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

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Extra info for Elementary Linear Algebra, Students Solutions Manual (e-only)

Example text

For the Type (II) operation R : i ←− c j + i : Now R−1 is i ←− −c j + i . Again, R and R−1 change only the ith row of A, and we need to show that R−1 R leaves A i unchanged. But R−1 (R(A)) i = −c R(A) j + R(A) i = −c A j + R(A) i = −c A j + c A j + A i = A i . For the Type (III) operation R : i ←→ j : Now, R−1 = R. Also, R changes only the ith and jth rows of A, and these get swapped. Obviously, a second application of R swaps them back to where they were, proving that R is indeed its own inverse.

N) False. In fact, using an argument similar to the one used in part (m), AD is the matrix whose ith 1 2 3 0 3 10 column is the ith column of A multiplied by dii . For example, = . 3 4 0 5 9 20 (o) False. If A is m × n and B is n × m, then both AB and BA are deﬁned, with AB being m × m and BA being n × n. If m = n, then, clearly, AB = BA. 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.

Substituting both values into the ﬁrst equation produces x1 + 25 (3)− 25 (5) = − 14 5 , which leads to x1 = −2. Hence, the solution set for the system is {(−2, 3, 5)}. ⎤ ⎡ −54 3 −2 4 20 ⎦. The ﬁrst pivot is the (1,1) entry. ” We do this using the following type (II) row operation: Type (II) operation: 2 ← (1) × 1 + 2 : Side Calculation (1)× (row 1) 1 − 23 (row 2) 4 3 −18 −1 1 −2 20 0 1 3 − 23 2 (sum) Resultant Matrix ⎤ 4 1 − 23 −18 3 ⎥ ⎢ 1 2 ⎦ − 23 ⎣ 0 3 5 −4 8 −83 ⎡ Next, we target the (3,1) entry.