Math254FinalSols

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ].
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 = = = =
Background image of page 2
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 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online