Question

Fit by the method of least squares the plane \(z=a+b x+c y\) to the five points \((x, y, z):(-1,-2,5),(0,-2,4),(0,0,4),(1,0,2),(2,1,0)\).

Short answer

The steps lead to the equation of the plane that best fits the given points. This equation will be in the form \(z=a+b x+c y\), where a, b, and c are the coefficients found in step 4.

Step-by-step solution

Set up the matrix equation

We can set up the matrix equation in the form \(AZ = B\), where A is a matrix consists of the xy-values of the five points and Z includes the unknowns a, b, and c. B is a vector whose elements are the z-values of the five points. This gives: \[A = \begin{bmatrix} 1 & -1 & -2\ 1 & 0 & -2\1 & 0 & 0\ 1 & 1 & 0\ 1 & 2 & 1\end{bmatrix}\] , \[Z =\begin{bmatrix} a\ b\ c\end{bmatrix}\] and \[B =\begin{bmatrix} 5\ 4\ 4\ 2\ 0\end{bmatrix}\]

Solve for Z using the normal equation

From this, we can use the normal equation to calculate the unknowns a, b, and c. The normal equation is given by \((A^T A)Z = (A^T)B\). It is important to note that \(A^T\) is the transpose of matrix A. The goal here is to solve for Z which includes the variables a, b, and c.

Compute the matrices \(A^T A\) and \(A^T B\)

Using the given values from A and B, we calculate \(A^T A\) and \(A^T B\), by multiplying matrix A by its transpose, and then multiplying the resulting matrix by Z. Similarly, we multiply the transpose of A by B.

Solve for Z

We now have all the pieces to solve the normal equation for Z. This step involves performing the operation \((A^T A)^{-1}(A^T B)\) to find Z which consists of a, b, and c. This can be done using matrix algebra, Gauss-Jordan elimination or other methods.

Interpret the results

The resulting values of a, b, and c provide the coefficients for the equation of the plane that best fits the given points in the least squares sense. Each coefficient is indicative of the degree to which x, y, or the constant term contribute to the z-value of a point on the plane.