Question

Find a least-squares solution of $Ax = b$ by (a) constructing the normal equations for $x$ and (b) solving for $x$. $A = \begin{bmatrix} 1 & -3 \ -1 & 3 \ 0 & 4 \ 3 & 9 \ 2 & 2 \end{bmatrix}$ $b = \begin{bmatrix} 1 \ 1 \ -5 \ 2 \ 1 \end{bmatrix}$ a. Construct the normal equations for $x$ without solving. $x = \begin{bmatrix} \\ \end{bmatrix}$ (Simplify your answers.)

          Find a least-squares solution of $Ax = b$ by (a) constructing the normal equations for $x$ and (b) solving for $x$.
$A = \begin{bmatrix} 1 & -3 \ -1 & 3 \ 0 & 4 \ 3 & 9 \ 2 & 2 \end{bmatrix}$  $b = \begin{bmatrix} 1 \ 1 \ -5 \ 2 \ 1 \end{bmatrix}$
a. Construct the normal equations for $x$ without solving.
$x = \begin{bmatrix} \\ \end{bmatrix}$ (Simplify your answers.)
        
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Find a least-squares solution of Ax = b by (a) constructing the normal equations for x and (b) solving for x.
A = 
    < b m a t r i x >  b = 
    < b m a t r i x >
a. Construct the normal equations for x without solving.
x = 
    < b m a t r i x > (Simplify your answers.)

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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A = [6 3; -1 1-3] a. Construct the normal equations for A without solving. b = [2; -5; 7; 2] Find a least-squares solution of Ax=b by constructing the normal equations for A and solving for b.
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Transcript

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00:01 Hi, let's start the solution.
00:02 In this question, we have given matrix a is minus 1 2 minus 1 2 minus 3 3 and b is 4 2 3.
00:16 Here order of matrix a is 3 cross 2.
00:21 Then we have to find the least square solution of ax equal to b.
00:40 So first by constructing normal equation.
00:44 Now compute a transpose a which is equal to a transpose means interchange row and columns.
01:00 Then we get minus 1 2 minus 1 2 minus 3 3 and a is minus 1 2 minus 1 2 minus 3 3.
01:15 So by multiplying these two matrix here we get which is equal to 6 minus 11 minus 11 22.
01:24 So we get here a transpose a is 6 minus 11 minus 11 22.
01:34 Now we compute a transpose into b.
01:38 A transpose is minus 1 2 minus 1 2 minus 3 3 and b is 4 2 3.
01:47 By multiplying this here we get which is equal to minus 3 11.
01:53 So we get here a transpose b is minus 3 11.
01:58 So that means normal equation is a transpose a into x equal to a transpose b.
02:18 So we get here vector x cap x cap that is.
02:23 So a transpose a is 6 minus 11 minus 11 22 and this one is x 1 x 2 which is equal to a transpose b is minus 3 11...
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