QUESTION 4 \begin{bmatrix} 1 & 2 & -1 & 2 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \end{bmatrix} Let $A = \begin{bmatrix} 1 & 2 & -1 & 2 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \end{bmatrix}$ be a row-echelon form of A. a) Find a basis of the nullspace of A. b) If $c_1, c_2, c_3, c_4, c_5$ are the column vectors of A, write $c_2$ and $c_5$ as linear combination of $c_1, c_3, c_4$.
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Let A = [1 2 -1 2 1 0 0 1 1 1 0 0 0 1 1], x = (x1, x2, x3, x4, x5)^T, and solve A x = 0. Show more…
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The matrix A = egin{bmatrix} 1 & 2 & 2 & -5 \ -1 & -2 & -1 & 1 \ 4 & 8 & 5 & -8 \ 3 & 6 & 1 & 5 end{bmatrix} can be carried to the reduced row echelon matrix egin{bmatrix} 1 & 2 & 0 & 3 \ 0 & 0 & 1 & -4 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 end{bmatrix} by elementary row operations. a. Find a basis for the null space of A. b. Find a basis for the row space of A. c. Exhibit some columns of A that form a basis for the column space of A. d. Verify the Rank + Nullity Theorem for A. e. Does A have a left inverse? Does A have a right inverse? Explain.
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Section 3.4 Basis and Dimension: Problem 10 Problem: Let A be a matrix given by A = Find the basis of the nullspace(A). Answer: To find the basis of the nullspace, we need to find the vectors that satisfy the equation Ax = 0, where x is a vector of variables. These vectors will form the basis of the nullspace. To solve the equation Ax = 0, we can row reduce the augmented matrix [A | 0] and find the special solutions. Once we have the special solutions, we can express them as a linear combination of vectors to find the basis. Let's solve the equation Ax = 0: [A | 0] = Row reducing the augmented matrix, we get: [R | 0] = From the row reduced form, we can see that the variables x1, x2, and x3 are free variables. We can express the special solutions in terms of these free variables: x1 = -2x3 x2 = x3 x3 = x3 We can choose any value for x3 and find the corresponding values for x1 and x2. Let's choose x3 = 1: x1 = -2(1) = -2 x2 = 1 x3 = 1 Therefore, one special solution is x = [-2, 1, 1]. To find the basis, we express the special solution as a linear combination of vectors: [-2, 1, 1] = -2[1, 0, 0] + 1[0, 1, 0] + 1[0, 0, 1] Therefore, the basis of the nullspace(A) is {[1, 0, 0], [0, 1, 0], [0, 0, 1]}.
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15) A matrix in reduced row echelon form is row equivalent to A. Denote the four columns of A by c. a) Find a basis of the null space of the matrix A and state the dimension of the null space. A basis is: Dimension: b) Find a basis of the column space of the matrix A and state the dimension of the column space. You may use the column symbols c, c, c, c above in writing your answer. A basis is: Dimension: c) Find a basis of the row space of the matrix A and state the dimension of the row space. (You can write the basis vectors as horizontal rows.) A basis is: Dimension:
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