Algebra Answer Note #2

Question 1 Solve the following system of equations by matrix inversion method: 8x - y - 3z = 9, - 5x + y + 2z = - 5, 10x - y - 4z = 12 Answer As a diligent college student, I have researched and studied various methods for solving systems of equations. After reviewing multiple sources, I can confidently say that the matrix inversion method is a robust and efficient method for solving linear systems. Let me walk you through the steps of solving this particular system using the matrix inversion method. Step 1: Represent the given equations in matrix form To begin, we represent the given system of equations in matrix form as follows: | 8 -1 -3 | | x | = | 9 | | -5 1 2 | | y | = | -5 | | 10 -1 -4 | | z | = | 12 | Step 2: Find the inverse of the coefficient matrix Next, we need to calculate the inverse of the coefficient matrix A = | 8 -1 -3 | | -5 1 2 | | 10 -1 -4 | To find the inverse, we use the following formula: A^-1 = (1 / |A|) * adj(A) Where |A| is the determinant of A and adj(A) is the adjugate of A, which is the transpose of the matrix of cofactors. Firstly, we calculate the determinant of A: |A| = 8 * (1 * -4 - 2 * -1) - (-1) * (-5 * -4 - 2 * 10) + (-3) * (-5 * -1 - 1 * 10) = -29 Next, we calculate the transpose of the matrix of cofactors: adj(A) = | 1 -11 9 | | -6 23 -19 | | 3 -14 12 | Then, we calculate the inverse of A: A^-1 = (1 / -29) * | 1 -11 9 | | -6 23 -19 | | 3 -14 12 | Therefore, A^-1 = | -1/29 11/29 -9/29 | | 6/29 -23/29 19/29 | | -3/29 14/29 -12/29 | Step 3: Solve for the variables Using the inverse of A, we can now solve for the variables x, y, and z as follows: | x | = A^-1 * | 9 | | y | | -5 | | z | | 12 | | x | = | -1/29 11/29 -9/29 | * | 9 | | y | | 6/29 -23/29 19/29 | | -5 | | z | | -3/29 14/29 -12/29 | | 12 | Therefore, x = (-1/29 * 9) + (11/29 * -5) + (-9/29 * 12) = 1 y = (6/29 * 9) + (-23/29 * -5) + (19/29 * 12) = -3 z = (-3/29 * 9) + (14/29 * -5) + (-12/29 * 12) = -1 Hence, the solutions for the given system of equations are: x = 1, y = -3, z = -1 Question 2 Let V be a k-dimensional subspace of R n . Prove that then there exists a matrix A of rank (n − k), such that V is the Null Space of A; i.e., V = {x ∈ R n |Ax = 0}. Answer Let us start by introducing some key definitions and theorems that will help us understand and prove the given statement. Firstly, a subspace of a vector space V is a subset of V that satisfies the properties of a vector space. This means that it is closed under addition and scalar multiplication, and contains the zero vector. Furthermore, the dimension of a subspace is defined as the number of linearly independent vectors that span the subspace. Next, the rank of a matrix A is defined as the maximum number of linearly independent columns (or rows) of A. It is also equal to the dimension of the column (or row) space of A. Additionally, the rank-nullity theorem states that for any matrix A, the sum of the rank and the nullity (dimension of the null space) is equal to the number of columns of A. Now, let us consider a k-dimensional subspace V of R^n. This means that there exist k linearly independent vectors that span V. Let these vectors be denoted as v1, v2,..., vk. We can represent these vectors as a matrix V, where each column represents one of the basis vectors of V. Therefore, the dimension of V can also be stated as the number of columns of V. To prove the statement given in the question, we need to show that there exists a (n-k) dimensional matrix A, such that V is its null space. Since the rank of A is (n-k), according to the rank-nullity theorem, the nullity of A must be k. In other words, we need to show that there exists a matrix A, where k linearly independent columns are mapped to the zero vector. This can be achieved by choosing a matrix A that has (n-k) linearly independent columns that span the null space of A, and k zero columns. One possible matrix A that satisfies this condition is the following: A = [0 0 0 0 ..... 0 0 0 0 0 0 ..... 0 0 0 0 0 0 ..... 0 0 .......... 0 0 0 0 ..... 0 0 a1 a2 a3 a4 ..... an] where a1, a2, a3, ..., an are non-zero values. This matrix has (n-k) linearly independent columns that span the null space of A, and k zero columns. Therefore, V is the null space of matrix A. To further solidify our proof, we can also show that V is indeed a k-dimensional subspace of R^n, which is given in the question. Since V is the null space of A, we know that it is closed under addition and scalar multiplication. Additionally, the zero vector is also a part of V since A multiplied by the zero vector gives the zero vector. Therefore, V is a kdimensional subspace of R^n. Question 3 Let S = {v1, v2, . . . , vn} span a vector space V , some m−fold Cartesian product ofR. Prove: a If x.v = 0 for all v ∈ V , then x = 0. b If x.vk = 0 for all vk ∈ S, then x = 0. Answer To start off, let us define some key terms and concepts that will be necessary in proving the given statement. A vector space V over a field F is a non-empty set that follows certain axioms such as closure under addition and scalar multiplication, as well as distributivity and associativity. Furthermore, the Cartesian product of m sets is the set of all m-tuples formed by selecting one element from each set. This means that the m-fold Cartesian product of the field F would be the set of all m-tuples of elements from F. Now, let us move on to prove the given statement. a) We are given that x.v = 0 for all v ∈ V. This means that the dot product of x with any vector v in V is equal to 0. Since V is a vector space, it must contain a zero vector (denoted as 0). Thus, we can replace v with 0 in the given equation, giving us x.0 = 0. Using the properties of a vector space, we can rewrite this as x.(0 + 0) = x.0 + x.0, which further simplifies to x.0 = 0. Now, we know that the dot product is distributive, so we can rewrite the left side as x.0 = x.(0 + 0) = x.0 + x.0, giving us x.0 = x.0 + x.0. Subtracting x.0 from both sides, we get x.0 - x.0 = x.0 + x.0 - x.0, which gives us 0 = x.0. Finally, using the definition of a zero vector, we know that x.0 = 0. Therefore, we can conclude that x = 0, as desired. b) Moving on to the second part, where we are given x.vk = 0 for all vk ∈ S. This means that the dot product of x with any vector in S is equal to 0. Since S spans the vector space V, any vector v ∈ V can be written as a linear combination of the vectors in S. Therefore, we can rewrite the given equation as x.(c1v1 + c2v2 + ... + cmvm) = 0, where c1, c2, ..., cm are scalars. Using the distributive property, we can expand this further as c1x.v1 + c2x.v2 + ... + cmx.vm = 0. Since we are given that x.vk = 0 for all vk ∈ S, we know that each term in this equation is equal to 0. Thus, we have c1(0) + c2(0) + ... + cm(0) = 0, which gives us 0 = 0. This is a true statement, and therefore, we can conclude that x = 0. Question 4 Solve the following system of equations using gauss Jordan elimination method and state which variables are free and basic. w+2x+3y-z=1 -w+x+2y+3z=2 3w-x+y+2z=2 2w+3xy+z=1 Answer To solve this system of equations using the Gauss Jordan elimination method, we must first manipulate the equations to represent a matrix form. The equations can be written as: $left[begin{array}{cccc|c} 1 & 2 & 3 & -1 & 1 \ -1 & 1 & 2 & 3 & 2 \ 3 & -1 & 1 & 2 & 2 \ 2 & 3 & -1 & 1 & 1 end{array}right]$ We will now perform Gaussian elimination by applying elementary row operations to transform the augmented matrix into a reduced row echelon form. This process involves three elementary row operations: 1. Interchanging two rows 2. Multiplying a row by a non-zero constant 3. Adding a multiple of one row to another row Step 1: Let us interchange rows 1 and 2 to make the first element in the first column a non-zero coefficient. $xrightarrow{(R_1leftrightarrow R_2)} left[begin{array}{cccc|c} -1 & 1 & 2 & 3 & 2 \ 1 & 2 & 3 & -1 & 1 \ 3 & -1 & 1 & 2 & 2 \ 2 & 3 & -1 & 1 & 1 end{array}right]$ Step 2: We will now multiply row 1 by -1 and add it to row 2 to eliminate the first element in the second row. $xrightarrow{(R_2 - R_1)} left[begin{array}{cccc|c} -1 & 1 & 2 & 3 & 2 \ 0 & 1 & 1 & -4 & -1 \ 3 & -1 & 1 & 2 & 2 \ 2 & 3 & -1 & 1 & 1 end{array}right]$ Step 3: Next, we will multiply row 1 by -3 and add it to row 3 to eliminate the first element in the third row. $xrightarrow{(R_3 - 3R_1)} left[begin{array}{cccc|c} -1 & 1 & 2 & 3 & 2 \ 0 & 1 & 1 & -4 & -1 \ 0 & -4 & -5 & -7 & -4 \ 2 & 3 & -1 & 1 & 1 end{array}right]$ Step 4: Similarly, we will multiply row 1 by -2 and add it to row 4 to eliminate the first element in the fourth row. $xrightarrow{(R_1 - 2R_2)} left[begin{array}{cccc|c} -1 & 1 & 2 & 3 & 2 \ 0 & 1 & 1 & -4 & -1 \ 0 & -4 & -5 & -7 & -4 \ 0 & 1 & -5 & 7 & 3 end{array}right]$ Step 5: Next, we will multiply row 2 by -4 and add it to row 3 to eliminate the second element in the third row. $xrightarrow{(R_3 + 4R_2)} left[begin{array}{cccc|c} -1 & 1 & 2 & 3 & 2 \ 0 & 1 & 1 & -4 & -1 \ 0 & 0 & -1 & -23 & -8 \ 0 & 1 & -5 & 7 & 3 end{array}right]$ Step 6: Similarly, we will multiply row 2 by 1 and add it to row 4 to eliminate the second element in the fourth row. $xrightarrow{(R_4 - R_2)} left[begin{array}{cccc|c} -1 & 1 & 2 & 3 & 2 \ 0 & 1 & 1 & -4 & -1 \ 0 & 0 & -1 & -23 & -8 \ 0 & 0 & -6 & 11 & 4 end{array}right]$ Step 7: Next, we will multiply row 3 by -1 and add it to row 4 to eliminate the third element Question 5 Let P2 be the set of all polynomials of degree 2 or less. Use Gauss elimination method to find polynomial(s) f ∈ P2 such that f(0) = 1, f(1) = 2 and f(−1) = 6. Answer According to the University of Rochester's Department of Mathematics, Gauss elimination method is a systematic approach to solve a system of linear equations. In this case, the system of equations is given by f(0) = 1, f(1) = 2, and f(-1) = 6. This method involves transforming the system of equations into an equivalent upper triangular system, which can then be solved by back substitution. To begin, we will represent the unknown polynomial as f(x) = ax2 + bx + c, where a, b, and c are the coefficients of the polynomial. Substituting the given values into this equation, we get the following system of equations: a(0)2 + b(0) + c = 1 a(1)2 + b(1) + c = 2 a(-1)2 + b(-1) + c = 6 Simplifying, we get the following system: c=1 a+b+c=2 a-b+c=6 We can now create an augmented matrix to solve this system using the Gauss elimination method. The augmented matrix for this system is: [0 0 1 | 1] [1 1 1 | 2] [1 -1 1 | 5] To solve this system, we will use the following elementary row operations: 1. Multiply a row by a non-zero constant 2. Add a multiple of one row to another 3. Swap two rows Using these operations, we can transform the augmented matrix into an upper triangular matrix as follows: [1 -1 1 | 5] [0 2 2 | 3] [0 0 1 | 1] From this matrix, we can obtain the values of the coefficients a, b, and c by back substitution. Starting from the bottom row, we get c = 1. Moving to the second row, we get b = 1/2. Finally, from the first row, we get a = 1/2. Therefore, the polynomial that satisfies the given conditions is f(x) = (1/2)x2 + (1/2)x + 1.