If a multiply two real numbers I get the product. If I divide the product by one of those real numbers I get the other.

By my reasoning since, $ { |A|^{-1} * |b|= |Solution|} $ , if I know the solution vector ( Using RRef) and divide by $ {|b| } $ I should get $ {|A|^{-1}} $.

But my online adventures tell me you can't divide vectors.

So what gives? Can I deduce the inverse of a matrix If I know ${|b|}$ and the solution vector or can't I?

  • $\begingroup$ No, unless the matrices are 1-by-1, there are always many matrices $A$ such that $Ax = b$ for given vectors $x$ and $b$. (The division $b/x$ is not well-defined.) $\endgroup$ – Shaun Ault Nov 4 '13 at 13:19
  • $\begingroup$ You cannot. You can if (for an $n\times n$) you know this for $n$ linearly independent vectors. $\endgroup$ – André Nicolas Nov 4 '13 at 13:20
  • $\begingroup$ More concretely, consider the vector $v = ( 1; 1 )^T$. Then $ ( 1; 1) \cdot v = (2; 0) \cdot v = 2$. $\endgroup$ – Johannes Kloos Nov 4 '13 at 13:21
  • $\begingroup$ I already have |A|.(presumably) as I completed RRef to get the solution vector. What I want is the INVERSE of |A|. Does it matter? $\endgroup$ – Chris Nov 4 '13 at 13:26
  • $\begingroup$ Thanks for the input. Guess I'll have to go into row operators for Rref for my next adventure. $\endgroup$ – Chris Nov 4 '13 at 13:37

It works like that if and only if $A$ is invertible. If $A$ is invertible and we're trying to solve $A\mathbf{x}=\mathbf{b}$, then the solution is unique and is $A^{-1}\mathbf{b}$.

For example, if we want to solve $$\overbrace{\begin{bmatrix} -3 & -2 \\ -1 & 0 \\ \end{bmatrix}}^A \overbrace{\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}}^{\mathbf{x}}=\overbrace{\begin{bmatrix} 2 \\ 3 \end{bmatrix}}^{\mathbf{b}}$$ then the unique solution is $$\mathbf{x}=\overbrace{\begin{bmatrix} 0 & -1 \\ -1/2 & 3/2 \\ \end{bmatrix}}^{A^{-1}} \overbrace{\begin{bmatrix} 2 \\ 3 \end{bmatrix}}^{\mathbf{b}}=\begin{bmatrix} -3 \\ 7/2 \\ \end{bmatrix}.$$

But it's typically not the easiest approach:

  • Computing $A^{-1}$ could be more involved than finding $\mathbf{x}$ the usual way (row operations and back substitution).

  • $A$ might not be invertible, but there still might be solutions to $A\mathbf{x}=\mathbf{b}$.


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