Advanced Engineering Mathematics
6th Edition
ISBN: 9781284105902
Author: Dennis G. Zill
Publisher: Jones & Bartlett Learning
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Chapter 11, Problem 12CR
(a)
To determine
The geometric nature of the given general solutions to the linear system
(b)
To determine
The geometric nature of the given general solutions to the linear system
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2. What is the solution set to the system of equations y = 2x + 3 and y = g(x) where g(x) is defined by the
function below?
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A.
Which of the following is a solution to the given system of equations found below?
(1) x = – /Y
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(2) y? – x² = 6
O (-v2, 2)
O (3, V3)
O (-v2, –2)
O (-v3, 3)
2) Find the General Solution of the System of Equations.
[1 0 0]
X' = 2 1
-2 X
3 2
1
Chapter 11 Solutions
Advanced Engineering Mathematics
Ch. 11.1 - Prob. 1ECh. 11.1 - Prob. 2ECh. 11.1 - Prob. 3ECh. 11.1 - Prob. 4ECh. 11.1 - Prob. 5ECh. 11.1 - Prob. 6ECh. 11.1 - Prob. 7ECh. 11.1 - Prob. 8ECh. 11.1 - Prob. 9ECh. 11.1 - Prob. 10E
Ch. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 23ECh. 11.1 - Prob. 24ECh. 11.1 - Prob. 25ECh. 11.1 - Prob. 26ECh. 11.1 - Prob. 27ECh. 11.1 - Prob. 28ECh. 11.1 - Prob. 29ECh. 11.1 - Prob. 30ECh. 11.2 - Prob. 17ECh. 11.2 - Prob. 18ECh. 11.2 - Prob. 19ECh. 11.2 - Prob. 20ECh. 11.2 - Prob. 21ECh. 11.2 - Prob. 22ECh. 11.2 - Prob. 23ECh. 11.2 - Prob. 24ECh. 11.2 - Prob. 25ECh. 11.2 - Prob. 26ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.3 - Prob. 11ECh. 11.3 - Prob. 12ECh. 11.3 - Prob. 13ECh. 11.3 - Prob. 14ECh. 11.3 - Prob. 15ECh. 11.3 - Prob. 16ECh. 11.3 - Prob. 17ECh. 11.3 - Prob. 18ECh. 11.3 - Prob. 19ECh. 11.3 - Prob. 20ECh. 11.3 - Prob. 21ECh. 11.3 - Prob. 22ECh. 11.3 - Prob. 23ECh. 11.3 - Prob. 24ECh. 11.3 - Prob. 25ECh. 11.3 - Prob. 26ECh. 11.3 - Prob. 27ECh. 11.3 - Prob. 28ECh. 11.3 - Prob. 29ECh. 11.3 - Prob. 30ECh. 11.3 - Prob. 31ECh. 11.3 - Prob. 32ECh. 11.3 - Prob. 33ECh. 11.3 - Prob. 34ECh. 11.3 - Prob. 35ECh. 11.3 - Prob. 36ECh. 11.3 - Prob. 37ECh. 11.3 - Prob. 38ECh. 11.3 - Prob. 39ECh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 11ECh. 11.4 - Prob. 12ECh. 11.4 - Prob. 13ECh. 11.4 - Prob. 14ECh. 11.4 - Prob. 15ECh. 11.4 - Prob. 16ECh. 11.4 - Prob. 17ECh. 11.4 - Prob. 18ECh. 11.4 - Prob. 19ECh. 11.4 - Prob. 20ECh. 11.4 - Prob. 21ECh. 11.4 - Prob. 22ECh. 11 - Prob. 1CRCh. 11 - Prob. 2CRCh. 11 - Prob. 3CRCh. 11 - Prob. 4CRCh. 11 - Prob. 5CRCh. 11 - Prob. 6CRCh. 11 - Prob. 7CRCh. 11 - Prob. 8CRCh. 11 - Prob. 11CRCh. 11 - Prob. 12CRCh. 11 - Prob. 13CRCh. 11 - Prob. 15CRCh. 11 - Prob. 16CRCh. 11 - Prob. 17CR
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- Find a system of two equations in two variables, x1 and x2, that has the solution set given by the parametric representation x1=t and x2=3t4, where t is any real number. Then show that the solutions to the system can also be written as x1=43+t3 and x2=t.arrow_forwardFind a system of two equations in three variables, x1, x2 and x3 that has the solution set given by the parametric representation x1=t, x2=s and x3=3+st, where s and t are any real numbers. Then show that the solutions to the system can also be written as x1=3+st,x2=s and x3=t.arrow_forwardThe break-even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system (x) = 500,000 + 400x |R(x) = 600x [y = 500,000 + 400x or ly = 600x. Using the substitution method, we can substitute 600x for y in the first equation. 600x = 500,000 + 400x Substitute 600x for y in y= 500,000 + 400x. 200x 500,000 Subtract 400x from both sides. x = 2500 Divide both sides by 200. Back-substituting 2500 for x in either of the system's equations (or functions), we obtain R(2500) = 600(2500) = 1,500,000. We used R[x) = 60ox.arrow_forward
- 2. Solve the following system of nonlinear equation using Newton's Method. (you can use MS Excel) X6 f.(x) = 0.5x1 + x2 + 0.5x3 = 0 -- %3D X7 2 f2(x) = x3 + X4 + 2x5 -- - X7 1 f3(x) = x1 + X2 + x5 - - = f.(x) = -28837x1 – 139009x2 – 78213x3 + 18927x4 + 8427x5 %3D 13492 10690*é = 0 fs(x) = x1 + x2 + X3 + X4 + X5 – 1 = 0 fo(x) = 400x,x? – 1.7837×10$x3X5 = 0 f,(x) = x1X3 – 2.6058x4 = 0 %3Darrow_forward2) Find the General Solution of the System of Equations. i 1 1 X' =|2 1 -1X 0 -1 1arrow_forward(a) Find the general solution of the given system of equations. 3 -1 = ( ³ = ₂ ) x 4 -2 X x' x(t) = C₁ (3) ? +C₂ (?)arrow_forward
- 1. (a) Determine which one of the following equations are linear equation. If nonlinear identify the nonlinear terms. 2x – y+ z - t = sin- x +2+z + w = 4 (i) (ii) y (b) Solve the system of linear equations using Gauss-Jordan elimination technique 2x + y + 2z + 3t = 13 -x + 2y + 4z – t = -5 3x + 2y – 6z +t = 4 4x – 3y + 2z – 2t = 6 Write the general solution in vector form. Find the cubic polynomial function p(x) = a + bx + cx² + dx³ such that p(1) = 1, p'(1) = 5, p(-1) = 3, and p'(-1) = 1. 2. 3. (a) If A be an n xn invertible matrix. Prove that the inverse of A is unique. (b) Find the inverse of the matrix A using row operation if A is a non-singular matrix, where, [1 A = |2 3 Lo -2] 1 -1] Hence compute (3A)-1. 4. (a) Use row reduction to evaluate the determinant of the matrix, [2 3 -1 4 3 B = 1 -1 1 1 2 [2 1 -3 3] 1. 1 (b) Determine the values of x so that the matrix A = |1 x is invertible. Lx 5. (a) Find two vectors of norm 1 that are orthogonal to the vectors u = -2 v = (b) If u =…arrow_forwardSolve the system t d X1 - dt x2 -1 with x1(0) = 1 and x2(0)arrow_forward2. Solve the following system of nonlinear equation using Newton's Method. (you can use MS Excel) x6 f₁(x) = 0.5x₁ + x₂ +0.5x3- = 0 f₂(x) = x3 + x4 + 2x5 f3(x) = x₁ + x₂ + x5 f4(x) = -28837x₁ - 139009x2 - 78213x3 + 18927x4 +8427x5 + 2 -1=0 X7 1 -1=0 fs(x) = x₁ + x₂ + x3 + x4 + x5-1=0 400x₁x2-1.7837x105x3x5 = 0 f(x) = f(x) = x₁x3 - 2.6058x4 = 0 13492 x7 106906 = 0 x7arrow_forward
- I need the answer as soon as possiblearrow_forwardConsider the ecuation ky" – ky' – 210y = 0, with constant real k + 0. If one - %3D solution of the ecuation is: Y1(x) = e then another solution linearly %3D dependent of the equation with y1 is: O a) y2(z) = el7z O b) y2(x) = ekz Oc) y2(x) = e¬1llz %3D O d) y2(x) = e-5z %3Darrow_forward1. Find steady states of the equation: 2xn (а) хр+1 Xn+1 K (b) Xn+1 where k1, k2 and K are constants. ki + k2/Xnarrow_forward
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