University:
California State University, Northridge
Course:
MATH 262 | Introduction to Linear Algebra
Academic year:

2024
Views:

163

Pages:

9
Author:

Beckett Gray

Question 1 I need to find a unique description of Nul A, namely by listing the vectors that measure the null space. 𝐴 = [ 1 5 − 4 − 3 1 0 1 − 2 1 0 0 0 0 0 0 ] A= 1 0 0 5 1 0 −4 −2 0 −3 1 0 1 0 0 Answer 1 First, find the general solution of 𝐴 𝑥 = 0 Ax=0 in terms of the free variables. Since [ 𝐴 ∣ 0 ] = [ 1 5 − 4 − 3 1 0 0 1 − 2 1 0 0 0 0 0 0 0 0 ] → 𝑅 1 → 𝑅 1 − 5 𝑅 2 [ 1 0 6 − 8 1 0 0 1 − 2 1 0 0 0 0 0 0 0 0 ] . [A∣0]= 1 0 0 5 1 0 −4 −2 0 −3 1 0 1 0 0 0 0 0 R 1 →R 1 −5R 2 1 0 0 0 1 0 6 −2 0 −8 1 0 1 0 0 0 0 0 . The general solution is: 𝑥 1 = − 6 𝑥 3 + 8 𝑥 4 − 𝑥 5 , 𝑥 2 = 2 𝑥 3 − 𝑥 4 , x 1 =−6x 3 +8x 4 −x 5 ,x 2 =2x 3 −x 4 , with 𝑥 3 , 𝑥 4 , 𝑥 5 x 3 ,x 4 ,x 5 free. So, 𝑥 = [ 𝑥 1 𝑥 2 𝑥 3 𝑥 4 𝑥 5 ] = 𝑥 3 [ − 6 2 1 0 0 ] + 𝑥 4 [ 8 − 1 0 1 0 ] + 𝑥 5 [ − 1 0 0 0 1 ] . x= x 1 x 2 x 3 x 4 x 5 =x 3 −6 2 1 0 0 +x 4 8 −1 0 1 0 +x 5 −1 0 0 0 1 . A spanning set for Nul A is: { [ − 6 2 1 0 0 ] , [ 8 − 1 0 1 0 ] , [ − 1 0 0 0 1 ] } . ⎩ ⎨ ⎧ −6 2 1 0 0 , 8 −1 0 1 0 , −1 0 0 0 1 ⎭ ⎬ ⎫ . Answer 2 To find a unique description of the null space of matrix 𝐴 A, we need to list the vectors that measure the null space. The matrix 𝐴 A is given by: 𝐴 = [ 1 5 − 4 − 3 1 0 1 − 2 1 0 0 0 0 0 0 ] . A= 1 0 0 5 1 0 −4 −2 0 −3 1 0 1 0 0 . To determine the null space, we solve the equation 𝐴 𝑥 = 0 Ax=0, where 𝑥 x is a vector in the null space and 0 0 is the zero vector. Writing this in augmented form, we have: [ 1 5 − 4 − 3 1 0 0 1 − 2 1 0 0 0 0 0 0 0 0 ] . 1 0 0 5 1 0 −4 −2 0 −3 1 0 1 0 0 0 0 0 . Simplifying the matrix equation, we get the following system of equations: 𝑥 1 + 5 𝑥 2 − 4 𝑥 3 − 3 𝑥 4 + 𝑥 5 = 0 , 𝑥 2 − 2 𝑥 3 + 𝑥 4 = 0. x 1 +5x 2 −4x 3 −3x 4 +x 5 =0,x 2 −2x 3 +x 4 =0. To find the null space, we solve this system of equations. First, we can express 𝑥 1 , 𝑥 2 x 1 ,x 2 , and 𝑥 5 x 5 in terms of the remaining variables: 𝑥 1 = − 6 𝑥 3 + 8 𝑥 4 − 𝑥 5 , 𝑥 2 = 2 𝑥 3 − 𝑥 4 . x 1 =−6x 3 +8x 4 −x 5 ,x 2 =2x 3 −x 4 . Now, we can write the general solution for the null space vector 𝑥 x as follows: 𝑥 = [ 𝑥 1 𝑥 2 𝑥 3 𝑥 4 𝑥 5 ] = 𝑥 3 [ − 6 2 1 0 0 ] + 𝑥 4 [ 8 − 1 0 1 0 ] + 𝑥 5 [ − 1 0 0 0 1 ] , x= x 1 x 2 x 3 x 4 x 5 =x 3 −6 2 1 0 0 +x 4 8 −1 0 1 0 +x 5 −1 0 0 0 1 , where 𝑥 3 , 𝑥 4 , x 3 ,x 4 , and 𝑥 5 x 5 are arbitrary constants. Thus, the unique description of the null space of matrix 𝐴 A is given by the set of vectors: { [ − 6 2 1 0 0 ] , [ 8 − 1 0 1 0 ] , [ − 1 0 0 0 1 ] } . ⎩ ⎨ ⎧ −6 2 1 0 0 , 8 −1 0 1 0 , −1 0 0 0 1 ⎭ ⎬ ⎫ . Answer 3 The null space of a matrix 𝐴 A is the set of all solutions 𝑥 x to the equation 𝐴 𝑥 = 0 Ax=0, where 0 0 is the zero vector. For the given matrix 𝐴 = [ 1 5 − 4 − 3 1 0 1 − 2 1 0 0 0 0 0 0 ] A= 1 0 0 5 1 0 −4 −2 0 −3 1 0 1 0 0 , we need to find the null space. Writing the equation in augmented form and reducing the matrix to echelon form, we get: [ 1 5 − 4 − 3 1 0 0 1 − 2 1 0 0 0 0 0 0 0 0 ] → 𝑅 1 → 𝑅 1 − 5 𝑅 2 [ 1 0 6 − 8 1 0 0 1 − 2 1 0 0 0 0 0 0 0 0 ] . 1 0 0 5 1 0 −4 −2 0 −3 1 0 1 0 0 0 0 0 R 1 →R 1 −5R 2 1 0 0 0 1 0 6 −2 0 −8 1 0 1 0 0 0 0 0 . From this, we can see that the first and fourth columns of the matrix contain the leading variables 𝑥 1 x 1 and 𝑥 2 x 2 , respectively. The second and third columns contain the free variables 𝑥 3 x 3 and 𝑥 5 x 5 , respectively. Using the leading and free variables, we can express the general solution for the null space vector 𝑥 x as: 𝑥 = 𝑥 3 [ − 6 2 1 0 0 ] + 𝑥 4 [ 8 − 1 0 1 0 ] + 𝑥 5 [ − 1 0 0 0 1 ] , x=x 3 −6 2 1 0 0 +x 4 8 −1 0 1 0 +x 5 −1 0 0 0 1 , where 𝑥 3 , 𝑥 4 , x 3 ,x 4 , and 𝑥 5 x 5 are arbitrary constants. Therefore, the null space of matrix 𝐴 A can be described as the span of the set of vectors: { [ − 6 2 1 0 0 ] , [ 8 − 1 0 1 0 ] , [ − 1 0 0 0 1 ] } , ⎩ ⎨ ⎧ −6 2 1 0 0 , 8 −1 0 1 0 , −1 0 0 0 1 ⎭ ⎬ ⎫ , which consists of three linearly independent vectors and thus forms a basis for the null space.