The pseudo-inverse of a. Adjoint and Inverse of a Matrix - GeeksforGeeks Python is crazy accurate, and rounding allows us to compare to our human level answer. Equation 3 is equivalent to Equation 1, with the variables substituted. A=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{5em} I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}. Applying Polynomial Features to Least Squares Regression using Pure Python without Numpy or Scipy, AX=B,\hspace{5em}\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix}=\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, X=A^{-1}B,\hspace{5em} \begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix} =\begin{bmatrix}ai_{11}&ai_{12}&ai_{13}\\ai_{21}&ai_{22}&ai_{23}\\ai_{31}&ai_{32}&ai_{33}\end{bmatrix}\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, I= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, AX=IB,\hspace{5em}\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix}= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, IX=A^{-1}B,\hspace{5em} \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix} =\begin{bmatrix}ai_{11}&ai_{12}&ai_{13}\\ai_{21}&ai_{22}&ai_{23}\\ai_{31}&ai_{32}&ai_{33}\end{bmatrix}\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, S = \begin{bmatrix}S_{11}&\dots&\dots&S_{k2} &\dots&\dots&S_{n2}\\S_{12}&\dots&\dots&S_{k3} &\dots&\dots &S_{n3}\\\vdots& & &\vdots & & &\vdots\\ S_{1k}&\dots&\dots&S_{k1} &\dots&\dots &S_{nk}\\ \vdots& & &\vdots & & &\vdots\\S_{1 n-1}&\dots&\dots&S_{k n-1} &\dots&\dots &S_{n n-1}\\ S_{1n}&\dots&\dots&S_{kn} &\dots&\dots &S_{n1}\\\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\3&9&4\\1&3&5\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\0&1&0\\0&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&7.2&3.4\\1&3&5\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.6&1&0\\0&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&7.2&3.4\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.6&1&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&1&0.472\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.083&0.139&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&0&3.667\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\0&-0.333&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\0&-0.091&0.273\end{bmatrix}, A_M=\begin{bmatrix}1&0&0\\0&1&0.472\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.091&0.023\\-0.083&0.139&0\\0&-0.091&0.273\end{bmatrix}, A_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.091&0.023\\-0.083&0.182&-0.129\\0&-0.091&0.273\end{bmatrix}, A \cdot IM=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, Gradient Descent Using Pure Python without Numpy or Scipy, Clustering using Pure Python without Numpy or Scipy, Least Squares with Polynomial Features Fit using Pure Python without Numpy or Scipy, use the element thats in the same column as, replace the row with the result of [current row] multiplier * [row that has, this will leave a zero in the column shared by. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e.
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