The solution set of the system of linear equations
Answer to Problem 1RE
The solution set of linear equations
Explanation of Solution
Given:
The system of linear equations is
Procedure used:
“Solving system of linear equations by elimination:
(1) If the equations are not in the standard form, then they are converted accordingly into a standard form.
(2) The variable to be eliminated is recognized from the given equations in the system.
(3) After obtaining the LCM (lowest common multiple) of coefficients of the variable to be eliminated, the equations are multiplied with appropriate multipliers so that the coefficients in each equation become additive inverses and adding the new equations will result in a fresh equation with the decided variable to been eliminated.
(4) The equation in step (3) is solved obtaining value of remaining variable.
(5) The value of variable in step (4) is substituted to the equation in any of the given system equations.
(6) Solving the equation obtained in step (5) gives the value of variable eliminated in step (3).
(7) The correctness of solution is asserted by substituting back the values in given equations of the system.
Calculation:
The given equations are, as follows:
Step 1:
Since the given equations are already in standard form, nothing has to be done.
Step 2:
The coefficient of y in equation (1) is
Step 3:
Multiply equation (1) with number
Now, add resultant equation of equation (1) and equation (2) of given system of linear equations:
The above addition gives the equation
Step 4:
Solve the equation obtained in step 3 to obtain the value of
Thus, the value of x is
Step-5:
Now, substitute
Step-6:
The equation obtained in Step (5) is solved:
Thus, the value of y is
Step-7:
In order to check whether the solution is correct or not, substitute
Substitute
Therefore, the point
Substitute
Therefore, the point
Hence, it is asserted that the value of x is
Thus, the solution set for the system of linear equations
Want to see more full solutions like this?
Chapter 12 Solutions
Algebra And Trigonometry (11th Edition)
- Question 3 over a field K. In this question, MË(K) denotes the set of n × n matrices (a) Suppose that A Є Mn(K) is an invertible matrix. Is it always true that A is equivalent to A-¹? Justify your answer. (b) Let B be given by 8 B = 0 7 7 0 -7 7 Working over the field F2 with 2 elements, compute the rank of B as an element of M2(F2). (c) Let 1 C -1 1 [4] [6] and consider C as an element of M3(Q). Determine the minimal polynomial mc(x) and hence, or otherwise, show that C can not be diagonalised. [7] (d) Show that C in (c) considered as an element of M3(R) can be diagonalised. Write down all the eigenvalues. Show your working. [8]arrow_forwardR denotes the field of real numbers, Q denotes the field of rationals, and Fp denotes the field of p elements given by integers modulo p. You may refer to general results from lectures. Question 1 For each non-negative integer m, let R[x]m denote the vector space consisting of the polynomials in x with coefficients in R and of degree ≤ m. x²+2, V3 = 5. Prove that (V1, V2, V3) is a linearly independent (a) Let vi = x, V2 = list in R[x] 3. (b) Let V1, V2, V3 be as defined in (a). Find a vector v € R[×]3 such that (V1, V2, V3, V4) is a basis of R[x] 3. [8] [6] (c) Prove that the map ƒ from R[x] 2 to R[x]3 given by f(p(x)) = xp(x) — xp(0) is a linear map. [6] (d) Write down the matrix for the map ƒ defined in (c) with respect to the basis (2,2x + 1, x²) of R[x] 2 and the basis (1, x, x², x³) of R[x] 3. [5]arrow_forwardQuestion 4 (a) The following matrices represent linear maps on R² with respect to an orthonormal basis: = [1/√5 2/√5 [2/√5 -1/√5] " [1/√5 2/√5] A = B = [2/√5 1/√5] 1 C = D = = = [ 1/3/5 2/35] 1/√5 2/√5 -2/√5 1/√5' For each of the matrices A, B, C, D, state whether it represents a self-adjoint linear map, an orthogonal linear map, both, or neither. (b) For the quadratic form q(x, y, z) = y² + 2xy +2yz over R, write down a linear change of variables to u, v, w such that q in these terms is in canonical form for Sylvester's Law of Inertia. [6] [4]arrow_forward
- part b pleasearrow_forwardQuestion 5 (a) Let a, b, c, d, e, ƒ Є K where K is a field. Suppose that the determinant of the matrix a cl |df equals 3 and the determinant of determinant of the matrix a+3b cl d+3e f ГЪ e [ c ] equals 2. Compute the [5] (b) Calculate the adjugate Adj (A) of the 2 × 2 matrix [1 2 A = over R. (c) Working over the field F3 with 3 elements, use row and column operations to put the matrix [6] 0123] A = 3210 into canonical form for equivalence and write down the canonical form. What is the rank of A as a matrix over F3? 4arrow_forwardQuestion 2 In this question, V = Q4 and - U = {(x, y, z, w) EV | x+y2w+ z = 0}, W = {(x, y, z, w) € V | x − 2y + w − z = 0}, Z = {(x, y, z, w) € V | xyzw = 0}. (a) Determine which of U, W, Z are subspaces of V. Justify your answers. (b) Show that UW is a subspace of V and determine its dimension. (c) Is VU+W? Is V = UW? Justify your answers. [10] [7] '00'arrow_forward
- Tools Sign in Different masses and Indicated velocities Rotational inert > C C Chegg 39. The balls shown have different masses and speeds. Rank the following from greatest to least: 2.0 m/s 8.5 m/s 9.0 m/s 12.0 m/s 1.0 kg A 1.2 kg B 0.8 kg C 5.0 kg D C a. The momenta b. The impulses needed to stop the balls Solved 39. The balls shown have different masses and speeds. | Chegg.com Images may be subject to copyright. Learn More Share H Save Visit > quizlet.com%2FBoyE3qwOAUqXvw95Fgh5Rw.jpg&imgrefurl=https%3A%2F%2Fquizlet.com%2F529359992%2Fc. Xarrow_forwardSimplify the below expression. 3 - (-7)arrow_forward(6) ≤ a) Determine the following groups: Homz(Q, Z), Homz(Q, Q), Homz(Q/Z, Z) for n E N. Homz(Z/nZ, Q) b) Show for ME MR: HomR (R, M) = M.arrow_forward
- 1. If f(x² + 1) = x + 5x² + 3, what is f(x² - 1)?arrow_forward2. What is the total length of the shortest path that goes from (0,4) to a point on the x-axis, then to a point on the line y = 6, then to (18.4)?arrow_forwardموضوع الدرس Prove that Determine the following groups Homz(QZ) Hom = (Q13,Z) Homz(Q), Hom/z/nZ, Qt for neN- (2) Every factor group of adivisible group is divisble. • If R is a Skew ficald (aring with identity and each non Zero element is invertible then every R-module is free.arrow_forward
- Algebra and Trigonometry (6th Edition)AlgebraISBN:9780134463216Author:Robert F. BlitzerPublisher:PEARSONContemporary Abstract AlgebraAlgebraISBN:9781305657960Author:Joseph GallianPublisher:Cengage LearningLinear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning
- Algebra And Trigonometry (11th Edition)AlgebraISBN:9780135163078Author:Michael SullivanPublisher:PEARSONIntroduction to Linear Algebra, Fifth EditionAlgebraISBN:9780980232776Author:Gilbert StrangPublisher:Wellesley-Cambridge PressCollege Algebra (Collegiate Math)AlgebraISBN:9780077836344Author:Julie Miller, Donna GerkenPublisher:McGraw-Hill Education