Solving $Ax=B$ where $x$ is a $3\times 3$ matrix

Question

I've often seen questions regarding how to find matrix $x$ in $Ax = B$, but almost all of these include $x$ as a $3\times 1$ matrix. I'm trying to solve this question:

Find a matrix x that satisfies the following equation: $x ⋅ \begin{bmatrix}2 \\ 0 \\2\end{bmatrix} = \begin{bmatrix}2 \\ 0.5 \\ 1.5\end{bmatrix}$

So my thought was that x is supposed to be a $3\times 3$ matrix, otherwise you won't get $B$. I am however at a loss on how to solve the equation for such a matrix.

I tried writing this as a system of equations, which gives me:

$$ \begin{bmatrix}a & b & c \\ d & e& f \\g & h & i\end{bmatrix} ⋅ \begin{bmatrix}2 \\ 0 \\2\end{bmatrix} =\begin{bmatrix}2 \\ 0.5 \\ 1.5\end{bmatrix}$$

$$2a + 2c = 2 \\ 2d + 2f = 0.5 \\ 2g + 2i = 1.5$$

but I really don't think this is right. I also wouldn't know how to go from here. I tried row reduction on the augmented matrix of this system but it gave me the wrong answer.

If anyone could guide me in te right way, I would really appreciate it!

Answer 1

You have $9$ unknowns, and $3$ equations for the system: $$ \begin{bmatrix} 2 \\ 1/2 \\ 3/2 \end{bmatrix} = x \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \\ e \\ f \\ g \\ h \\ i \end{bmatrix} $$ This means you have $6$ free variables.

There are a lot of $3\times 3$ matrices which satisfy this. So let us pick an easy one.

$$ x = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \end{bmatrix} $$

The middle column unknowns $x_{i2}$ do not show up, this means we can choose any value for them, so we set them to zero. $3$ free variables of the $6$ original ones determined, $3$ free variables to go.

Otherwise we have equations $$ 2 x_{i1} + 2 x_{i3} = b_i \quad (i \in \{1,2,3\}) $$ E.g. if we wish $x_{i1} = x_{i3}$ we end up with $$ x_{i1} = x_{i3} = b_i/4 $$ This gives: $$ x = \begin{bmatrix} 1/2 & 0 & 1/2 \\ 1/8 & 0 & 1/8 \\ 3/8 & 0 & 3/8 \end{bmatrix} $$

Or if we instead wish $x_{i3} = 0$ we end up with $$ x_{i1} = b_i/2 $$ This gives: $$ x = \begin{bmatrix} 1 & 0 & 0 \\ 1/4 & 0 & 0 \\ 3/4 & 0 & 0 \end{bmatrix} $$

Answer 2

You are correct in finding $$2a + 2c = 2 \\ 2d + 2f = 0.5 \\ 2g + 2i = 1.5$$

Do not worry if the solutions are not unique.

pick some values for $a$,$d$,and $g$ and solve for $c$,$f$,and $i$.

You will find many marices to work for your equation.