# “Dividing” matrix by a vector

If I have $$\mathbf{Ax}=\mathbf{c}$$ where $$\mathbf{x} = \left[\begin{array}{r} x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{array}\right]$$ and $$\mathbf{c} = \left[\begin{array}{r} 2x_1-4x_2-x_3-3x_4+2x_5\\ -x_1+2x_2+x_3+x_5\\ x_1-2x_2-x_3-3x_4-x_5\\ -x_1+4x_2-x_3+5x_5 \end{array}\right]$$. To solve the equation for $$\mathbf{A}$$, I would therefore like to isolate $$\mathbf{A}$$ - is there a way to do this? Like "dividing" by $$\mathbf{a}$$. I know that $$\mathbf{A}$$ is obviously the coefficient matrix but I would like some justification to actually show this like by mathematically solving the equation for $$\mathbf{A}$$.

• math.stackexchange.com/questions/768667/… – Eff May 15 at 11:32
• One solution for $x\ne0$ takes the $i$th row of $A$ to be $(b_i/x^2)x^T$. – J.G. May 15 at 11:39
• Arriving at $A$ as the result of some algorithm doesn't make it any more correct than an intuitive one. I think the point of a question like this is to see if you get the concept of matrix-vector multiplication. – Matthew Leingang May 15 at 12:12

Notice that $$Ax = A \cdot x =\left[\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\ \cdots & \cdots & \cdots & \cdots & \cdots\\ a_{41} & \cdots & \cdots & \cdots & a_{45} \end{array}\right]\cdot\left[\begin{array}{c} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5} \end{array}\right]=\left[\begin{array}{c} a_{11} x_{1}+a_{12}x_{2}+a_{13}x_{3} +a_{14}x_{4} +a_{15}x_{5}\\ a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}+a_{24}x_{4}+a_{25}x_{5}\\ a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}+a_{34}x_{4}+a_{35}x_{5}\\ a_{41}x_{1}+a_{42}x_{2}+a_{43}x_{3}+a_{44}x_{4}+a_{45}x_{5} \end{array}\right]$$

$$\left[\begin{array}{c} a_{11} x_{1}+a_{12}x_{2}+a_{13}x_{3} +a_{14}x_{4} +a_{15}x_{5}\\ a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}+a_{24}x_{4}+a_{25}x_{5}\\ a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}+a_{34}x_{4}+a_{35}x_{5}\\ a_{41}x_{1}+a_{42}x_{2}+a_{43}x_{3}+a_{44}x_{4}+a_{45}x_{5} \end{array}\right] = \left[\begin{array}{r} 2x_1-4x_2-x_3-3x_4+2x_5\\ -x_1+2x_2+x_3+x_5\\ x_1-2x_2-x_3-3x_4-x_5\\ -x_1+4x_2-x_3+5x_5 \end{array}\right]$$

And you can continue from here.

• But still, how did you get to $mathbf{A}$? Is it just guess-work that leads you to $\mathbf{A}$ from the beginning? Because you do not actually solve the equation by isolating $mathbf{A}$ – nichl nss May 15 at 11:48
• well, it's pretty obvious. if $a_{11}x_{1} = 2x_1$ then $a_{11} = 2$ and so on – Jneven May 15 at 11:53
• this is the meaning of a linear system equation, where the difference between the equations and the matrix is the multiplication by a vector of variables. – Jneven May 15 at 11:55
• Once you understand this solution, you will see that the first line of your question is, in fact, a way of specifying the matrix $A$. The coefficients you see inside of $c$ are the entries of the matrix $A$. – GEdgar May 15 at 12:57

Write the coefficient of $$x_j$$ in $$i^{th}$$ row of $$x$$ as elements $$a_{ij}$$ in c.

$$\mathbf{c} = \left[\begin{array}{r} 2x_1-4x_2-x_3-3x_4+2x_5\\ -x_1+2x_2+x_3+x_5\\ x_1-2x_2-x_3-3x_4-x_5\\ -x_1+4x_2-x_3+5x_5 \end{array}\right]$$

$$\mathbf{A} = \left[\begin{matrix} 2& -4& -1& -3& 2\\ -1& 2& 1& 0& 1\\ 1& -2& -1& -3& -1\\ -1& 4& -1& 0& 5 \end{matrix}\right]$$

• @Jneven $A_{4\times 5} x_{5\times 1}=c_{4\times 1}$ – Abhiram V P May 16 at 11:22