Recall, the discussion (Section 5.1) on using matrices to describe the labour market flows as below
$$\begin{align}E_{t+1}&=0.962 E_t+0.276 U_t+0.044 N_t\\&\\U_{t+1}&=0.013 E_t+0.501 U_t+0.027 N_t\\&\\N_{t+1}&=0.025 E_t+0.223 U_t+0.929 N_t\end{align}\quad\Leftrightarrow\quad \begin{pmatrix}E_{t+1}\\ \\U_{t+1}\\ \\N_{t+1}\end{pmatrix}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\begin{pmatrix}E_{t}\\ \\U_{t}\\ \\N_{t}\end{pmatrix}$$We have also seen (Section 5.2) that
$$\text{Given }\quad \mathbf{A}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\quad\text{then}\quad \mathbf{A^{-1}}=\begin{pmatrix}1.048&-0.563&-0.033\\ & &\\-0.026&2.036&-0.058\\& &\\ -0.022&-0.474&1.091\end{pmatrix}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$This can be done by solving a system of linear equations. Note
$$\begin{align}E_{t+1}&=0.962 E_t+0.276 U_t+0.044 N_t\\&\\U_{t+1}&=0.013 E_t+0.501 U_t+0.027 N_t\\&\\N_{t+1}&=0.025 E_t+0.223 U_t+0.929 N_t\end{align}\color{white}{\quad\Rightarrow\quad \begin{align}E_{t+1}&=0.962 E_t+0.276 U_t+0.044 N_t\\&\\U_{t+1}&=0.013 E_t+0.501 U_t+0.027 N_t\\&\\N_{t+1}&=0.025 E_t+0.223 U_t+0.929 N_t\end{align}}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$This can be done by solving a system of linear equations. Note
$$\begin{align}E_{t+1}&=0.962 E_t+0.276 U_t+0.044 N_t\\&\\U_{t+1}&=0.013 E_t+0.501 U_t+0.027 N_t\\&\\N_{t+1}&=0.025 E_t+0.223 U_t+0.929 N_t\end{align}\color{black}{\quad\Rightarrow\quad \begin{align}E_{0}&=0.962 E_{-1}+0.276 U_{-1}+0.044 N_{-1}\\&\\U_{0}&=0.013 E_{-1}+0.501 U_{-1}+0.027 N_{-1}\\&\\N_{0}&=0.025 E_{-1}+0.223 U_{-1}+0.929 N_{-1}\end{align}}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$This can be done by solving a system of linear equations. Note
$$\begin{align}E_{0}&=0.962 E_{-1}+0.276 U_{-1}+0.044 N_{-1}\\&\\U_{0}&=0.013 E_{-1}+0.501 U_{-1}+0.027 N_{-1}\\&\\N_{0}&=0.025 E_{-1}+0.223 U_{-1}+0.929 N_{-1}\end{align}\color{black}{\quad\Rightarrow\quad \begin{align}0.570&=0.962 E_{-1}+0.276 U_{-1}+0.044 N_{-1}\\&\\0.038&=0.013 E_{-1}+0.501 U_{-1}+0.027 N_{-1}\\&\\0.392&=0.025 E_{-1}+0.223 U_{-1}+0.929 N_{-1}\end{align}}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$This can be done by solving the system of linear equations below for $E_{-1}$, $U_{-1}$ and $N_{-1}$
$$\begin{align}0.570&=0.962 E_{-1}+0.276 U_{-1}+0.044 N_{-1}\\&\\0.038&=0.013 E_{-1}+0.501 U_{-1}+0.027 N_{-1}\\&\\0.392&=0.025 E_{-1}+0.223 U_{-1}+0.929 N_{-1}\end{align}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$But it may be more convenient to use the matrix representation
$$\begin{pmatrix}E_{t+1}\\ \\U_{t+1}\\ \\N_{t+1}\end{pmatrix}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\begin{pmatrix}E_{t}\\ \\U_{t}\\ \\N_{t}\end{pmatrix}\color{white}{\quad\Rightarrow\quad\begin{pmatrix}E_{t+1}\\ \\U_{t+1}\\ \\N_{t+1}\end{pmatrix}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\begin{pmatrix}E_{t}\\ \\U_{t}\\ \\N_{t}\end{pmatrix}}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$But it may be more convenient to use the matrix representation
$$\begin{pmatrix}E_{t+1}\\ \\U_{t+1}\\ \\N_{t+1}\end{pmatrix}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\begin{pmatrix}E_{t}\\ \\U_{t}\\ \\N_{t}\end{pmatrix}\color{black}{\quad\Rightarrow\quad\begin{pmatrix}E_{0}\\ \\U_{0}\\ \\N_{0}\end{pmatrix}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\begin{pmatrix}E_{-1}\\ \\U_{-1}\\ \\N_{-1}\end{pmatrix}}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$But it may be more convenient to use the matrix representation
$$\begin{pmatrix}E_{0}\\ \\U_{0}\\ \\N_{0}\end{pmatrix}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\begin{pmatrix}E_{-1}\\ \\U_{-1}\\ \\N_{-1}\end{pmatrix}\color{black}{\quad\Rightarrow\quad\begin{pmatrix}0.570\\ \\0.038\\ \\0.392\end{pmatrix}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\begin{pmatrix}E_{-1}\\ \\U_{-1}\\ \\N_{-1}\end{pmatrix}}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$But it may be more convenient to use the matrix representation
$$\begin{pmatrix}0.570\\ \\0.038\\ \\0.392\end{pmatrix}=\begin{pmatrix}0.962&0.276&0.044\\ & &\\0.013&0.501&0.027\\& &\\ 0.025&0.223&0.929\end{pmatrix}\begin{pmatrix}E_{-1}\\ \\U_{-1}\\ \\N_{-1}\end{pmatrix}\color{black}{\quad\Rightarrow\quad\mathbf{x_0}=\mathbf{A}\mathbf{x_{-1}}}$$Suppose that we know that the current state of the economy is
$$\mathbf{x_0}=\begin{pmatrix}E_0\\U_0\\N_0\end{pmatrix}=\begin{pmatrix}0.570\\0.038\\0.392\end{pmatrix}$$and we would like to know the state of the economy last month
$$\mathbf{x_{-1}}=\begin{pmatrix}E_{-1}\\U_{-1}\\N_{-1}\end{pmatrix}$$But it may be more convenient to use the matrix representation
$$\color{black}{\mathbf{x_0}=\mathbf{A}\mathbf{x_{-1}}}$$and solve the above equation for $\mathbf{x_{-1}}$
Given
$$\color{black}{\mathbf{x_0}=\mathbf{A}\mathbf{x_{-1}}}$$and since an inverse of $\mathbf{A}$, $\mathbf{A^{-1}}$, exists, premultiply both sides of the equation by $\mathbf{A^{-1}}$
Given
$$\color{black}{\mathbf{x_0}=\mathbf{A}\mathbf{x_{-1}}}$$and since an inverse of $\mathbf{A}$, $\mathbf{A^{-1}}$, exists, premultiply both sides of the equation by $\mathbf{A^{-1}}$
$$\color{black}{\mathbf{A^{-1}}\mathbf{x_0}=\mathbf{A^{-1}}\mathbf{A}\mathbf{x_{-1}}}$$Given
$$\color{black}{\mathbf{x_0}=\mathbf{A}\mathbf{x_{-1}}}$$and since an inverse of $\mathbf{A}$, $\mathbf{A^{-1}}$, exists, premultiply both sides of the equation by $\mathbf{A^{-1}}$
$$\color{black}{\mathbf{A^{-1}}\mathbf{x_0}=\mathbf{A^{-1}}\mathbf{A}\mathbf{x_{-1}}}$$and since $\mathbf{A^{-1}}\mathbf{A}=\mathbf{I}$ (the 3x3 identity matrix)
$$\color{black}{\mathbf{A^{-1}}\mathbf{x_0}=\mathbf{I}\mathbf{x_{-1}}}$$Given
$$\color{black}{\mathbf{x_0}=\mathbf{A}\mathbf{x_{-1}}}$$and since an inverse of $\mathbf{A}$, $\mathbf{A^{-1}}$, exists, premultiply both sides of the equation by $\mathbf{A^{-1}}$
$$\color{black}{\mathbf{A^{-1}}\mathbf{x_0}=\mathbf{A^{-1}}\mathbf{A}\mathbf{x_{-1}}}$$and since $\mathbf{A^{-1}}\mathbf{A}=\mathbf{I}$ (the 3x3 identity matrix)
$$\color{black}{\mathbf{A^{-1}}\mathbf{x_0}=\mathbf{I}\mathbf{x_{-1}}}$$and since the identity matrix multiplied by a matrix gives the matrix itself
$$\color{black}{\mathbf{A^{-1}}\mathbf{x_0}=\mathbf{x_{-1}}}$$Given
$$\color{black}{\mathbf{x_0}=\mathbf{A}\mathbf{x_{-1}}}$$we have
$$\mathbf{x_{-1}}=\mathbf{A^{-1}}\mathbf{x_0}$$and we are done - this is the solution.
We have
$$\mathbf{A^{-1}}=\begin{pmatrix}1.048&-0.563&-0.033\\ & &\\-0.026&2.036&-0.058\\& &\\ -0.022&-0.474&1.091\end{pmatrix}\quad\text{and}\quad \mathbf{x_0}=\begin{pmatrix}0.570\\ \\0.038\\ \\0.392\end{pmatrix}$$so
$$\mathbf{x_{-1}}=\mathbf{A^{-1}}\mathbf{x_0}=\begin{pmatrix}1.048&-0.563&-0.033\\ & &\\-0.026&2.036&-0.058\\& &\\ -0.022&-0.474&1.091\end{pmatrix}\begin{pmatrix}0.570\\ \\0.038\\ \\0.392\end{pmatrix}=\begin{pmatrix}0.56\\ \\0.04\\ \\0.40\end{pmatrix}$$