If a matrix A has linearly independent columns, then what is the formula for the least-squares solution to Ax=b? Why does the formula for your previous answer not work if A does not have linearly independent columns?

If a matrix A has linearly independent columns, then what is the formula for the least-squares solution to Ax=b? Why does the formula for your previous answer not work if A does not have linearly independent columns?

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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From my Linear Algebra class (ungraded) pratice work: "If AB is a defined product of matrices A and B, and the columns of AB are linearly independent, is it necessarily the case that the columns of B are also linearly independent?" So I got this far: (AB)x = 0 has only the trivial solution by the associative law of multiplication, (AB)x = A(Bx) = 0 from here, I'm not sure how to show the columns of B must be linearly independent.
Ex. 2 By use of the augmented matrix, solve (3x+Sy=9 2x+3y=5
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For matrix A = if g(X) = X2 + 4x + 2, what is the answer to g(A)? %|
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.
Please don't provide handwritten solution ....
By using the method of least squares, find the best line through the points: (−2,−1), (2,1), (0,–3). Step 1. The general equation of a line is c + ₁ = y. Plugging the data points into this formula gives a matrix equation Ac = y. Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation ATA ĉ = A¹y ATA= ATy = Step 3. Solving the normal equation gives the answer Ĉ= which corresponds to the formula y = Analysis. Compute the predicted y values: y ŷ = Compute the error vector: e=y-ŷ. = Aĉ. Compute the total error: SSE = e² + e² + ez. SSE =
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Please help me to solve the problem on the screenshots.