# Find a least square solution of the inconsistent system Ax =  b forA = $$\begin{bmatrix}-1 & 2\\ 2 & -3\\ -1 & 3\end{bmatrix}$$, b = $$\begin{bmatrix}4\\ 2\\ 1\end{bmatrix}$$

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

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

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

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

= $$\begin{bmatrix}3\\ \frac{19}{11}\end{bmatrix}$$