The document discusses reduced row-echelon matrices and provides examples of elementary row operations used to transform matrices into reduced row-echelon form. It gives the initial matrix, shows the elementary row operations used, and provides the resulting reduced row-echelon matrix. Several examples are provided with the goal of getting a 1 in the upper left corner and clearing out the rest of the first column through row operations.
5. 1 0 2 1 3
R 3−2 R1
1 0 2 1 3
R 2− R1
→ 0 3 −6 −9 3 → 0 3 −6 −9 3
2 7 −10 −19 13
0 7 −14 −21 7
(1/3) R 2
1 0 2 1 3
R 3−7 R 2
1 0 2 1 3
→ 0 1 −2 −3 1 → 0 1 −2 − 3 1
0 7 −14 −21 7
0 0 0 0 0
3 6 1 7 13 1 2 −4 −2 −13
R1− R 3
20. 5 10 8 18 47 → 5 10 8 18 47
2 4 5 9 26
2 4 5 9 26
1 2 −4 −2 −13 1 2 −4 −2 −13
R 2−5 R1 R 3− 2 R1
→ 0 0 28 28 112 → 0 0 28 28 112
2 4 5 9 26
0 0 13 13 52
(1/ 28) R 2
1 2 −4 −2 −13 R 3−13 R 2
1 2 0 2 3
→ 0 0 1 1 4 → → 0 0 1 1 4
0 0 13 13 52
0 0 0 0 0
In each of Problems 21-30, we give just the first two or three steps in the reduction. Then we
display the resulting reduced echelon form E of the augmented coefficient matrix A of the
given linear system, and finally list the resulting solution (if any).
21. Begin by interchanging rows 1 and 2 of A. Then subtract twice row 1 both from row 2
and from row 3.
1 0 0 3
E = 0 1 0 −2 ;
x1 = 3, x2 = −2, x3 = 4
0 0 1 4
22. Begin by subtracting row 2 of A from row 1. Then subtract twice row 1 both from row
2 and from row 3.
1 0 0 5
E = 0 1 0 −3 ;
x1 = 5, x2 = −3, x3 = 2
0 0 1 2
6. 23. Begin by subtracting twice row 1 of A both from row 2 and from row 3. Then add row
2 to row 3.
1 0 −3 14
E = 0 1 2 3 ;
x1 = 4 + 3t , x2 = 3 − 2t , x3 = t
0 0 0 0
24. Begin by interchanging rows 1 and 3 of A. Then subtract twice row 1 from row 2, and
three times row 1 from row 3.
1 −2 0 5
E = 0 0 1 7 ;
x1 = 5 + 2t , x2 = t , x3 = 7
0 0 0 0
25. Begin by interchanging rows 1 and 2 of A. Then subtract three times row 1 from row 2,
and five times row 1 from row 3.
1 0 −2 0
E = 0 1 3 0 .
The system has no solution.
0 0 0 1
26. Begin by subtracting row 1 from row 2 of A. Then interchange rows 1 and 2. Next
subtract twice row 1 from row 2, and five times row 1 from row 3.
1 0 1 0
E = 0 1 2 0 .
The system has no solution.
0 0 0 1
27. x1 − 4 x2 − 3 x3 − 3 x4 = 4
2 x1 − 6 x2 − 5 x3 − 5 x4 = 5
3x1 − x2 − 4 x3 − 5 x4 = − 7
[The first printing of the textbook had a misprinted 2 (instead of 4) as the right-hand
side constant in the first equation.] Begin by subtracting twice row 1 from row 2 of A,
and three times row 1 from row 3.
1 0 0 2 3
E = 0 1 0 −1 −4 ;
x1 = 3 − 2t , x2 = −4 + t , x3 = 5 − 3t , x4 = t
0 0 1 3 5
7. 28. Begin by subtracting row 3 from row 1 of A. Then subtract 3 times row 1 from row 2,
and twice row 1 from row 3.
1 −2 0 3 4
E = 0 0 1 4 3 ;
x1 = 4 + 2 s − 3t , x2 = s, x3 = 3 − 4t , x4 = t
0 0 0 0 0
29. Begin by interchanging rows 1 and 2 of A. Then subtract three times row 1 from row 2,
and four times row 1 from row 3.
1 0 1 1 3
E = 0 1 −2 3 5 ;
x1 = 3 − s − t , x2 = 5 + 2s − 3t , x3 = s, x4 = t
0 0 0 0 0
30. Begin by interchanging rows 1 and 2 of A. Then subtract twice row 1 from row 2, and
five times row 1 from row 3.
1 0 0 0 −3 2
E = 0 1 0 −1 2 1 ;
x1 = 2 + 3t , x2 = 1 + s − 2t , x3 = 2 + 2s, x4 = s, x5 = t
0 0 1 −2 0 2
1 2 3 (1/6) R3
1 2 3 R 2−5 R 3
1 2 3 (1/ 4) R 2
1 2 3
31. 0 4 5 → 0 4 5 → 0 4 0 → 0 1 0
0 0 6
0 0 1
0 0 1
0 0 1
1 0 3 1 0 0
R1−2 R 2 R1−3 R 3
→ 0 1 0 → 0 1 0
0 0 1
0 0 1
32. If ad − bc ≠ 0, then not both a and b can be zero. If, for instance, a ≠ 0, then
a b (1/ a ) R1 1 b / a R 2−c R1 1 b / a a R 2 1 b/a
c d → c d → 0 d − bc / a → 0 ad − bc
(1/( ad −bc )) R 2 1 b / a R1−(b / a ) R 2 1 0
→ 0 1 → 0 1 .
33. If the upper left element of a 2 × 2 reduced echelon matrix is 1, then the possibilities are
1 0 1 *
0 1 and 0 0 , depending on whether there is a nonzero element in the second
8. row. If the upper left element is zero — so both elements of the second row are also 0,
0 1 0 0
then the possibilities are and 0 0 .
0 0
34. If the upper left element of a 3 × 3 reduced echelon matrix is 1, then the possibilities are
1 0 0 1 0 * 1 * 0 1 * *
0 1 0 , 0 1 * , 0 0 1 , and 0 0 0 ,
0 0 1
0 0 0
0 0 0
0 0 0
depending on whether the second and third row contain any nonzero elements. If the
upper left element is zero — so the first column and third row contain no nonzero
elements — then use of the four 2 × 2 reduced echelon matrices of Problem 33 (for the
upper right 2 × 2 submatrix of our reduced 3 × 3 matrix) gives the additional possibilities
0 1 0 0 1 * 0 0 1 0 0 0
0 0 1 , 0 0 0 , 0 0 0 , and 0 0 0 .
0 0 0
0 0 0
0 0 0
0 0 0
35. (a) If ( x0 , y0 ) is a solution, then it follows that
a (kx0 ) + b(ky0 ) = k (ax0 + by0 ) = k ⋅ 0 = 0,
c( kx0 ) + d ( ky0 ) = k (cx0 + dy0 ) = k ⋅ 0 = 0
so (kx0 , ky0 ) is also a solution.
(b) If ( x1 , y1 ) and ( x2 , y2 ) are solutions, then it follows that
a ( x1 + x2 ) + b( y1 + y2 ) = (ax1 + by1 ) + (ax2 + by2 ) = 0 + 0 = 0,
c( x1 + x2 ) + d ( y1 + y2 ) = (cx1 + dy1 ) + (cx2 + dy2 ) = 0 + 0 = 0
so ( x1 + x2 , y1 + y2 ) is also a solution.
36. By Problem 32, the coefficient matrix of the given homogeneous 2 × 2 system is row-
equivalent to the 2 × 2 identity matrix. Therefore, Theorem 4 implies that the given
system has only the trivial solution.
37. If ad − bc = 0 then, much as in Problem 32, we see that the second row of the reduced
echelon form of the coefficient matrix is allzero. Hence there is a free variable, and thus
the given homogeneous system has a nontrivial solution involving a parameter t.
9. 38. By Problem 37, there is a nontrivial solution if and only if
(c + 2)(c − 3) − (2)(3) = c 2 − c − 12 = (c − 4)(c + 3) = 0,
that is, either c = 4 or c = –3.
39. It is given that the augmented coefficient matrix of the homogeneous 3 × 3 system has the
form
a1 b1 c1 0
a b2 c2 0 .
2
pa1 + qa2 pb1 + qb2 pc1 + qc2 0
Upon subtracting both p times row 1 and q times row 2 from row 3, we get the matrix
a1 b1 c1 0
a b2 c2 0
2
0
0 0 0
corresponding to two homogeneous linear equations in three unknowns. Hence there is at
least one free variable, and thus the system has a nontrivial family of solutions.
40. In reducing further from the echelon matrix E to the matrix E*, the leading entries of
E become the leading ones in the reduced echelon matrix E*. Thus the nonzero rows of
E* come precisely from the nonzero rows of E. We therefore are talking about the same
rows — and in particular about the same number of rows — in either case.