Find a least square solution of the inconsistent system Ax =  b for

A = \(\begin{bmatrix}-1 & 2\\ 2 & -3\\ -1 & 3\end{bmatrix}\), b = \(\begin{bmatrix}4\\ 2\\ 1\end{bmatrix}\)

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Solution

Given,

A = \(\begin{bmatrix}-1 & 2\\ 2 & -3\\ -1 & 3\end{bmatrix}\) and b = \(\begin{bmatrix}4\\ 2\\ 1\end{bmatrix}\)

Now,

\(\hat{x} = (A^{T}A)^{-1} (A^{T}b)\)  ———– (1)

AT = \(\begin{bmatrix}-1 & 2 & -1\\ 2 & -3 & 3\end{bmatrix}\)

ATA = \(\begin{bmatrix}-1 & 2 & -1\\ 2 & -3 & 3\end{bmatrix}\) \(\begin{bmatrix}-1 & 2\\ 2 & -3\\ -1 & 3\end{bmatrix}\)

= \(\begin{bmatrix}1+4+1 & -2-6-3\\ -2-6-3 & 4+9+9\end{bmatrix}\)

= \(\begin{bmatrix}6 & -11\\ -11 & 22\end{bmatrix}\)

and

ATb = \(\begin{bmatrix}-1 & 2 & -1\\ 2 & -3 & 3\end{bmatrix}\) \(\begin{bmatrix}4\\ 2\\ 1\end{bmatrix}\)

= \(\begin{bmatrix}-4+4-1\\ 8-6+3\end{bmatrix}\)

= \(\begin{bmatrix}-1\\ 5\end{bmatrix}\)

Since we have definition of least square; the set of least square solutions of Ax = b coincides with the non-empty set of solutions of ATAx = ATb

therefore

\(\hat{x} = (A^{T}A)^{-1} (A^{T}b)\)

Here,

|ATA| = \(\begin{bmatrix}6 & -11\\ -11 & 22\end{bmatrix}\) = -11 ≠ 0

then

(ATA)-1 = \(\frac{1}{-11} \begin{bmatrix}22 & 11\\ 11 & 6\end{bmatrix}\)

then equation(1) becomes

\(\hat{x}\) = \(\frac{1}{-11} \begin{bmatrix}22 & 11\\ 11 & 6\end{bmatrix}\) \(\begin{bmatrix}-1\\ 5\end{bmatrix}\)

= \(\begin{bmatrix}2 & 1\\ 1 & \frac{6}{11}\end{bmatrix}\) \(\begin{bmatrix}-1\\ 5\end{bmatrix}\)

= \(\begin{bmatrix}-2-5\\ -1 + \frac{6}{11}\end{bmatrix}\)

= \(\begin{bmatrix}3\\ \frac{19}{11}\end{bmatrix}\)

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