Problem 3 . Find the matrix A that solves the equation [} 0&2&1 0&0&-1 2&1&0

To solve the given matrix equation, we need to find the matrix A that satisfies the equation:<br /><br />$[\begin{matrix} 0&2&1\\ 0&0&-1\\ 2&1&0\end{matrix} ]A+A=[\begin{matrix} 1&3&1\\ 2&2&2\\ 3&1&-1\end{matrix} ]^{T}$<br /><br />First, let's rewrite the equation in a more standard form:<br /><br />$[\begin{matrix} 0&2&1\\ 0&0&-1\\ 2&1&0\end{matrix} ]A = [\begin{matrix} 1&3&1\\ 2&2&2\\ 3&1&-1\end{matrix} ]^{T} - A$<br /><br />Now, we can solve for matrix A by multiplying both sides of the equation by the inverse of the matrix on the left-hand side:<br /><br />$A = ([\begin{matrix} 0&2&1\\ 0&0&-1\\ 2&1&0\end{matrix} ]^{-1}) \cdot ([\begin{matrix} 1&3&1\\ 2&2&2\\ 3&1&-1\end{matrix} ]^{T} - A)$<br /><br />To find the inverse of the matrix $[\begin{matrix} 0&2&1\\ 0&0&-1\\ 2&1&0\end{matrix} ]$, we can use the following formula:<br /><br />$[\begin{matrix} a&b&c\\ d&e&f\\ g&h&i\end{matrix} ]^{-1} = \frac{1}{\text{det}([\begin{matrix} a&b&c\\ d&e&f\\ g&h&i\end{matrix} ])} \cdot [\begin{matrix} ei-fh&-bg+ch&-ai+bg\\ -di+fh&ad-bi&-ag+dh\\ cg-de&-ah+be&-ad+bg\end{matrix} ]$<br /><br />where $\text{det}([\begin{matrix} a&b&c\\ d&e&f\\ g&h&i\end{matrix} ])$ is the determinant of the matrix $[\begin{matrix} a&b&c\\ d&e&f\\ g&h&i\end{matrix} ]$.<br /><br />Calculating the determinant, we get:<br /><br />$\text{det}([\begin{matrix} 0&2&1\\ 0&0&-1\\ 2&1&0\end{matrix} ]) = 0 \cdot (0 \cdot 0 - (-1) \cdot 1) - 2 \cdot (0 \cdot 0 - (-1) \cdot 2) + 1 \cdot (0 \cdot 1 - 0 \cdot 2) = 2$<br /><br />Now, we can calculate the inverse of the matrix $[\begin{matrix} 0&2&1\\ 0&0&-1\\ 2&1&0\end{matrix} ]$:<br /><br />$[\begin{matrix} 0&2&1\\ 0&0&-1\\ 2&1&0\end{matrix} ]^{-1} = \frac{1}{2} \cdot [\begin{matrix} 0&-2&1\\ 0&2&1\\ -2&1&0\end{matrix} ]$<br /><br />Finally, we can substitute the inverse matrix and the given matrix $[\begin{matrix} 1&3&1\\ 2&2&2\\ 3&1&-1\end{matrix} ]^{T} - A$ into the equation for A:<br /><br />$A = \frac{1}{2} \cdot [\begin{matrix} 0&-2&1\\ 0&2&1\\ -2&1&0\end{matrix} ] \cdot ([\begin{matrix} 1&3&1\\ 2&2&2\\ 3&1&-1\end{matrix} ]^{T} - A)$<br /><br />Simplifying the right-hand side, we get:<br /><br />$A = \frac{1}{2} \cdot [\begin{matrix} 0&-2&1\\ 0&2&1\\ -2&1&0\end{matrix} ] \cdot [\begin{matrix} 1&2&3\\ 3&2&1\\ 1&2&-1\end{matrix} ] - \frac{1}{2} \cdot [\begin{matrix} 0&-2&1\\ 0&2&1\\ -2&