# How many matrices satisfy this equality?

How many matrices $$A\in\mathcal{M}_{3\times 3} (\mathbb{N})$$ satisfy this equality?

$$\begin{pmatrix} 1 \ \ 2 \ \ 4 \end{pmatrix}\cdot A=\begin{pmatrix} 3 \ \ 2 \ \ 1 \end{pmatrix}$$

I tried with examples and I found just one but I want to know how to approach this exercise.The right answer is $$3$$

• What are the entries of $A$? Integers, real numbers...? Commented Jul 13, 2019 at 12:46
• natural numbers Commented Jul 13, 2019 at 12:46
• I forgot about that.I edited my question Commented Jul 13, 2019 at 12:47
• @Yanko the matrix is multiplied from the left with a row vector, which makes perfect sense.
– daw
Commented Jul 13, 2019 at 13:20

Let $$A=[x_{i,j}]$$ Then we get

$$x_{1,1} +2 x_{2,1} +4 x_{3,1} = 3$$

$$x_{1,2} + 2x_{2,2} + 4 x_{3,2} = 2$$

$$x_{1,3} + 2 x_{2,3} + 4 x_{3,3} = 1$$

We shall count the number of solutions for each equation.

First equation has 2 solutions: Since $$4>3$$ we have that $$x_{3,1}=0$$. Similarly $$x_{2,1}<2$$. If $$x_{2,1}=1$$ then $$x_{1,1}=1$$ and if $$x_{2,1}=0$$ then $$x_{1,1}=3$$. Therefore there are two solutions for the first equation.

Second equation has 2 solutions: As before, $$x_{3,2}=0$$. Also $$x_{2,2}$$ is at most $$1$$, hence if $$x_{2,2}=1$$ then $$x_{1,2}=0$$ and if not then $$x_{2,2}=0$$ and $$x_{1,2}=1$$.

Third equation has 1 solution: $$4,2>1$$ so $$x_{2,3},x_{3,3}=0$$ and so the only solution is $$x_{1,3}=1$$.

Therefore there is a total of $$4$$ solutions for all of the equations.

$$A_1 =\begin{bmatrix} 3 & 0 &1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}, A_2=\begin{bmatrix} 1 & 0 &1 \\ 1 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ $$A_3 =\begin{bmatrix} 3 & 2 &1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}, A_4=\begin{bmatrix} 1 & 2 &1 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$

• The OP forgot to mention that their definition of $\mathbb{N}$ includes $0$ because otherwise there are zero solutions. Commented Jul 13, 2019 at 12:59
• Thank for your answer.If $x_{2,1}=1$ then $x_{1,1}=1$, right? Because otherwise the result will be 4 Commented Jul 13, 2019 at 13:12
• @DaniVaja Correct. Thank you Commented Jul 13, 2019 at 13:16