Math254FinalSols

# Math254FinalSols - FINAL SOLUTIONS MATH 254 SUMMER 2002...

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FINAL - SOLUTIONS MATH 254 - SUMMER 2002 - KUNIYUKI GRADED OUT OF 100 POINTS ¥ 2 = 200 POINTS TOTAL Assume that n represents a positive integer. 1) Find A 10 if A = - - È Î Í Í Í ˘ ˚ ˙ ˙ ˙ 10 0 020 00 2 . (3 points) A A 10 10 10 10 10 00 02 0 002 10 0 0 1024 0 1024 = - - È Î Í Í Í ˘ ˚ ˙ ˙ ˙ = () - È Î Í Í Í ˘ ˚ ˙ ˙ ˙ = È Î Í Í Í ˘ ˚ ˙ ˙ ˙ -1 (This works, because is a diagonal matrix.) 2) Find the determinant below. (10 points) 23 30 112 0 05 00 34 13 1 4 0 2 10 20 1 -- - Let's expand along the third row, since it has many zeros. (The fourth column is also good.) Its only nonzero entry is the " 5 "; the corresponding sign from the sign matrix is " - " [or you could observe that - () =- ++ 111 1 32 5 ij ]. To get the submatrix for the corresponding minor, we delete the row and the column containing the " 5 ". 23 30 0 1 4 2 10 201 15 2301 12 0 4 31 2 1 201 / - / - // / / - / =- ( ) - - 05 000 3 from sign matrix { Let’s expand along the third column. Its only nonzero entry is the " 3 "; the corresponding sign from the sign matrix is "+" [or you could observe that - () =+ 1 33 6 ].

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To get the submatrix for the corresponding minor, we delete the row and the column containing the " 3 ". =- () / - / /// - / / + ( ) - - 5 23 1 12 4 31 2 1 51 3 231 4 121 15 4 0 0 3 0 from sign matrix { You could then use Sarrus's Rule or expansion by minors/cofactors to compute the determinant of the resulting 33 ¥ matrix, which turns out to be - 7. Using Sarrus’s Rule: Determinant of ¥ matrix = 21 6341 22 7 -- - ++ = - . Original determinant - = 15 7 105 3) If A is a 44 ¥ matrix, and A = 10 , then find 3 A . (3 points) A has order n = 4. See Theorem 3.6 on p.130. (Idea: When you multiply A by 3, you multiply each row by 3; each of these 4 row- multiplications has the effect of multiplying the determinant by 3.) 0 81 10 810 4 AA n = = = =
4) Let v 1 = 0360 ,,, () , v 2 = 0246 , and v 3 = 112 1 ,, , -- . Express 4 121 1 , -- - as a linear combination of v 1 , v 2 , and v 3 . The weights in this linear combination must be given as specific real numbers; don’t just leave them as "" c i s . (12 points) Solve vvv 123 4 1 2 11 - - - È Î Í Í Í Í Í ˘ ˚ ˙ ˙ ˙ ˙ ˙ for c 1 , c 2 , and c 3 , the weights for the linear combination. 0 3 6 0 0 2 4 6 1 1 2 1 4 1 2 11 - - - - - È Î Í Í Í Í Í ˘ ˚ ˙ ˙ ˙ ˙ ˙ Reorder the rows. Wrap around the first row to the bottom. 3 6 0 0 2 4 6 0 1 2 1 1 1 2 11 4 - - - - - È Î Í Í Í Í Í ˘ ˚ ˙ ˙ ˙ ˙ ˙ RR R 21 2 2 +- Æ .

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Math254FinalSols - FINAL SOLUTIONS MATH 254 SUMMER 2002...

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