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PROBLEM 2. Let the following matrix and vector be given. [2 2 -1 -3] 1 2 -1 -4 A = and y= 2 -1 1 1 -2 1 6 1. Determine the rank of the matrix. 2. Compute bases for the column space, row space and null space of the matrix. 3. Determine if y is in the null space or column space of A.

PROBLEM 2. Let the following matrix and vector be given. [2 2 -1 -3] 1 2 -1 -4 A = and y= 2 -1 1 1 -2 1 6 1. Determine the rank of the matrix. 2. Compute bases for the column space, row space and null space of the matrix. 3. Determine if y is in the null space or column space of A.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.2: Linear Independence, Basis, And Dimension
Problem 5AEXP
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Hello bartle

I marked the 2 questions i need help for.

in question 2 i dont undestand how my teacher reach the solution
x4 [ -1 2 -1 1 ] 

and in question 3 solution i dont understand whats happening either in the red marked circle. how does it get Ay = [ 51 57 -52 -69 ] 

This entire picture contains the question and my teachers solution

PROBLEM 2. Let the following matrix and vector be given. [2 2 -1 1 2 A = -1 -4 and y = 2 -1 1 5 1 1. Determine the rank of the matrix. 2. Compute bases for the column space, row space and null space of the matrix. 3. Determine if y is in the null space or column space of A. PROBLEM 2. Solution The rank is determined by row reducing the matrix and counting the number of pivots (encircled below). O 0 0 [2 2 1 2 2 -1 -2 1 -1 1. -1 -4 -2 1 0 0 0 1 6. As seen, rank A=3. The pivot columns of A form a basis for col A. basis for col A = The row columns of the row reduced matrix forms a basis for row A. basis for row A = {[1 0 0 1],[0 1 0 -2] , [0 0 1 1]}. The basis for null A is found by solving Ax = 0. This can be done based on the above row reduction, where it is seen that x4 is a free variable. The solution is Therefore basis for null A = It can be checked if y is in the null space by direct computation: 51 57 Ay = -52 69
Transcribed Image Text:PROBLEM 2. Let the following matrix and vector be given. [2 2 -1 1 2 A = -1 -4 and y = 2 -1 1 5 1 1. Determine the rank of the matrix. 2. Compute bases for the column space, row space and null space of the matrix. 3. Determine if y is in the null space or column space of A. PROBLEM 2. Solution The rank is determined by row reducing the matrix and counting the number of pivots (encircled below). O 0 0 [2 2 1 2 2 -1 -2 1 -1 1. -1 -4 -2 1 0 0 0 1 6. As seen, rank A=3. The pivot columns of A form a basis for col A. basis for col A = The row columns of the row reduced matrix forms a basis for row A. basis for row A = {[1 0 0 1],[0 1 0 -2] , [0 0 1 1]}. The basis for null A is found by solving Ax = 0. This can be done based on the above row reduction, where it is seen that x4 is a free variable. The solution is Therefore basis for null A = It can be checked if y is in the null space by direct computation: 51 57 Ay = -52 69
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