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

By Stephen Andrilli, David Hecker

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The final matrix corresponds to the linear system ⎧ ⎨ x − y + z = 20 y − 2z = 0 . ⎩ z = 31 The third equation in this system yields z = 31. Plugging this value into the second equation produces y − 2(31) = 0, or y = 62. Finally, we substitute these values into the first equation to get x − 62 + 31 = 20, or x = 51. Hence, there are 51 nickels, 62 dimes, and 31 quarters. (4) First, plug each of the three given linear system: ⎧ ⎨ 18 = 9 = ⎩ 13 = points into the equation y = ax2 + bx + c. This yields the following 9a + 3b + c ←− using (3, 18) 4a + 2b + c ←− using (2, 9) .

The last equation states that x5 = 2. Hence, the complete solution set is {(2b − d − 4, b, 2d + 5, d, 2) | b, d ∈ R}. 1 is ⎡ 5 ⎤ 1 − 12 − 74 4 ⎥ ⎢ 1 − 14 − 74 ⎥ ⎢ 0 ⎢ ⎥. ⎣ 0 0 1 3 ⎦ 0 0 0 0 We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2,2) entry as the pivot and target the (1,2) entry. 2 Next, we use the (3,3) entry as a pivot and target the (1,3) and (2,3) entries. ⎡ ⎤ 1 0 0 6 ⎢ 0 1 0 1 ← ( 15 −1 ⎥ 8 )× 3 + 1 ⎢ ⎥ ⎣ 0 0 1 3 ⎦ 2 ← ( 14 ) × 3 + 2 0 0 0 0 Ignoring the last row, which gives the equation 0 = 0, this matrix yields the linear system ⎧ = 6 ⎨ x1 = −1 , x2 ⎩ 3 x3 = producing the unique solution (6, −1, 3).

Hence, the pivot moves to the (2,3) entry. We convert that to 1. ⎡ ⎤ 10 19 4 0 1 4 − 27 7 7 7 ⎢ ⎥ ⎢ 0 0 1 0 ⎥ 3 −15 −32 ⎢ ⎥ 2 ← (7) 2 ⎢ 0 0 6 12 20 2 0 ⎥ ⎣ ⎦ 7 7 7 7 10 26 0 0 0 − 37 − 97 7 7 Next, we target the (1,3), (3,3), and (4,3) ⎡ 1 4 1 ← (− 47 ) × 2 + 1 ⎢ ⎢ 0 0 3 ← (− 27 ) × 2 + 3 ⎣ 0 0 4 ← ( 37 ) × 2 + 4 0 0 entries. 0 −2 1 3 0 0 0 0 10 21 −15 −32 6 12 −5 −10 ⎤ 0 0 ⎥ ⎥ 0 ⎦ 0 We try to move the pivot to the third column but cannot due to the zeroes in the (3,4) and (4,4) entries. Hence, the pivot moves to the (3,5) entry.

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