0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $A\mathbf{x}=\mathbf{b}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.