#show link: set text(blue) #set text(font: "Calibri") #show raw: set text(font: "Fira Code") #set table.cell(breakable: false) #set table(stroke: (x, y) => ( left: if x > 0 { .1pt }, top: if y == 1 { 0.5pt } else if y > 1 { 0.1pt }, )) #set math.mat(delim: "[") #set page("us-letter") #let solve(solution) = { block( inset: 5pt, stroke: blue + .3pt, fill: rgb(0, 149, 255, 15%), radius: 4pt, )[#solution] } #let solvein(solution) = { let outset = 3pt h(outset) box( outset: outset, stroke: blue + .3pt, fill: rgb(0, 149, 255, 15%), radius: 4pt, )[#solution] } #let note(content) = { block( inset: (left: 5pt, right: 5pt, top: 10pt, bottom: 10pt), stroke: luma(20%) + .3pt, fill: luma(95%), radius: 4pt, )[#content] } #let notein(content) = { let outset = 3pt h(outset) box( outset: outset, stroke: luma(20%) + .3pt, fill: luma(95%), radius: 4pt, )[#content] } #align(center)[ = CS 3333 Mathematical Foundations Homework 5 (100 pts)\ #underline[Price Hiller] | #underline[zfp106] ] #line(length: 100%, stroke: .25pt) *Questions*\ Please write down the major intermediate steps. 1. Calculate the rank of the following matrices. (12 pts) #grid( columns: 2, align: center, gutter: 2em, [ $A = mat( 2, -1, 3; 1, 0, 1; 0, 2, -1; 1, 1, 4; )$ ], [ (6 pts) ], grid.cell( colspan: 2, [ #note[ #grid( column-gutter: (1em, 4em, 1em, 1em), columns: 2, align: left + horizon, row-gutter: 2em, [$R_4 = R_4 - R_2$], [$mat( 2, -1, 3; 1, 0, 1; 0, 2, -1; 0, 1, 3; )$], [$R_1 = 1 / 2 R_1$], [$mat( 1, -1/2, 3/2; 1, 0, 1; 0, 2, -1; 0, 1, 3; )$], [$R_2 = R_2 - R_1$], [$mat( 1, -1/2, 3/2; 0, 1/2, -1/2; 0, 2, -1; 0, 1, 3; )$], [$R_2 = 2R_2$], [$mat( 1, -1/2, 3/2; 0, 1, -1; 0, 2, -1; 0, 1, 3; )$], [$R_4 = R_4 - R_2$], [$mat( 1, -1/2, 3/2; 0, 1, -1; 0, 2, -1; 0, 0, 4; )$], [$R_3 = R_3 - 2R_2$], [$mat( 1, -1/2, 3/2; 0, 1, -1; 0, 0, 1; 0, 0, 4; )$], [$R_4 = R_4 - 3R_3$], [$mat( 1, -1/2, 3/2; 0, 1, -1; 0, 0, 1; 0, 0, 0; )$], grid.cell(colspan: 2, align: center, solve[The rank of $A$ is 3.]) ) ] ], ), colbreak(), colbreak(), [ $B = mat( 3, 2, -1; 2, -3, -5; -1, -4, -3; )$ ], [ (6 pts) ], grid.cell( colspan: 2, [ #note[ #grid( column-gutter: (1em, 4em, 1em, 1em), columns: 2, align: left + horizon, row-gutter: 2em, [$R_2 = R_2 + 2R_4$], [$mat( 3, 2, -1; 0, -11, -11; -1, -4, -3; )$], [$R_2 = -1 / 11 R_2$], [$mat( 3, 2, -1; 0, 1, 1; -1, -4, -3; )$], [$R_4 ↔ R_2$], [$mat( 3, 2, -1; -1, -4, -3; 0, 1, 1; )$], [$R_2 = R_2 + 1 / 3R_1$], [$mat( 3, 2, -1; 0, -10/3, 10/3; 0, 1, 1; )$], [$R_2 = -3 / 10 R_2$], [$mat( 3, 2, -1; 0, 1, 1; 0, 1, 1; )$], [$R_3 = R_3 - R_2$], [$mat( 3, 2, -1; 0, 1, 1; 0, 0, 0; )$], grid.cell(colspan: 2, align: center, solve[The rank of $B$ is 2.]) ) ] ], ), ) 2. Solve the following systems using the inverse of a matrix. (12 pts) #grid( columns: 2, align: center, gutter: 2em, [ $ -7x + 3y &= -34\ 8x -4y &= 44 $ ], [ (4 pts) ], grid.cell( colspan: 2, [ #note[ #columns(2)[ $ cases( -7x + 3y = -34\ 8x -4y = 44 ) -> mat( -7, 3, -34; 8, -4, 44; augment: #2)\ $ $ R_1 &= -1 / 7 R_1& &-> mat( 1, -3/7, -34/7; 8, -4, 44; augment: #2)\ R_2 &= R_2 - 8R_1& &-> mat( 1, -3/7, -34/7; 0, -4/7, 36/7; augment: #2)\ R_2 &= -7 / 4R_2& &-> mat( 1, -3/7, -34/7; 0, 1, -9; augment: #2)\ R_1 &= R_1 + 3 / 7R_2& &-> mat(1, 0, 1; 0, 1, -9; augment: #2)\ $ #solve[$x = 1, y = -9$] ] ] ], ), [ $ 5x + 15y + 56z &= 35\ -4x - 11y -41z &= -26\ -x - 3y - 11z &= -7 $ ], [ (8 pts) ], grid.cell( colspan: 2, [ #note[ #block(inset: (left: 30pt, right: 30pt))[ $ cases( 5x + 15y + 56z& =& 35\ -4x - 11y -41z& =& -26\ -x - 3y - 11z& =& -7 ) -> mat(5, 15, 56, 35; -4,-11,-41,-26;-1,-3,-11,-7; augment: #3)\ $ $ R_1 &= 1 / 5 R_1& &-> mat(1, 3, 56/5, 7; -4,-11,-41,-26;-1,-3,-11,-7; augment: #3)\ R_2 &= R_2 + 4R_1& &-> mat(1, 3, 56/5, 7; 0, 1, 19/5, 2;-1,-3,-11,-7; augment: #3)\ R_3 &= R_3 + R_1& &-> mat(1, 3, 56/5, 7; 0, 1, 19/5, 2;0,0,1/5,0; augment: #3)\ R_1 &= R_1 - 3R_2& &-> mat(1,0,-1/5,1; 0, 1, 19/5, 2;0,0,1/5,0; augment: #3)\ R_3 &= 5R_3& &-> mat(1,0,-1/5,1; 0, 1, 19/5, 2;0,0,1,0; augment: #3)\ R_1 &= R_1 +1 / 5 R_3& &-> mat(1,0,0,1; 0, 1, 19/5, 2;0,0,1,0; augment: #3)\ R_2 &= R_2 - 19 / 5 R_3& &-> mat(1,0,0,1; 0, 1, 0, 2;0,0,1,0; augment: #3)\ $ #solve[$x = 1, y = 2, z=0$] ] ] ], ), ) #align(center + bottom, note[#text(size: 3em)[⇊ SEE NEXT PAGE ⇊]]) #pagebreak() 3. Find the eigenvalues and eigenvectors of the following matrices. (12 pts) #grid( columns: 2, align: center, gutter: 2em, [ $A = mat( 2, 3; -3, -5; )$ ], [ (4 pts) ], grid.cell( colspan: 2, [ #note[ #block(inset: (left: 30pt, right: 30pt))[ #underline[Calculating Eigenvalues] $ A - λ ⋅"I" &= mat(2 - λ, 3;-3, -5 - λ)&\ mat(delim: "|", 2 - λ, 3;-3, -5 - λ) &= 0&\ (2 - λ)(-5-λ) - 3(-3) &=0&\ (2 - λ)(-5-λ) + 9 &=0&\ (-3 + sqrt(13)) / 2,(-3 - sqrt(13)) / 2&= λ&\ $ #underline[Calculating Eigenvectors] #solve[ Eigen Values: $(-3 + sqrt(13)) / 2,(-3 - sqrt(13)) / 2$ ] ] ] #align(left)[#text( red, weight: "black", )[And it is at this point that I have run out of time :(. I forgot this was due tonight, I thought today was the 13th until I saw that football was on at 9pm. My stomach dropped a few miles as you might imagine. At this point, I've winged it and scraped what I could together. May the dice roll in my favor.]] ], ), [ $B = mat( 2, 0, 0; 0, 4, 5; 0, 4, 3; )$ ], [ (8 pts) ], grid.cell(colspan: 2, []), ) 4. List all the permutations of ${P,Q, R, S}$ (6 pts). #solve[ #grid( align: center, columns: 4, [ ${P,Q,R,S}$ ${P,Q,S,R}$ ${P,R,Q,S}$ ${P,R,S,Q}$ ${P,S,Q,R}$ ${P,S,R,Q}$ ], [ ${Q,P,R,S}$ ${Q,P,S,R}$ ${Q,R,P,S}$ ${Q,R,S,P}$ ${Q,S,P,R}$ ${Q,S,R,P}$ ], [ ${R,P,Q,S}$ ${R,P,S,Q}$ ${R,Q,P,S}$ ${R,Q,S,P}$ ${R,S,P,Q}$ ${R,S,Q,P}$ ], [ ${S,P,Q,R}$ ${S,P,R,Q}$ ${S,Q,P,R}$ ${S,Q,R,P}$ ${S,R,P,Q}$ ${S,R,Q,P}$ ], )] #align(center + bottom, note[#text(size: 3em)[⇊ SEE NEXT PAGE ⇊]]) #pagebreak() 5. Calculate the value of each of these quantities (10 pts). #grid( columns: 3, align: center, gutter: 2em, [ a. $P(7,2)$ #solve[$7! / (7 - 2)! = 42$] ], [ b. $P(6,3)$ #solve[$6! / (6 - 3)! = 120$] ], [ c. $P(12,9)$ #solve[$12! / (12 - 9)! = 79,833,600$] ], [ d. $C(9,5)$ #solve[$9! / (5! ⋅ (9 - 5)!) = 120$] ], [ f. $C(12,7)$ #solve[$12! / (7! ⋅ (12 - 7)!) = 792$] ], ) 6. How many ways are there to select a first-prize winner, a second-prize winner, and a third-prize winner from 180 different people who have entered a contest? (6 pts) #note[$180 ⋅ 179 ⋅ 178 = #solvein[5,735,160]$] 7. How many bit strings of length 16 contain... (12 pts) #grid( columns: 2, align: center, gutter: 2em, [ a. Exactly five 0s? #solve[$16! / (5! ⋅ (16 - 5)!) = 4,368$] ], [ b. At most five 0s? #note[ $ &C(16,0) &= &16! / (0! ⋅ (16 - 0)!)& &= &1\ + &C(16,1) &= &16! / (1! ⋅ (16 - 1)!)& &= &16\ + &C(16,2) &= &16! / (2! ⋅ (16 - 2)!)& &= &120\ + &C(16,3) &= &16! / (3! ⋅ (16 - 3)!)& &= &560\ + &C(16,4) &= &16! / (4! ⋅ (16 - 4)!)& &= &1,820\ + &C(16,5) &= &16! / (5! ⋅ (16 - 5)!)& &= &4,368\ $ #align(center)[#solve[$6,885$]] ] ], [ c. At least five 0s? #note[ $ 2^16 - 6,885 = #solvein[58,651] $ ] ], [ d. An equal number of 0s and 1s? #note[ $ C(16,8) = 16!/(8! ⋅ (16 - 8)!) = #solvein[12,870] $ ] ], ) #align(center + bottom, note[#text(size: 3em)[⇊ SEE NEXT PAGE ⇊]]) #pagebreak() 8. How many permutations of letters ${"A", "B", "C", "D", "E", "F", "G", "H", "I", "J"}$ contain (no letters repeated)... (12 pts) #grid( columns: 2, gutter: 1em, inset: (left: 2em, right: 2em), [ a. The string AJ? #note[ Our set is ${{"AJ"}, "B", "C", "D", "E", "F", "G", "H", "I"}$. Total of 9 permutations: $9! = #solvein[362,880]$ ] ], [ b. The string BIG? #note[ Our set is ${{"BIG"}, "A", "C", "D", "E", "F", "H", "J"}$. Total of 8 permutations: $8! = #solvein[40,320]$ ] ], [ c. The strings AEG and DB? #note[ Our set is ${{"AEG"}, {"DB"}, "C", "F", "H", "I", "J"}$. Total of 7 permutations: $7! = #solvein[5,040]$ ] ], [ d. The strings BG, FC, and AE? #note[ Our set is ${{"AE"}, {"BG"}, {"FC"}, "D", "H", "I", "J"}$. Total of 7 permutations: $7! = #solvein[5,040]$ ] ], [ e. The strings FIG and GIF? #note[ FIG = GIF Our set is ${"A", "B", "C", "D", "E", {"GIF"}, "H", "J"}$. Total of 8 permutations: $8! = #solvein[40,320]$ ] ], [ f. The strings CD and DJ? #note[ CD and DJ overlap, thus the string can be considered CDJ. Our set is ${"A", "B", {"CDJ"}, "E", "F", "G", "H", "I"}$. Total of 8 permutations: $8! = #solvein[40,320]$ ] ], ) 9. Suppose that a department contains 9 dentists and 15 optometrists. How many ways are there to form a committee with 7 members if it must have more dentists than optometrists? (8 pts) #note[ $ "1. 4 dentists and 3 optometrists" -> C(9,4) ⋅ C(15,3) &= 126 ⋅ 455 &=& 57,330\ "2. 5 dentists and 2 optometrists" -> C(9,5) ⋅ C(15,2) &= 126 ⋅ 105 &=& 13,230\ "3. 6 dentists and 1 optometrists" -> C(9,6) ⋅ C(15,1) &= 84 ⋅ 15 &=& 1,260\ "4. 7 dentists and 0 optometrists" -> C(9,7) ⋅ C(15,0) &= 36 ⋅ 1 &=& 36\ \ \ 57,330 + 13,230 + 1,260 + 63 = #solvein[71,883] $ ] #align(center + bottom, note[#text(size: 3em)[⇊ SEE NEXT PAGE ⇊]]) #pagebreak() 10. What is the coefficient of $x^17 ⋅ y^14$ in $(2x - 3y)^31$? (You do not need to calculate the final value. Just write down the formula of the coefficient) (10 pts) #note[ $x = 2x, y = -3y, (X + Y)^31$ $ &= mat(31; 17) X^17 ⋅ Y^14\ &= mat(31; 17) (2x)^17 ⋅ (-3y)^14\ &= mat(31; 17) (2)^17 ⋅ (-3)^14 x^17 ⋅ y^13\ &= #solve[$-mat(31; 17) 2^17 ⋅ 3^14$] $ ]