SAQ 1 What are the slope and y intercept of the following functions:

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1 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions SAQ Wht re the slope nd y intercept of the following functions: () y = x + 7 (b) y = -½ x (c) y = -x + The generl form of the eqution of line is y = mx + c, where m is the slope of the line c is the y intercept. () y = x + 7 slope = : y intercept = 7 (i.e. the point (0,7) to be precise) (b) y = -½ x slope = -½ : y intercept = -½. (c) y = -x + slope = - : y intercept =. SAQ () Find the slope of the line representing equtions (ii)nd (iii) bove. (b) Plot equtions (ii) nd (iii) on the sme grph, nd from the grph find the x nd y vlues t their point of intersection. () (ii) x y = 5 y = x 5 thus slope = +: (iii) 6x y = -y = 6x + y = 6 x + y = -x thus slope = -: - - (b) To plot line we just find the coordintes of ny points on the line, nd join them up. Bsiclly, you cn pick ny vlues you like for x, nd then find the y vlue from the eqution. (i) x y = 5 I choose x = 0 x y = 5 0 y = 5 y = -5 the point (0, -5) lies on the line I choose x = 0 the eqution becomes 0 y = 5 y = 0 5 = 5.. (0,5) lies on the line. Now I plot the points nd join them, to get the grph of the line: (-, 8) (0, 5) (0, -) Point of intersection = (, -) (0, -5) DCU 00

2 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions Similrly for the line -6x y = I use points to drw the line. x = 0-6x y = 0 y = y = - the point (0, -) lies on the line x = - (-6)(-) y = 8 y = -y = - 8 = -6 y = -6/- = +8 the point (-, 8) lso lies on the line.. note I cn choose ny vlue for x tht I wnt. I choose x = - Thus, the point of intersection where the lines meet is (, -) SAQ Solve the following liner systems: () x + y = 8 x y = (b) b = - b = -5 () (i) x + y = 8 (ii) x y = I choose to eliminte x so I will multiply (i) by nd subtrct x + 6y = 6 x y = 0 + 0y = 0 0y = 0 y = 0/0 = y = Now put this y =, into either eqution (i) or (ii) to find x. I choose eqn (i) (i) x + y = 8 x + () = 8 x + = 8 x = 8 = x =. So finlly, the point of intersection (x, y) of the lines is: (, ) (b) (i) b = - (ii) b = -5 I choose to eliminte s its esier to do, so I will multiply (ii) by nd subtrct b = - b = b = - b = - Now, I find by putting b = - into either of the equtions. I choose to use eqn (ii) b = -5 (-) = -5 + = -5 = -5 = -9 = -9. Thus, these lines intersect t the point (-9, -) DCU 00

3 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions SAQ Solve the following liner systems: () (i) x + y z = 6 (ii) x y + z = -8 (iii) x y + z = -7 We eliminte one vrible. I choose to eliminte z, s this will be esiest! I will dd (iii) + (i) to eliminte z from these. (i) x + y z = 6 (iii) x y + z = -7 5x + y + 0 = - We end up with new eqution tht only hs vribles: (iv) 5x + y = - Now, I will eliminte z using eqn (ii) nd one other eqution. I choose to use eqn (i), but I must multiply it by to mke the coefficient of z = x + 6y z = (ii) x y + z = -8 (i) x + 6y z = 6x + 5y + 0 = If I dd the equtions we will eliminte z so we end up with nother new eqution involving vribles: (v) 6x + 5y = I hve gone from system of equtions in unknowns, to new system involving equtions with unknowns. I now solve this new system of equtions in the stndrd wy. (iv) 5x + y = - (v) 6x + 5y = I choose to eliminte y here. To mke the y coefficient equl in both equtions I multiply (iv) x 5, nd (v) x 5 x (iv) 5x + 0y = -5 x (v) x + 0y = 8 x + 0 = - Thus, x = - x = /- = -. x = -. Now I will put this x vlue bck into either eqn (iv) or (v) (I choose (iv) here) (iv) 5x + y = - 5(-) + y = y = - y = = y = / y =. Now, I put my x nd y vlue into ny of the originl equtions (i), (ii) or (iii) to find the z vlue. I choose (iii) here (iii) x y + z = -7 (-) + z = z = -7 z = = - z = -. So, the solution to the bove system of equtions is (-,, -) DCU 00

4 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions (b) (i) x y + z = - (ii) x y + z = -5 (iii) x + y + z = Here I choose to eliminte x. I will use (i) (ii) first. (i) x y + z = - x (ii) x y + z = y + z = - Thus we get n eqution with vribles. (iv) y + z = - We now must eliminte x from eqn (iii) nd other of our equtions. I choose (ii). This I will use (iii) (ii) to mke the x coefficients equl. (iii) x + y + z = x (ii) x 6y + z = y z = 6 Thus, we get the second eqution in unknowns: (v) 7y z = 6 We cn now solve the new system of equtions in unknowns s usul. (iv) y + z = - (v) 7y z = 6 I choose to eliminte z using the opertion (iv) (v) Thus, y = 0/5 = y =. y + z = - y z = 5y + 0 = 0 Now, put this bck into (v) or (iv) to find z. I choose (iv) y + z = - + z = - z = - Now, we out our y nd z vlues into ny of equtions (i), (ii) or (iii) to find the x vlue. I choose (ii) x y + z = -5 x () + - = -5 x 6 = -5 x = =. x = Thus, the solution to this system of equtions is (,, -) DCU 00

5 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions SAQ 5 If B = () B (b) C 7 6 5, C = Find: 0 () Here we re clculting the determinnt of x mtrix. We multiply the numbers in the min digonl, nd then subtrct the product of the other digonl: b For the mtrix the determinnt is d bc c d B = 7 6 = 7() (6) = = - C = 5 = 5() (0) = 0 0 = 0 0 SAQ 6 In ech of the following system, determine if solution exists nd if so, find it using Crmer s rule. x 7y = x y = 0 x + y = - x + 7y = Crmers rule is: x x b = b is given by b b x = nd x = b b We hve similr system of liner equtions, so we cn use Crmers rule. () The system if equtions: x 7y = x y = 0 cn be written s: -7 x = -y 0 If the determinnt of the mtrix exists, nd is non-zero, solution exists DCU 00 5

6 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions Thus the detminnt is: 7 = (-) (-7)() = - + = thus, solution exists. 7 0 ( ) ( 7)(0) 68 x = = = = 7 ( ) ( 7)() 7 x 0 (0) - ()() 7 = = = = 7 ( ) - (-7)() 7 Thus, the solution (x, x ) = (, ) (b) x + y = - x + 7y = cn be written x - = 7y Lets check if solution exists. 7 = (7) ()() = 7 0, so solution exists. x 7 = = 7 ( )(7) ()() 5 = (7) ()() 7 x = = 7 ()() ( )() 9 = (7) ()() 7 Thus, the solution is (-5/7, 9/7) DCU 00 6

7 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions SAQ 7 Evlute the determinnt of the following mtrices. () (b) 5 6 () When evluting determinnt of mtrices of order x, or greter, we generlly expnd long row (or column). We cn pick ny row (or column) we like. I will expnd long row. Thus, we need to find the minor determinnts, for ech element in this row. Recll, to get the minor, we just cross out the row nd column contining the element of interest, nd we end up with x mtrix whose determinnt is the required number. The first row contins the elements. We must determine the minor for ech of these elements. For the in row: = () () = - For the in row : = () () = For the in row : = () () = Now, we multiply ech element of row, by its corresponding minor, nd then dd ll the resulting numbers together. However, we must be creful of the sign convention for determinnts (see p 6 of notes). We must chnge the sign of the nd element in row before we evlute the determinnt. Thus we get, Det = (-) - () + () = = 6 Thus, the determinnt of the mtrix is 6. DCU 00 7

8 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions (b) 5 6 I will expnd long the first row gin, so Lets find the minor determinnts for the elements in the first row = (-)(6) (5)() = - 0 = = ()(6) (5)() = 8 0 = = ()() (-)() = + = 6 Det = (-) (8) + (6) = -8 + = - 0 SAQ 8 () Evlute the following determinnt: Here we will convert the determinnt to tringulr form using row opertions, nd then find the determinnt. In r, the first non-zero element is, so this is fine. Now, in column, I must chnge the first element of r nd r to zero by dding multiples of nother row. I first subtrct times r from r to get the leding element in row = r r Similrly, if I dd times r to r I end up with r + r Thus r nd column re done!! DCU 00 8

9 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions Now, I will extrct the from r by dividing cross by to get r / 6 7 ( ) 0 5/7 0 7 note: -0/- = 5/7 if divide bove nd below by Now, we must get ll zeros below the leding in column, so we must get rid of the leding 7 in r. To do this I subtrct 7 times r from r. r - 7r 6 7 ( ) 0 5/ Now the determinnt is digonl, so we just multiply the elements in the min digonl to get the vlue of the determinnt. Det = (-)()()(7) = -98 (b) If A = nd B = 5 0 show tht det(ab) = detadetb AB = 5 0 = () + ( ) (5) + (0) ( )() + ()( ) ( )(5) + (0) = DetAB = = (-5)(-0) 5(-0) = = 00 Now DetA = Det B = 5 0 = () ()(-) = + 6 = 0 = (0) 5(-) = 0 Thus DetA x DetB = 0 x 0 = 00 Thus, DetAB = DetADetB = 00 DCU 00 9

10 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions SAQ9 Which of the following mtrices re non-singulr A = B = 5 6 We cn expnd the determinnt long row (or column) to find the determinnt. Similrly, we cn reduce it to tringulr form, nd then get the product of the min digonl to find the determinnt. Thus we hve wys to do this problem. I will tringulte it. A = r r 5 r r = () r r () Multiply the min digonl elements to get the determinnt. Thus, the determinnt = ()()(0) = 0 This mtrix is singulr mtrix, so it hs no inverse. B = 5 6 We determined B in SAQ 7 to be 0 so this mtrix is non-singulr s its determinnt 0 SAQ0 Find the inverse of the following mtrices: () 6 (b) 0 In this problem we will use third method to find the inverse of the mtrices given. The new method involves forming n ugmented mtrix consisting of the mtrix of interest nd the identity mtrix I. Then, we perform row opertions to convert the given mtrix into the identity mtrix I. If we perform the opertions on the ugmented mtrix, then the identity mtrix gets converted into the inverse during the process. Our mtrix is 6, so lets form the ugmented mtrix DCU 00 0

11 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions Now lets use row opertions on row Well the leding element is so this row is fine. Lets now look t column. We need to convert the in r into zero. Lets dd times r to r to do this r + r 0 0 Now, for row, the leding element must be zero, so we divide by r / 0 0 /6 / Lstly, we must get rid of the in column, so lets subtrct times r from r. r / 0/ / 0 /6 / Thus, the inverse mtrix is / / /6 / (b) 0 This is slightly hrder problem, s we re deling with x mtrix, but the bsic ide is the sme. Lets form the ugmented mtrix r r r + r r + r r - r r + r x r DCU 00

12 Certificte in Plsm nd Vcuum Studies Introductory Mths Unit -SAQ solutions Thus, the inverse mtrix is: 0 SAQ Find the solution of the following liner system using mtrices: x + y = 8 x 5y = - This cn be written in mtrix form s: x 8 = 5y - Here, we find the inverse of the system mtrix, nd multiply the mtrix eqution by the inverse, we find the solution (x, y). Let A =, so we will find A -. 5 DetA = (-5) ()() = -5 = -9 Thus, A - = 5 9 If we now multiply the mtrix eqution by A -, we will get: x = y 5 8 ( 5)(8) + (-)(-) = 9 9 (8) + (-) 8 = = 9 9 Thus, the solution to the system of liner equtions is the point (, ) DCU 00