Find the least-square solution of Ax = b for A = \(\begin{bmatrix}1 & 3 & 5\\ 1 & 1 & 0\\ 1 & 1 & 2\\ 1 & 3 & 3\end{bmatrix}\) and b= \(\begin{pmatrix}3\\ 5\\ 7\\ 3\end{pmatrix}\)

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Least Square Solution of Ax = b

Where A= \(\begin{bmatrix}1 & 3 & 5\\ 1 & 1 & 0\\ 1 & 1 & 2\\1 & 3 & 3\end{bmatrix}\), b = \(\begin{bmatrix}3\\5\\7\\3\end{bmatrix}\)

Here, Ax = b

ATAx = ATb

∴ x = (ATA)-1 ATB   ——— (i)

Now,

ATA = \(\begin{bmatrix}1 & 1 & 1 & 1\\ 3 & 1 & 1 & 3\\ 5 & 0 & 2 & 3\end{bmatrix}\)\(\begin{bmatrix}1 & 3 & 5\\ 1 & 1 & 0\\ 1 & 1 & 2\\1 & 3 & 3\end{bmatrix}\)

= \(\begin{bmatrix}1+1+1+1 & 3+1+1+3 & 5+0+2+3\\ 3+1+1+3 & 9+1+1+9 & 15+0+2+9\\ 5+0+2+3 & 15+0+2+9 & 25+0+4+9\end{bmatrix}\)

= \(\begin{bmatrix}4 & 8 & 10\\ 8 & 20 & 26\\ 10 & 26 & 38\end{bmatrix}\)

ATB = \(\begin{bmatrix}1 & 1 & 1 & 1\\ 3 & 1 & 1 & 3\\ 5 & 0 & 2 & 3\end{bmatrix}\) \(\begin{bmatrix}3\\5\\7\\3\end{bmatrix}\)

= \(\begin{bmatrix}18\\20\\3\end{bmatrix}\)

Now (ATA)-1

det(ATA) = \(\begin{vmatrix}4 & 8 & 10\\ 8 & 20 & 26\\ 10 & 26 & 38\end{vmatrix}\)

= 4\(\begin{vmatrix}20 & 26\\ 26 & 38\end{vmatrix}\) – 8\(\begin{vmatrix}8 & 26\\ 10 & 38\end{vmatrix}\) + 10\(\begin{vmatrix}8 & 20\\ 10 & 26\end{vmatrix}\)

= 4 × 84 – 44 × 8 + 10 × 8

= 64

Now,

a11 = \(\begin{vmatrix}20 & 26\\ 26 & 38\end{vmatrix}\) = 84

a12 = -\(\begin{vmatrix}8 & 26\\ 10 & 38\end{vmatrix}\) = -44

a13 = \(\begin{vmatrix}8 & 20\\ 10 & 26\end{vmatrix}\) = 8

a21 = \(\begin{vmatrix}8 & 10\\ 26 & 38\end{vmatrix}\) = -44

a22 = \(\begin{vmatrix}4 & 10\\ 10 & 38\end{vmatrix}\) = 52

a23 = -\(\begin{vmatrix}4 & 8\\ 10 & 26\end{vmatrix}\) = -24

a31 = \(\begin{vmatrix}8 & 10\\ 26 & 38\end{vmatrix}\) = 44

a32 = -\(\begin{vmatrix}4 & 10\\ 8 & 26\end{vmatrix}\) = -24

a33 = \(\begin{vmatrix}4 & 8\\8 & 20\end{vmatrix}\) = 16

i.e. \(\begin{bmatrix}84 & -44 & 8\\ -44 & 52 & -24\\ 44 & -24 & 16\end{bmatrix}\)

Its transpose is

\(\begin{bmatrix}84 & -44 & 44\\ -44 & 52 & -24\\ 8 & -24 & 16\end{bmatrix}\)

∴ (ATA)-1 = \(\frac{1}{64} \begin{bmatrix}84 & -44 & 44\\ -44 & 52 & -24\\ 8 & -24 & 16\end{bmatrix}\)

= \(\begin{bmatrix}\frac{21}{16} & \frac{-11}{16} & \frac{1}{8}\\ \frac{-11}{16} & \frac{13}{16} & \frac{-3}{8}\\ \frac{1}{8} & \frac{-3}{8} & \frac{1}{4}\end{bmatrix}\)

From (i)

x = (ATA)-1 × ATB

= \(\begin{bmatrix}\frac{21}{16} & \frac{-11}{16} & \frac{1}{8}\\ \frac{-11}{16} & \frac{13}{16} & \frac{-3}{8}\\ \frac{1}{8} & \frac{-3}{8} & \frac{1}{4}\end{bmatrix}\) \(\begin{bmatrix}18\\20\\3\end{bmatrix}\)

∴ x = \(\begin{bmatrix}\frac{31}{4}\\\frac{-9}{4}\\\frac{1}{2}\end{bmatrix}\) is the solution

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