 particular, if t = 0, then c = d, and if t = 1, then
. k = l.
x1 + 2 x2 + 4 x3 =1
True/False 1.1
⎡1 0 0 0 7⎤
⎢ 0 −2 ⎥ corresponds to (a) True; x1 = x2 = = xn = 0 will be a solution.
(d) ⎢0 1 0
⎥
⎢0 0 1 0 3⎥
⎢⎣0 0 0 1 4 ⎥⎦ (b) False; only multiplication by nonzero constants
is acceptable.
x1 = 7
x2 = −2 . (c) True; if k = 6 the system has infinitely many
x3 = 3 solutions, while if k ≠ 6, the system has no
x4 = 4 solution.
(d) True; the equation can be solved for one variable
13. (a) The augmented matrix for −2 x1 = 6
in terms of the other(s), yielding parametric
3x1 = 8 equations that give infinitely many solutions.
9 x1 = −3
(e) False; the system 3x − 5 y = −7 has the
⎡ −2 6 ⎤
2 x + 9 y = 20
is ⎢ 3 8⎥ .
⎢ ⎥ 6 x − 10 y = −14
⎢⎣ 9 −3⎥⎦
solution x = 1, y = 2.
(b) The augmented matrix for (f) False; multiplying an equation by a nonzero
⎡ ⎤ constant c does not change the solutions of the
6 x1 − x2 + 3x3 = 4 is ⎢6 −1 3 4 ⎥ . system.
5x − x = 1 ⎣0 5 −1 1⎦
2 3
(g) True; subtracting one equation from another is
(c) The augmented matrix for the same as multiplying an equation by −1 and
2 x2 − 3x4 + x5 = 0 adding it to another.
−3x1 − x2 + x3 = −1
(h) False; the second row corresponds to the
6 x1 + 2 x2 − x3 + 2 x4 − 3x5 = 6
equation 0 x1 + 0 x2 = −1 or 0 = −1 which is false.
⎡ 0 2 0 −3 1 0⎤
is ⎢ −3 −1 1 0 0 −1⎥ . Section 1.2
⎢ ⎥
⎢⎣ 6 2 −1 2 −3 6 ⎥⎦
Exercise Set 1.2
(d) The augmented matrix for x1 − x5 = 7 is 1. (a) The matrix is in both row echelon and
[1 0 0 0 −1 7]. reduced row echelon form.
(b) The matrix is in both row echelon and
reduced row echelon form.
2
,SSM: Elementary Linear Algebra Section 1.2
(c) The matrix is in both row echelon and (d) The last line of the matrix corresponds to the
reduced row echelon form. equation 0 x1 + 0 x2 + 0 x3 = 1, which is not
satisfied by any values of x1 , x2 , and x3 ,
(d) The matrix is in both row echelon and
reduced row echelon form. so the system is inconsistent.
(e) The matrix is in both row echelon and ⎡ 1 1 2 8⎤
reduced row echelon form. 5. The augmented matrix is ⎢⎢ −1 −2 3 1⎥⎥ .
(f) The matrix is in both row echelon and ⎢⎣ 3 −7 4 10 ⎥⎦
reduced row echelon form. Add the first row to the second row and add −3
times the first row to the third row.
(g) The matrix is in row echelon form.
⎡1 1 2 8⎤
⎢0 −1 5 9⎥
3. (a) The matrix corresponds to the system ⎢ ⎥
x1 − 3x2 + 4 x3 = 7 x1 = 7 + 3x2 − 4 x3 ⎢⎣0 −10 −2 −14⎥⎦
x2 + 2 x3 = 2 or x2 = 2 − 2 x3 . Multiply the second row by −1.
x3 = 5 x3 = 5 ⎡1 1 2 8⎤
Thus x3 = 5, x2 = 2 − 2(5) = −8, and ⎢0 1 −5 −9⎥⎥
⎢
x1 = 7 + 3(−8) − 4(5) = −37. The solution is ⎢⎣0 −10 −2 −14⎥⎦
x1 = −37, x2 = −8, x3 = 5. Add 10 times the second row to the third row.
⎡1 1 2 8⎤
(b) The matrix corresponds to the system ⎢ 0 1 −5 −9 ⎥
x1 + 8 x3 − 5 x4 = 6 ⎢ ⎥
x2 + 4 x3 − 9 x4 = 3 or ⎣⎢0 0 −52 −104 ⎦⎥
x3 + x4 = 2 1
Multiply the third row by − .
x1 = 6 − 8 x3 + 5 x4 52
x2 = 3 − 4 x3 + 9 x4 . ⎡1 1 2 8⎤
x3 = 2 − x4 ⎢0 1 −5 −9 ⎥
⎢ ⎥
Let x4 = t , then x3 = 2 − t , ⎢⎣0 0 1 2 ⎥⎦
x2 = 3 − 4(2 − t ) + 9t = 13t − 5, and Add 5 times the third row to the second row and
x1 = 6 − 8(2 − t ) + 5t = 13t − 10. The solution −2 times the third row to the first row.
is x1 = 13t − 10, x2 = 13t − 5, x3 = −t + 2, ⎡ 1 1 0 4⎤
⎢0 1 0 1⎥
x4 = t . ⎢ ⎥
⎢⎣0 0 1 2⎥⎦
(c) The matrix corresponds to the system Add −1 times the second row to the first row.
x1 + 7 x2 − 2 x3 − 8 x5 = −3 ⎡ 1 0 0 3⎤
x3 + x4 + 6 x5 = 5 or ⎢0 1 0 1⎥
x4 + 3x5 = 9 ⎢ ⎥
⎢⎣0 0 1 2⎥⎦
x1 = −3 − 7 x2 + 2 x3 + 8 x5
The solution is x1 = 3, x2 = 1, x3 = 2.
x3 = 5 − x4 − 6 x5
x4 = 9 − 3 x5
7. The augmented matrix is
Let x2 = s and x5 = t , then x4 = 9 − 3t , ⎡ 1 −1 2 −1 −1⎤
x3 = 5 − (9 − 3t ) − 6t = −3t − 4, and ⎢ 2 1 −2 −2 −2 ⎥
⎢ ⎥.
x1 = −3 − 7 s + 2(−3t − 4) + 8t = −7 s + 2t − 11. ⎢ −1 2 −4 1 1⎥
The solution is x1 = −7 s + 2t − 11, x2 = s, ⎣⎢ 3 0 0 −3 −3⎦⎥
x3 = −3t − 4, x4 = −3t + 9, x5 = t . Add −2 times the first row to the second row, 1
times the first row to the third row, and −3 times
the first row to the fourth row.
3
, Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
⎡ 1 −1 2 −1 −1⎤ 13. Since the system has more unknowns (4) than
⎢0 3 −6 0 0 ⎥ equations (3), it has nontrivial solutions.
⎢ ⎥
⎢0 1 −2 0 0 ⎥ 15. Since the system has more unknowns (3) than
⎢⎣0 3 −6 0 0 ⎥⎦ equations (2), it has nontrivial solutions.
1
Multiply the second row by and then add −1
3 ⎡2 1 3 0⎤
times the new second row to the third row and 17. The augmented matrix is ⎢⎢ 1 2 0 0 ⎥⎥ .
add −3 times the new second row to the fourth ⎢⎣ 0 1 1 0 ⎥⎦
row.
Interchange the first and second rows, then add
⎡ 1 −1 2 −1 −1⎤ −2 times the new first row to the second row.
⎢0 1 −2 0 0 ⎥
⎢ ⎥ ⎡ 1 2 0 0⎤
⎢0 0 0 0 0 ⎥ ⎢0 −3 3 0 ⎥
⎣⎢0 0 0 0 0 ⎦⎥ ⎢ ⎥
⎢⎣0 1 1 0 ⎥⎦
Add the second row to the first row.
⎡ 1 0 0 −1 −1⎤ 1
Multiply the second row by − , then add −1
⎢ 0 1 −2 0 0 ⎥ 3
⎢ ⎥ times the new second row to the third row.
⎢0 0 0 0 0⎥
⎢⎣0 0 0 0 0⎥⎦ ⎡ 1 2 0 0⎤
The corresponding system of equations is ⎢0 1 −1 0 ⎥
⎢ ⎥
x − w = −1 x = w −1
or . ⎣⎢0 0 2 0 ⎦⎥
y − 2z =0 y = 2z
1
Let z = s and w = t. The solution is x = t − 1, Multiply the third row by , then add the new
y = 2s, z = s, w = t. 2
third row to the second row.
9. In Exercise 5, the following row echelon matrix ⎡ 1 2 0 0⎤
occurred. ⎢0 1 0 0⎥
⎢ ⎥
⎡1 1 2 8⎤ ⎢⎣0 0 1 0⎥⎦
⎢0 1 −5 −9 ⎥
⎢ ⎥ Add −2 times the second row to the first row.
⎢⎣0 0 1 2 ⎥⎦
⎡ 1 0 0 0⎤
The corresponding system of equations is ⎢0 1 0 0⎥
x1 + x2 + 2 x3 = 8 x1 = − x2 − 2 x3 + 8 ⎢ ⎥
x2 − 5 x3 = −9 or x2 = 5 x3 − 9 . ⎣⎢0 0 1 0⎦⎥
x3 = 2 x3 = 2 The solution, which can be read from the matrix,
is x1 = 0, x2 = 0, x3 = 0.
Since x3 = 2, x2 = 5(2) − 9 = 1, and
x1 = −1 − 2(2) + 8 = 3. The solution is x1 = 3, ⎡ ⎤
19. The augmented matrix is ⎢3 1 1 1 0⎥ .
x2 = 1, x3 = 2. ⎣ 5 −1 1 −1 0 ⎦
1
11. From Exercise 7, one row echelon form of the Multiply the first row by , then add −5 times
3
⎡ 1 −1 2 −1 −1⎤ the new first row to the second row.
⎢ ⎥
augmented matrix is ⎢0 1 −2 0 0 ⎥ . ⎡1 1 1 1 0⎤
⎢0 0 0 0 0 ⎥ ⎢ 3 3 3 ⎥
⎢⎣0 0 0 0 0 ⎥⎦ ⎢⎣0 − 83 − 23 − 83 0 ⎥⎦
The corresponding system is 3
x − y + 2 z − w = −1 x = y − 2z + w −1 Multiply the second row by − .
or . 8
y − 2z =0 y = 2z
⎡1 1 1 1 0⎤
Let z = s and w = t. Then y = 2s and ⎢ 3 3 3 ⎥
x = 2s − 2s + t − 1 = t − 1. The solution is ⎢⎣0 1 1 1 0 ⎥⎦
4
x = t − 1, y = 2s, z = s, w = t.
4
Systems of Linear Equations and Matrices
Section 1.1
Exercise Set 1.1
1. (a), (c), and (f) are linear equations in x1 , x2 , and x3 .
(b) is not linear because of the term x1 x3 .
(d) is not linear because of the term x −2 .
(e) is not linear because of the term x13/ 5 .
3. (a) and (d) are linear systems.
(b) is not a linear system because the first and second equations are not linear.
(c) is not a linear system because the second equation is not linear.
5. By inspection, (a) and (d) are both consistent; x1 = 3, x2 = 2, x3 = −2, x4 = 1 is a solution of (a) and
x1 = 1, x2 = 3, x3 = 2, x4 = 2 is a solution of (d). Note that both systems have infinitely many solutions.
7. (a), (d), and (e) are solutions.
(b) and (c) do not satisfy any of the equations.
9. (a) 7 x − 5 y = 3
5 3
x= y+
7 7
Let y = t. The solution is
5 3
x = y+
7 7
y=t
(b) −8 x1 + 2 x2 − 5 x3 + 6 x4 = 1
1 5 3 1
x1 = x2 − x3 + x4 −
4 8 4 8
Let x2 = r , x3 = s, and x4 = t . The solution is
1 5 3 1
x1 = r− s+ t−
4 8 4 8
x2 =r
x3 =s
x4 =t
⎡2 0 0⎤
11. (a) ⎢ 3 −4 0 ⎥ corresponds to
⎢ ⎥
⎢⎣ 0 1 1⎥⎦
2 x1 =0
3x1 − 4 x2 = 0.
x2 = 1
1
,Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
⎡ 3 0 −2 5⎤ 15. If (a, b, c) is a solution of the system, then
(b) ⎢7 1 4 −3⎥⎥ corresponds to ax12 + bx1 + c = y1 , ax22 + bx2 + c = y2 , and
⎢
⎢⎣ 0 −2 1 7 ⎥⎦ ax32 + bx3 + c = y3 which simply means that the
3x1 − 2 x3 = 5 points are on the curve.
7 x1 + x2 + 4 x3 = −3.
−2 x2 + x3 = 7 17. The solutions of x1 + kx2 = c are x1 = c − kt ,
x2 = t where t is any real number. If these
⎡ ⎤ satisfy x1 + lx2 = d , then c − kt + lt = d or
(c) ⎢7 2 1 −3 5⎥ corresponds to
⎣ 1 2 4 0 1⎦ c − d = (k − l) t for all real numbers t. In
7 x1 + 2 x2 + x3 − 3x4 = 5 particular, if t = 0, then c = d, and if t = 1, then
. k = l.
x1 + 2 x2 + 4 x3 =1
True/False 1.1
⎡1 0 0 0 7⎤
⎢ 0 −2 ⎥ corresponds to (a) True; x1 = x2 = = xn = 0 will be a solution.
(d) ⎢0 1 0
⎥
⎢0 0 1 0 3⎥
⎢⎣0 0 0 1 4 ⎥⎦ (b) False; only multiplication by nonzero constants
is acceptable.
x1 = 7
x2 = −2 . (c) True; if k = 6 the system has infinitely many
x3 = 3 solutions, while if k ≠ 6, the system has no
x4 = 4 solution.
(d) True; the equation can be solved for one variable
13. (a) The augmented matrix for −2 x1 = 6
in terms of the other(s), yielding parametric
3x1 = 8 equations that give infinitely many solutions.
9 x1 = −3
(e) False; the system 3x − 5 y = −7 has the
⎡ −2 6 ⎤
2 x + 9 y = 20
is ⎢ 3 8⎥ .
⎢ ⎥ 6 x − 10 y = −14
⎢⎣ 9 −3⎥⎦
solution x = 1, y = 2.
(b) The augmented matrix for (f) False; multiplying an equation by a nonzero
⎡ ⎤ constant c does not change the solutions of the
6 x1 − x2 + 3x3 = 4 is ⎢6 −1 3 4 ⎥ . system.
5x − x = 1 ⎣0 5 −1 1⎦
2 3
(g) True; subtracting one equation from another is
(c) The augmented matrix for the same as multiplying an equation by −1 and
2 x2 − 3x4 + x5 = 0 adding it to another.
−3x1 − x2 + x3 = −1
(h) False; the second row corresponds to the
6 x1 + 2 x2 − x3 + 2 x4 − 3x5 = 6
equation 0 x1 + 0 x2 = −1 or 0 = −1 which is false.
⎡ 0 2 0 −3 1 0⎤
is ⎢ −3 −1 1 0 0 −1⎥ . Section 1.2
⎢ ⎥
⎢⎣ 6 2 −1 2 −3 6 ⎥⎦
Exercise Set 1.2
(d) The augmented matrix for x1 − x5 = 7 is 1. (a) The matrix is in both row echelon and
[1 0 0 0 −1 7]. reduced row echelon form.
(b) The matrix is in both row echelon and
reduced row echelon form.
2
,SSM: Elementary Linear Algebra Section 1.2
(c) The matrix is in both row echelon and (d) The last line of the matrix corresponds to the
reduced row echelon form. equation 0 x1 + 0 x2 + 0 x3 = 1, which is not
satisfied by any values of x1 , x2 , and x3 ,
(d) The matrix is in both row echelon and
reduced row echelon form. so the system is inconsistent.
(e) The matrix is in both row echelon and ⎡ 1 1 2 8⎤
reduced row echelon form. 5. The augmented matrix is ⎢⎢ −1 −2 3 1⎥⎥ .
(f) The matrix is in both row echelon and ⎢⎣ 3 −7 4 10 ⎥⎦
reduced row echelon form. Add the first row to the second row and add −3
times the first row to the third row.
(g) The matrix is in row echelon form.
⎡1 1 2 8⎤
⎢0 −1 5 9⎥
3. (a) The matrix corresponds to the system ⎢ ⎥
x1 − 3x2 + 4 x3 = 7 x1 = 7 + 3x2 − 4 x3 ⎢⎣0 −10 −2 −14⎥⎦
x2 + 2 x3 = 2 or x2 = 2 − 2 x3 . Multiply the second row by −1.
x3 = 5 x3 = 5 ⎡1 1 2 8⎤
Thus x3 = 5, x2 = 2 − 2(5) = −8, and ⎢0 1 −5 −9⎥⎥
⎢
x1 = 7 + 3(−8) − 4(5) = −37. The solution is ⎢⎣0 −10 −2 −14⎥⎦
x1 = −37, x2 = −8, x3 = 5. Add 10 times the second row to the third row.
⎡1 1 2 8⎤
(b) The matrix corresponds to the system ⎢ 0 1 −5 −9 ⎥
x1 + 8 x3 − 5 x4 = 6 ⎢ ⎥
x2 + 4 x3 − 9 x4 = 3 or ⎣⎢0 0 −52 −104 ⎦⎥
x3 + x4 = 2 1
Multiply the third row by − .
x1 = 6 − 8 x3 + 5 x4 52
x2 = 3 − 4 x3 + 9 x4 . ⎡1 1 2 8⎤
x3 = 2 − x4 ⎢0 1 −5 −9 ⎥
⎢ ⎥
Let x4 = t , then x3 = 2 − t , ⎢⎣0 0 1 2 ⎥⎦
x2 = 3 − 4(2 − t ) + 9t = 13t − 5, and Add 5 times the third row to the second row and
x1 = 6 − 8(2 − t ) + 5t = 13t − 10. The solution −2 times the third row to the first row.
is x1 = 13t − 10, x2 = 13t − 5, x3 = −t + 2, ⎡ 1 1 0 4⎤
⎢0 1 0 1⎥
x4 = t . ⎢ ⎥
⎢⎣0 0 1 2⎥⎦
(c) The matrix corresponds to the system Add −1 times the second row to the first row.
x1 + 7 x2 − 2 x3 − 8 x5 = −3 ⎡ 1 0 0 3⎤
x3 + x4 + 6 x5 = 5 or ⎢0 1 0 1⎥
x4 + 3x5 = 9 ⎢ ⎥
⎢⎣0 0 1 2⎥⎦
x1 = −3 − 7 x2 + 2 x3 + 8 x5
The solution is x1 = 3, x2 = 1, x3 = 2.
x3 = 5 − x4 − 6 x5
x4 = 9 − 3 x5
7. The augmented matrix is
Let x2 = s and x5 = t , then x4 = 9 − 3t , ⎡ 1 −1 2 −1 −1⎤
x3 = 5 − (9 − 3t ) − 6t = −3t − 4, and ⎢ 2 1 −2 −2 −2 ⎥
⎢ ⎥.
x1 = −3 − 7 s + 2(−3t − 4) + 8t = −7 s + 2t − 11. ⎢ −1 2 −4 1 1⎥
The solution is x1 = −7 s + 2t − 11, x2 = s, ⎣⎢ 3 0 0 −3 −3⎦⎥
x3 = −3t − 4, x4 = −3t + 9, x5 = t . Add −2 times the first row to the second row, 1
times the first row to the third row, and −3 times
the first row to the fourth row.
3
, Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
⎡ 1 −1 2 −1 −1⎤ 13. Since the system has more unknowns (4) than
⎢0 3 −6 0 0 ⎥ equations (3), it has nontrivial solutions.
⎢ ⎥
⎢0 1 −2 0 0 ⎥ 15. Since the system has more unknowns (3) than
⎢⎣0 3 −6 0 0 ⎥⎦ equations (2), it has nontrivial solutions.
1
Multiply the second row by and then add −1
3 ⎡2 1 3 0⎤
times the new second row to the third row and 17. The augmented matrix is ⎢⎢ 1 2 0 0 ⎥⎥ .
add −3 times the new second row to the fourth ⎢⎣ 0 1 1 0 ⎥⎦
row.
Interchange the first and second rows, then add
⎡ 1 −1 2 −1 −1⎤ −2 times the new first row to the second row.
⎢0 1 −2 0 0 ⎥
⎢ ⎥ ⎡ 1 2 0 0⎤
⎢0 0 0 0 0 ⎥ ⎢0 −3 3 0 ⎥
⎣⎢0 0 0 0 0 ⎦⎥ ⎢ ⎥
⎢⎣0 1 1 0 ⎥⎦
Add the second row to the first row.
⎡ 1 0 0 −1 −1⎤ 1
Multiply the second row by − , then add −1
⎢ 0 1 −2 0 0 ⎥ 3
⎢ ⎥ times the new second row to the third row.
⎢0 0 0 0 0⎥
⎢⎣0 0 0 0 0⎥⎦ ⎡ 1 2 0 0⎤
The corresponding system of equations is ⎢0 1 −1 0 ⎥
⎢ ⎥
x − w = −1 x = w −1
or . ⎣⎢0 0 2 0 ⎦⎥
y − 2z =0 y = 2z
1
Let z = s and w = t. The solution is x = t − 1, Multiply the third row by , then add the new
y = 2s, z = s, w = t. 2
third row to the second row.
9. In Exercise 5, the following row echelon matrix ⎡ 1 2 0 0⎤
occurred. ⎢0 1 0 0⎥
⎢ ⎥
⎡1 1 2 8⎤ ⎢⎣0 0 1 0⎥⎦
⎢0 1 −5 −9 ⎥
⎢ ⎥ Add −2 times the second row to the first row.
⎢⎣0 0 1 2 ⎥⎦
⎡ 1 0 0 0⎤
The corresponding system of equations is ⎢0 1 0 0⎥
x1 + x2 + 2 x3 = 8 x1 = − x2 − 2 x3 + 8 ⎢ ⎥
x2 − 5 x3 = −9 or x2 = 5 x3 − 9 . ⎣⎢0 0 1 0⎦⎥
x3 = 2 x3 = 2 The solution, which can be read from the matrix,
is x1 = 0, x2 = 0, x3 = 0.
Since x3 = 2, x2 = 5(2) − 9 = 1, and
x1 = −1 − 2(2) + 8 = 3. The solution is x1 = 3, ⎡ ⎤
19. The augmented matrix is ⎢3 1 1 1 0⎥ .
x2 = 1, x3 = 2. ⎣ 5 −1 1 −1 0 ⎦
1
11. From Exercise 7, one row echelon form of the Multiply the first row by , then add −5 times
3
⎡ 1 −1 2 −1 −1⎤ the new first row to the second row.
⎢ ⎥
augmented matrix is ⎢0 1 −2 0 0 ⎥ . ⎡1 1 1 1 0⎤
⎢0 0 0 0 0 ⎥ ⎢ 3 3 3 ⎥
⎢⎣0 0 0 0 0 ⎥⎦ ⎢⎣0 − 83 − 23 − 83 0 ⎥⎦
The corresponding system is 3
x − y + 2 z − w = −1 x = y − 2z + w −1 Multiply the second row by − .
or . 8
y − 2z =0 y = 2z
⎡1 1 1 1 0⎤
Let z = s and w = t. Then y = 2s and ⎢ 3 3 3 ⎥
x = 2s − 2s + t − 1 = t − 1. The solution is ⎢⎣0 1 1 1 0 ⎥⎦
4
x = t − 1, y = 2s, z = s, w = t.
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