Algebra in matrices

Question

$$\mathbf{A} = \begin{pmatrix} 2 & 1\\ -2 & 5 \end{pmatrix} and\ (\mathbf{A}+\mathbf{I})^{-1}=\frac{1}{20}\begin{pmatrix} 6 & -1\\ 2 & 3 \end{pmatrix}\\\mathbf{X}\text{ is such that }\mathbf{AX}+\mathbf{X}=\begin{pmatrix} 14\\4 \end{pmatrix}$$ How can I find $\mathbf{X}$ using $(\mathbf{A}+\mathbf{I})^{-1}$ ?

Answer 1

$$\begin{align} \mathbf{AX}+\mathbf{X}&=\begin{pmatrix} 14\\4 \end{pmatrix}\\ (\mathbf{A}+\mathbf{1})(\mathbf{X})&=\begin{pmatrix} 14\\4 \end{pmatrix}\\ (\mathbf{A}+\mathbf{1})^{-1}(\mathbf{A}+\mathbf{1})(\mathbf{X})&=(\mathbf{A}+\mathbf{1})^{-1}\begin{pmatrix} 14\\4 \end{pmatrix}\\ \mathbf{X}&=\frac{1}{20}\begin{pmatrix} 6 & -1\\ 2 & 3 \end{pmatrix}\begin{pmatrix} 14\\4 \end{pmatrix}\\ \mathbf{X}&=\begin{pmatrix}4\\2\end{pmatrix} \end{align}$$