Matrix inversion question with different dimensions

Question

Let $A=\left[\begin{matrix}0&0&0\\-1&0&0\\0&-1&0\end{matrix}\right]$ and $B=\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]$ verify the matrix equation $X=AX+B$

1. What are the dimensions of $X$?

2. Find the matrix $X$. Is the matrix you found the unique solution of the equation?

Can you please discuss how to answer this type of question, I already know that $X$ has to be $3\times 2$ and this is clear through the equation given $X = AX + B$

Please discuss the second part of the question about how we could determine Matrix $X$

Answer 1

Start by moving the term $AX$ to the other side of the equation, producing $(I-A)X=B$. Since each column of $B$ is the result of multiplying the corresponding column of $X$ by $I-A$, you have a series of systems of linear equations to solve for the columns of $X$. These can all be solved simultaneously by row-reducing an augmented matrix just as you would for each individual system of equations.

In this case, the augmented matrix to be reduced is $$\left[\begin{array}{ccc|cc} 1&0&0 & 4&1 \\ 1&1&0 & 5&2 \\ 0&1&1 & 6&3 \end{array}\right].$$ As always, if you end up with a row that’s zero on the left side but non-zero on the right, then there’s no solution. If there are no zero rows, the solution is unique, and if there is an all-zero row, there is an infinite number of solutions.

Answer 2

You have already noticed that $X$ must be a $3 \times 2$ matrix. Now, one way is to find each column $\mathbf{x_i}$ of $X$ separately. Thus, for the first column we are going to have the equation: $$ \mathbf{x_1} = A\mathbf{x_1} + \mathbf{b_1} \iff (I - A)\mathbf{x_1} = \mathbf{b_1}.$$

Since $\det(I-A) \neq 0$ it is implied that $\mathbf{x_1}$is uniquely determined. The same applies for the column $\mathbf{x_2}$.

The two columns $\mathbf{x_1},\mathbf{x_2}$ can be defined by: $$\mathbf{x_i} = (I-A)^{-1} \cdot \mathbf{b_i}, \quad i = 1,2.$$

Answer 3

We know $X=\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]$ so we can say

\begin{align}X&=AX+B\\ \left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}0&0&0\\-1&0&0\\0&-1&0\end{matrix}\right]\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]+\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]\\ \left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}0&0\\-a&-b\\-c&-d\end{matrix}\right]+\left[\begin{matrix}4&1\\5&2\\6&3\end{matrix}\right]\\ \left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]&=\left[\begin{matrix}4&1\\5-a&2-b\\6-c&3-d\end{matrix}\right] \end{align}

Therefore we have the following equalities

\begin{align}a&=4\\ b&=1\\ c&=5-a=1\\ d&=2-b=1\\ e&=6-c=5\\ f&=3-d=2\end{align}

and thus $$X=\left[\begin{matrix}4&1\\1&1\\5&2\end{matrix}\right]$$