I have TWO Matrices Problems for you. Help??
The first thing you should note is that they are asking you to find the inverse of the matrixes. To do that you should take the matrix that you are given and attach an identity matrix to it on the right. So: A = 6 4 1 0 3 2 0 1 B = 2 1 1 0 5 3 0 1 So now do row reduction. Essentially, add and subtract the rows to each other to arrive at the identity matrix being on the left. You may have to multiply the rows before you add. This is ok just make sure when you add or multiply or do any other operation that you apply it to all elements in that row. If you try this with A you will find out that it is singular. Notice that the first and second rows are multiples of each other. If you try this with B your resulting matrix is: 1 0 3 -1 0 1 -5 2 The right portion is the inverse of the matrix, so your answer would be 3 -1 -5 2 ..
Help with Inverse Matrices problems?
1. A, B and P are matrices. Suppose P is invertible and A=PB(P^-1). Solve for B in terms of A. 2. Suppose (B-C)D=0 where B and C are mxn matrices and D is invertible. Show that B=C. 3. Suppose A and B are mxn matrices. B is invertible and AB is invertible. Show that A is invertible. [Hint:let C=AB, and solve for A] Any help with any of these 3 problems would be very appreciated!
Help with this matrices word problem? 10 points!?
I've never known the winds aloft to blow that steadily, either in speed or direction, for that length of time, but assuming a first time for everything: Going: (A+W)(6) = 576 Coming: (A-W)(12) = 576 6A + 6W = 576 12A - 12W = 576 24A = 1278 A = 72 6*72 + 6W = 576 W = 576/6 - 72 = 24 Plane's airspeed = 72 mph Wind speed = 24 ...
Matrices word problem?
Let p be # of pepperoni pizzas Let s be # of sausage pizzas Let c be # of cheese pizzas Since sold 600 pizzas total: p + s + c = 600 Since made $5900 for all pizzas (with prices given in question): 12*p + 10*s + 8*c = 5900 And since 175 more cheese than sausage pizzas: c - s = 175, or c - s + 0*p = 175 Thus, the equations are: 12*p + 10*s + 8*c = 5900 1*p + 1*s + 1*c = 600 0*p - 1*s + 1*c = 175 Augmented matrix is: 12 10 08 | 5900 1 1 1 | 600 0 -1 1 | 175 To solve, you can convert to row reduced form (rrf). Basically, subtract a multiple of one row from another row. Keep doing this until you get an upper triangular shape (this will be demonstrated...) Note: For an algebra class, make sure you use the techniques they teach in the class... So, start with: 12 10 08 | 5900 1 1 1 | 600 0 -1 1 | 175 Subtract 11*(row 2) from (row 1) 1 -1 -3 | -700 1 1 1 | 600 0 -1 1 | 175 Subtract 1*(row 1) from (row 2) 1 -1 -3 | -700 0 2 4 | 1300 0 -1 1 | 175 Add 0.5*(row 2) to (row 3) 1 -1 -3 | -700 0 2 4 | 1300 0 0 3 | 825 Above is in row-reduced form (can see that the non-zero numbers form an upper triangle). This is called an upper triangular matrix. Anyways... Now to solve: Using row 3: 3c = 825 c = 275 Using row 2 (and c=275) 2s + 4c = 1300 s + 2c = 650 s = 650 - 2*(275) s = 650 - 550 s = 100 Using row 1 (and c=275, s =100) p - s - 3c = -700 p = -700 + s + 3c p = -700 + 100 + 825 p = 225 So there were 225 pepperoni, 100 sausage, and 275 cheese pizzas sold... Again, the above just shows the idea. Use the techniques taught in the class (I think a lot of teachers are particular about the process used to reduce the matrix) . .
Matrices word problem help!!!! please :)?
1) find the area of triangle whose vertices are located at (-4, 1/2), (-5/2, -1), and (6, -2). evaluate the determinant using diagonals. 2) a foundation spent all of its operating budget on administrative expenses, renewal grants, and low-interest loans. administrative expenses and renewal grants accounted for 78.4% of the total budget. renewable grants accounted for 5.1% more of the total budget than low-interest loans. let a, r, and l each represent the percent of the budget accounted for by administrative expenses, renewal grants, and low-interest loans, respectively write a matrix equation that describes this situation Thank you :)
Pauli spin matrices problem!!! Please help!!?
Note that since A = [0 1; 1 0], we have A^2 = I. Hence A = A^3 = A^5 = ... , and I = A^2 = A^4 = ... 1) sin(kA) = Σ(n = 0 to ∞) (-1)^n (kA)^(2n+1)/(2n+1)! ...............= Σ(n = 0 to ∞) (-1)^n k^(2n+1) A^(2n+1) / (2n+1)! ...............= Σ(n = 0 to ∞) (-1)^n k^(2n+1) A / (2n+1)!, by the remarks above ...............= A * Σ(n = 0 to ∞) (-1)^n k^(2n+1) / (2n+1)! ...............= A * sin k. 2) cos(kA) = Σ(n = 0 to ∞) (-1)^n (kA)^(2n)/(2n)! ...............= Σ(n = 0 to ∞) (-1)^n k^(2n) A^(2n) / (2n)! ...............= Σ(n = 0 to ∞) (-1)^n k^(2n) I / (2n)!, by the remarks above ...............= I * Σ(n = 0 to ∞) (-1)^n k^(2n) / (2n)! ...............= I * cos k. 3) e^(kA) = Σ(n = 0 to ∞) (-1)^n (kA)^n / n! = Σ(m = 0 to ∞) (-1)^(2m) (kA)^(2m) / (2m)! + Σ(n = 0 to ∞) (-1)^(2m+1) (kA)^(2m+1) / (2m+1)! by separating even and odd terms = Σ(m = 0 to ∞) (-1)^(2m) k^(2m) I / (2m)! + Σ(n = 0 to ∞) (-1)^(2m+1) k^(2m+1) A/ (2m+1)! = I * cosh k + A * sinh k. 4) Replace k with ik in the result from (3): e^(ikA) = I * cosh(ik) + A * sinh(ik) ...........= I * (1/2)(e^(ik) + e^(-ik)) + A * (1/2)(e^(ik) - e^(-ik)) ...........= I * (1/2)(e^(ik) + e^(-ik)) + A * i * (1/(2i))(e^(ik) - e^(-ik)) ...........= I * cos k + A * i sin k. (Alternately, imitate the process in (3), noting that i^2 = -1...) I hope this helps!
Need help with matrices math word problem?
Amanda Bryce and Corey enter a race in which they have to run, swim and cycle over a marked course.Their average speeds are given in the table below. Corey finishes first with a total time of 2hrs 21min. Amanda comes in second with a time of 3hrs 9min. Bryce finishes last with a time of 3hrs 42min Find the distance (in miles) for each part of the race. Running Swiming Cycling Amanda| 10 4 10| Bryce |7 (1/2) 6 15| Corey | 10 15 40| matrices format | 10 4 10| |7 (1/2) 6 15| |10 15 40| need help with question tried to solve it but not getting the answer run=_____miles swim=______miles cycle=_______miles please help
Help with a Algebra 2 word problem (Matrices)?
MIXED NUTS: Macadamia nuts cost $.90 per ounce, peanuts cost $.30 per ounce, and cashews cost $1.30 per ounce. You want a 20-ounce mixture of Macadamia nuts, Peanuts, and Cashews that cost $.68 per ounce. If combined weight of the macadamia nuts and cashews equals the weight of the peanuts, how many ounces of each nut should be used?
Inverse matrices help!!! production problem?
Let's say you have f pounds of fruit and n pounds of nuts. We're told that the number of pounds of pretzels has to be twice that of the fruit, so the weight of the pretzels is 2f. The total number of pounds you have is f+n+2f, or 3f+n. The total value of the mixture in dollars is 4(2f) + 5f + 9n, which is 13f + 9n. The price per pound is the value divided by the number of pounds, which is (13f+9n)/(3f+n). In each problem they give you the weight and the price per pound, and ask you to find the weights of the individual items. So set the two related expressions equal to the two numbers, and solve these as two equations with two unknowns. For example, in part "a" we have: 3f+n = 140 (13f+9n)/(3f+n) = 6
This is driving me up the WALL! :[ Matrices problem, help?
Let x be the number of $1 tickets sold, and y the number of $2 tickets sold. Since the total number of tickets sold is 239 we have x + y = 239 And since the total income is $322 we have x + 2y =322 There's your system of two equations in two variables. To solve using matrices, you just make the augmented coefficient matrix: [1 1 | 239] [1 2 | 322] Now reduce: subtract the first row from the second: [1 1 | 239] [0 1 | 83] Then subtract the second row from the first: [1 0 | 156] [0 1 | 83] Now just read off the answers: (x,y) = (156,83)