I have a question concerning Cramer's rule:

Let $A$ be a matrix and $A \cdot \vec x = \vec b$ a lineare equation. $A_i$ is the matrix $A$ where the i'th column is replaced by $\vec b$

  • if $det(A) \neq 0$, then we have a unique solution
  • if $det(A)=0$ and at least one $det(A_i) \neq 0$, we have no solution
  • if $det(A)=0$ and all $det(A_i)=0$ we have infinitely many solutions [false!]

I'm looking for a geometric interpretation of the rule. I know that $det(A)=$area of parallelepiped, but I'm not able draw a picture for Cramer's rule.

Anyone can help me here?

Thanks a lot in advance,

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    $\begingroup$ What is $A_i$? ' $\endgroup$ – M Turgeon Dec 21 '13 at 19:05
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    $\begingroup$ @MTurgeon My guess would be that $\mathbf{A}_{i}$ is the matrix $\mathbf{A}$ with the $i$th column replaced by $\vec{b}$. $\endgroup$ – Thomas Russell Dec 21 '13 at 19:12
  • $\begingroup$ @MTurgeon correct, I'll add it. $\endgroup$ – ulead86 Dec 21 '13 at 19:26
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    $\begingroup$ The last bullet point is incorrect, consider $$\begin{pmatrix} 1&1&1\\0&0&0\\0&0&0 \end{pmatrix} \vec{x} = \begin{pmatrix} 0\\1\\0\end{pmatrix}.$$ $\endgroup$ – Daniel Fischer Dec 21 '13 at 19:35
  • $\begingroup$ @Daniel Fischer hmm, I think you're right $\endgroup$ – ulead86 Dec 22 '13 at 10:06

One way to see Cramer's rule is that it simply makes use of a (very inefficient) way of calculating $A^{-1}$, specifically \begin{equation}A^{-1}=\frac{1}{\text{det}(A)}\text{Adj}(A),\end{equation} where Adj($A$) is known nowadays as the adjugate of $A$ (transpose of matrix of cofactors of $A$). If you substitute this equation into $X=A^{-1}B$ you basically get Cramer's rule, since for every component $x_i$ of $X$ you get the Laplace expansion by the ith column of $1/$det$(A)\cdot$det$(A_{C_i\leftrightarrow B})$ (you can check this...).

So then to interpret this geometrically it depends on how far you want to go. If you just see it as a way of calculating $A^{-1}B$ then it simply means that a vector $X$ which is transformed by $A$ to give the vector $B$, if this transformation is invertible then you can calculate $X$ by applying the inverse transformation to $B$. If you now want to go further you have to try and explain geometrically why \begin{equation}A^{-1}=\frac{1}{\text{det}(A)}\text{Adj}(A).\end{equation} So in terms of a geometrical interpretation, we know $A$ is a transformation, and det($A$) is the area of the parallelogram formed by the component vectors in $A$. So what can be interesting is to explore the role of the adjugate and how it determines the formulae for det$(A)$...if you want to...I'll leave this for you - in 2D it's simple, since the cofactors of each entry is just another entry (with a sign change where needed).

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    $\begingroup$ So maybe you can see the last part of the geometric interpretation as the answer to why the determinant is the area of the parallelogram spanned by component vectors (row/column) - there is another post where this is discussed: math.stackexchange.com/questions/29128/… $\endgroup$ – Christiaan Hattingh Dec 22 '13 at 14:49
  • $\begingroup$ Thanks for the interpretation, I'll to think about, how far I want to go :-) $\endgroup$ – ulead86 Dec 24 '13 at 9:15

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