Pseudoinverse of $$$\left[\begin{array}{cc}2 & 1\\3 & 4\end{array}\right]$$$

Find the Moore-Penrose pseudoinverse of $$$\left[\begin{array}{cc}2 & 1\\3 & 4\end{array}\right]$$$.

The pseudoinverse of a matrix $$$A$$$ is $$$A^{+} = A^{T} \left(A A^{T}\right)^{-1}$$$.

Find the transpose of the matrix: $$$\left[\begin{array}{cc}2 & 1\\3 & 4\end{array}\right]^{T} = \left[\begin{array}{cc}2 & 3\\1 & 4\end{array}\right]$$$ (for steps, see matrix transpose calculator).

Multiply the original matrix by its transpose:

$$$\left[\begin{array}{cc}2 & 1\\3 & 4\end{array}\right]\cdot \left[\begin{array}{cc}2 & 3\\1 & 4\end{array}\right] = \left[\begin{array}{cc}5 & 10\\10 & 25\end{array}\right]$$$ (for steps, see matrix multiplication calculator).

Find the inverse matrix: $$$\left[\begin{array}{cc}5 & 10\\10 & 25\end{array}\right]^{-1} = \left[\begin{array}{cc}1 & - \frac{2}{5}\\- \frac{2}{5} & \frac{1}{5}\end{array}\right]$$$ (for steps, see matrix inverse calculator).

Finally, multiply the matrices:

$$$\left[\begin{array}{cc}2 & 3\\1 & 4\end{array}\right]\cdot \left[\begin{array}{cc}1 & - \frac{2}{5}\\- \frac{2}{5} & \frac{1}{5}\end{array}\right] = \left[\begin{array}{cc}\frac{4}{5} & - \frac{1}{5}\\- \frac{3}{5} & \frac{2}{5}\end{array}\right]$$$ (for steps, see matrix multiplication calculator).

$$$\left[\begin{array}{cc}2 & 1\\3 & 4\end{array}\right]^{+} = \left[\begin{array}{cc}\frac{4}{5} & - \frac{1}{5}\\- \frac{3}{5} & \frac{2}{5}\end{array}\right] = \left[\begin{array}{cc}0.8 & -0.2\\-0.6 & 0.4\end{array}\right]$$$A