Suppose that Ax=b has no solution and the columns of A are linearly independent. Let t(A) denote the transpose of matrix A. Select all true statements. O If the A matrix and b vector represent data values, then the least-squares solution gives coefficients for a function that make this function the line that best fits the data values. The system has a least-squares solution. O The solution to t(A)Ax=t(A)b is the least-squares solution. O In the system Ax-b=e, it is possible for e to be the zero vector. O The system has a least squares solution but it may not be unique. O The inverse of t(A)A exists. O The least-squares solution (if it exists) yields a vector in the column space of A that is the "closest" to the vector b.

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.2: Direct Methods For Solving Linear Systems
Problem 3BEXP
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Suppose that Ax=b has no solution and the columns of A are linearly independent. Let t(A) denote
the transpose of matrix A.
Select all true statements.
O If the A matrix and b vector represent data values, then the least-squares solution gives coefficients for a
function that make this function the line that best fits the data values.
O The system has a least-squares solution.
The solution to t(A)Ax=t(A)b is the least-squares solution.
O In the system Ax-b=e, it is possible for e to be the zero vector.
O The system has a least squares solution but it may not be unique.
O The inverse of t(A)A exists.
O The least-squares solution (if it exists) yields a vector in the column space of A that is the "closest" to the
vector b.
Transcribed Image Text:Suppose that Ax=b has no solution and the columns of A are linearly independent. Let t(A) denote the transpose of matrix A. Select all true statements. O If the A matrix and b vector represent data values, then the least-squares solution gives coefficients for a function that make this function the line that best fits the data values. O The system has a least-squares solution. The solution to t(A)Ax=t(A)b is the least-squares solution. O In the system Ax-b=e, it is possible for e to be the zero vector. O The system has a least squares solution but it may not be unique. O The inverse of t(A)A exists. O The least-squares solution (if it exists) yields a vector in the column space of A that is the "closest" to the vector b.
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