APP. A] VECTORS AND MATRICES 427
(a) Form the matrixM=[A, I]and row reduceMto echelon form:M=⎡
⎣1 − 22 ... 100
2 − 36 ..... 010
117 .... 001⎤
⎦∼⎡
⎣1 − 2 ...2 100
01 ..... 2 − 210
03 ..... 5 − 101⎤
⎦∼⎡
⎣1 − 22 ... 100
012 .....− 210
00 − 1 ..... 5 − 31⎤
⎦In echelon form, the left half ofMis in triangular form; henceAhas an inverse. Further reduceMto row canonical
form:M=⎡
⎣1 − 20 ... 11 − 62
010 ..... 8 − 52
001 .....− 53 − 1⎤
⎦∼⎡
⎣100 ... 27 −16 6
010 ..... 8 − 52
001 .....− 53 − 1⎤
⎦The final matrix has the form[I,A−^1 ]; that is,A−^1 is the right half of the last matrix. ThusA−^1 =⎡
⎣27 −16 6
8 − 52
− 53 − 1⎤
⎦(b) Form the matrixM=[B, I]and row reduceMto echelom form:M=⎡
⎣13 − 4 ... 100
15 − 1 ..... 010
313 − 6 ..... 001⎤
⎦∼⎡
⎣13 − 4 ... 100
02 3.....− 110
04 6.....− 301⎤
⎦∼⎡
⎣13 − 4 ... 100
02 3.....− 110
00 0.....− 1 − 21⎤
⎦In echelon form,Mhas a zero row in its left half; that is,Bis not row reducible to triangular form. Accordingly,
Bhas no inverse.ECHELON MATRICES, ROW REDUCTION, GAUSSIAN ELIMINATION
A.19. Interchange the rows in each matrix to obtain an echelon matrix:(a)⎡
⎣01 − 346
4025 − 3
00 7− 28⎤
⎦; (b)⎡
⎣000 00
123 45
005 − 47⎤
⎦; (c)⎡
⎣02222
03100
00000⎤
⎦(a) Interchange the first and second rows.
(b) Bring the zero row to the bottom of the matrix.
(c) No number of row interchange can produce an echelon matrix.A.20. Row reduce the matrixA=⎡
⎣12 − 30
24 − 22
36 − 43⎤
⎦to echelon form.Use a 11 as a pivot to obtain zeros belowa 11 , that is, apply the row operations “Add− 2 R 1 toR 2 ” and “Add− 3 R 1
toR 3 ;” and then usea 23 =4 as a pivot to obtain a zero belowa 23 , that is, by applying the row operation “Add− 5 R 2
to 4R 3 .” These operations yield the following where the last matrix is in echelon form:A∼⎡
⎣12 − 30
00 42
00 53⎤
⎦∼⎡
⎣12 − 30
00 42
00 02⎤
⎦A.21. Which of the following matrices are in row canonical form?
⎡
⎣12 −30 1
00 52− 4
00 07 3⎤
⎦,⎡
⎣017 − 50
000 01
000 00⎤
⎦,⎡
⎣10502
01204
00017⎤
⎦The first matrix is not in row canonical form since, for example, two leading nonzero entries are 5 and 7, not 1.
Also, there are nonzero entries above the leading nonzero entries 5 and 7. The second and third matrices are in row
canonical form.