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By using the method of least squares, find the best line through the points: (2,-3), (-2,0), (1,-1). Step 1. The general equation of a line is co + c₁ = 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 ATAê= ATy 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 = Aê. Compute the error vector: ey ŷ. e= Compute the total error: SSE = e+ e + ez. SSE= 300

By using the method of least squares, find the best line through the points: (2,-3), (-2,0), (1,-1). Step 1. The general equation of a line is co + c₁ = 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 ATAê= ATy 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 = Aê. Compute the error vector: ey ŷ. e= Compute the total error: SSE = e+ e + ez. SSE= 300

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter6: Matrices And Determinants
Section: Chapter Questions
Problem 4CC
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By using the method of least squares, find the best line through the points: (2,-3), (-2,0), (1,-1). Step 1. The general equation of a line is co + C₁ = y. Plugging the data points into this formula gives a matrix equation Ac = y. 2- = Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation A¹A = A¹y ATA= A¹y = Step 3. Solving the normal equation gives the answer Ĉ= which corresponds to the formula y = Analysis. Compute the predicted y values: y = Aĉ. ŷ = Compute the error vector: e=y-ŷ. e= Compute the total error: SSE = e+ e + ez. SSE =
Transcribed Image Text:By using the method of least squares, find the best line through the points: (2,-3), (-2,0), (1,-1). Step 1. The general equation of a line is co + C₁ = y. Plugging the data points into this formula gives a matrix equation Ac = y. 2- = Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation A¹A = A¹y ATA= A¹y = Step 3. Solving the normal equation gives the answer Ĉ= which corresponds to the formula y = Analysis. Compute the predicted y values: y = Aĉ. ŷ = Compute the error vector: e=y-ŷ. e= Compute the total error: SSE = e+ e + ez. SSE =
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