Elementary Linear Algebra, Students Solutions Manual by Stephen Andrilli, David Hecker

By Stephen Andrilli, David Hecker

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Hence, we row reduce the matrix [b1 b2 |AT ], which yields ⎡ ⎤ 1 0 0 2 1 ⎢ 0 1 4 7 2 ⎥ ⎢ ⎥. ⎣ 0 0 0 0 0 ⎦ 0 0 0 0 0 In this case, each of the three linear systems has a unique solution. These three solutions are (0, 4), (2, 7), and (1, 2). Each gives the coefficients for expressing a row of A as a linear combination of the two nonzero rows of B. Hence, we have [0, 4, 12, 8] = 0[1, 0, −1, 2] + 4[0, 1, 3, 2], [2, 17, 19, 18] = 2[1, 0, −1, 2] + 7[0, 1, 3, 2], and [1, 2, 5, 6] = 1[1, 0, −1, 2] + 2[0, 1, 3, 2].

We do this using the following type (II) row operation: Type (II) operation: 2 ← (−8) × 1 + 2 : Side Calculation (−8)× (row 1) −8 − 32 3 (row 2) (sum) 4 3 400 3 8 4 −1 −80 0 − 20 3 1 3 160 3 Resultant Matrix ⎤ 4 1 − 50 − 16 3 3 ⎢ 1 160 ⎥ ⎣ 0 − 20 3 3 3 ⎦ 3 9 −1 −90 ⎡ Next, we target the (3,1) entry. 1 Type (II) operation: 3 ← (−3) × 1 + 3 : Side Calculation (−3)× (row 1) −3 −4 1 2 50 (row 3) 3 9 −1 −90 (sum) 0 5 − 12 −40 Resultant Matrix ⎤ 4 1 − 50 − 16 3 3 ⎢ 160 ⎥ 1 ⎣ 0 − 20 3 3 3 ⎦ 0 5 − 12 −40 ⎡ Now we move the pivot to the (2,2) entry.

Type (II) operation: 3 ← (−2) × 2 + 3 : Side Calculation (−2)× (row 2) 0 −2 (row 3) 0 2 (sum) 0 0 1 2 1 2 7 2 − 12 1 3 Matrix ⎡ Resultant 5 1 − 12 − 74 4 ⎢ 1 ⎢ 0 1 −4 − 74 ⎢ ⎢ 0 0 1 3 ⎣ 1 1 5 0 −4 −4 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ To finish the second column, we target the (4,2) entry. 1 The pivot now moves to the (3,3) entry. Luckily, the pivot already equals 1. Therefore, we next target the (4,3) entry. Type (II) operation: 4 ← ( 18 ) × 3 + 4 : Matrix ⎡ Resultant 5 1 − 12 − 74 4 ⎢ 1 ⎢ 0 1 −4 − 74 ⎢ ⎢ 0 1 0 3 ⎣ 0 0 0 0 Side Calculation ( 18 )× (row 3) 0 0 (row 4) 0 0 (sum) 0 0 1 8 − 18 3 8 − 38 0 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ This finishes the third column.

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