By Stephen Andrilli, David Hecker

ISBN-10: 0123814553

ISBN-13: 9780123814555

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**Additional resources for Elementary Linear Algebra, Students Solutions Manual (e-only) **

**Sample text**

For a speciﬁc counterexample, choose A = O23 and B = O32 . 25 Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition Chapter 1 Review (p) False. 5). In particular, if A = and B = , then (A + B)2 = , 1 1 0 −1 2 3 5 4 but A2 + 2AB + B2 = . 4 5 (q) False. For a counterexample, note that 0 2 −2 0 0 3 −3 0 = (r) True. 15. 26 6 0 0 6 . 1 (1) In each part, we ﬁrst set up the augmented matrix corresponding to the given system of linear equations. Then we perform the row operations designated by the Gaussian elimination method.

Plugging this value into the second equation yields x2 − 14 (3) = − 74 , or x2 = −1. Substituting the values we have for x2 and x3 into the ﬁrst equation produces x1 − 12 (−1) − 74 (3) = 54 , or x1 = 6. Hence, the original linear system has a unique solution. The full solution set is {(6, −1, 3)}. (2) (a) The system of linear equations corresponding to the given augmented matrix is ⎧ x1 − 5x2 + 2x3 + 3x4 − 2x5 = −4 ⎪ ⎪ ⎨ x2 − x3 − 3x4 − 7x5 = −2 5 x4 + 2x5 = ⎪ ⎪ ⎩ 0 = 0 The last equation, 0 = 0, provides no information regarding the solution set for the system because it is true for every value of x1 , x2 , x3 , x4 , and x5 .

We convert it to 1. ⎤ ⎡ 2 2 1 − 23 − 43 9 9 ⎣ 3 2 1 ← ( 19 ) 1 4 0 −3 ⎦ 27 12 22 12 8 We next target the (2,1) and (3,1) entries. 2 ⎤ ⎥ 1 ⎦ 44 The (2,2) entry is the second pivot. We convert it to 1. ⎤ ⎡ 2 − 23 − 43 1 29 9 ⎢ 3 ⎥ 5 3 2 ← ( 34 ) 2 ⎣ 0 1 2 2 4 ⎦ 0 6 16 30 44 Next, we target the (1,2) and (3,2) entries. ⎡ 1 0 − 13 2 1 ← (− 9 ) × 2 + 1 ⎢ 5 ⎣ 0 1 2 3 ← (−6) × 2 + 3 0 0 1 −1 3 2 21 − 32 3 4 79 2 ⎤ ⎥ ⎦ The (3,3) entry is the last pivot. Since it already equals 1, we target the (1,3) and (2,3) entries.

### Elementary Linear Algebra, Students Solutions Manual (e-only) by Stephen Andrilli, David Hecker

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