I was studying inverse matrix. Suddenly I stumbled on the inverse of 3×3 K. And it involved a division by the determinant (Well, only with numbers it was). And it was also said, about involving a division by the determinant. Then I got the inverse of a 2×2 matrix,with variables as entries(That's the general form of the 2×2 real number matrix). And it also involved a division by ad-bc(the determinant). But why is that? Maybe it's the result of something in the matrix?
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$\begingroup$Question: "Why the inverse of a matrix involves division by the determinant?"
Answer: We use the adjugate matrix and the determinant to prove existence of an inverse of a matrix as follows:
The "adjugate matrix" $ad(A)$ has the property that $ad(A)A=Aad(A)=det(A)I$ where $det(-): Mat(n,k) \rightarrow k$ is a map with $det(AB)=det(A)det(B)$. Here $Mat(n,k)$ is the set of $n\times n$-matrices with coefficients in $k$.$det(A)$ is the "determinant" of the matrix $A$ as defined in your linear algebra course.
Lemma: A square matrix $A$ has an inverse iff $det(A)\neq 0$.
Proof: If $det(A)\neq 0$ it follows $A^{-1}:=\frac{1}{det(A)}ad(A)$ is an inverse. Conversely assume there is a matrix $B$ with $AB=BA=I$. It follows $det(AB)=det(A)det(B)=1$ and hence $det(A) \neq 0$.
Hence the adjugate matrix and the determinant map implies the existence of an inverse of $A$: The matrix $A$ has a unique inverse $A^{-1}$ iff $det(A)\neq 0$.
Example: Let\begin{align*} A= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{align*}
and define the adjunct matrix $ad(A)$ by
\begin{align*} ad(A)= \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \end{align*}
It follows
\begin{align*} ad(A)A=Aad(A)= \begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix} =\end{align*}
\begin{align*} det(A)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{align*}
There is in general for any $n\times n$-matrix $A$ a unique matrix $ad(A)$ with $ad(A)A=Aad(A)=det(A)I$. This result is proved in any serious linear algebra course. Hence the above proves the Lemma explicitly for any $2\times 2$-matrix.
$\endgroup$ 3 $\begingroup$Here is an intuitive explanation: Remember that a matrix A regarded as a linear transformation scales up the area for 2x2 matrices, and volume for 3x3 matrices (and higher dimensional Lebesgue measure in those cases) by multiplicative factor of determinant of A.
Identity matrix preserves the area (volume, etc). So to undo the effect of A, its inverse has to involve 1/det(A).
$\endgroup$ $\begingroup$To understand it intuitively, you can consider the real number $x$. Its inverse is $\frac1x$. i.e. we are dividing by $x$. Similiary, to find the inverse of matrix $A$, we divide by the representation of matrix $A$, i.e. its determinant. The difference here being, instead of dividing the identity ($I$), we are dividing the adjoint matrix.
$\endgroup$ 5 $\begingroup$This is only a rough version of the answer.feel free to edit it.
Here is the visual explanation: First you need to understand what a matrix is and what a determinant is.
A matrix represents a Linear transformation. We can think about a Cartesian coordinate system and a linear transformation applied to it which lands $\hat{i}$ to (a,b) and $\hat{j}$ to (c,d). This transformation can be represented by a matrix $\begin{bmatrix} a & c\\ b & d \end{bmatrix}$ .
A vector which was originally at (x,y) lands to new postion$$x \begin{bmatrix} a \\ b \end{bmatrix} + y \begin{bmatrix} c \\ d \end{bmatrix} $$$$= \begin{bmatrix} ax+by \\ cx+dy \end{bmatrix}$$So the new coordinates of the vector is $(ax+by,cx+dy)$.
You see, matrix multiplication is just finding the new coordinates of a vector which has undergone a linear transformation.
So multiplication of a 2×2 matrix is just finding the composition matrix of two linear transformation. That is finding a matrix which when applied to a vector space give the same result as applying one transformation after the other.
Determinant
When we change the basis vectors and hence the coordinate system, any square or circle or any kind of area in the coordinate system expands or compresses or remain the same.
Determinant is the scale to which the area changes.If we multiply the initial area to the determinant, we get the final area.
Finding the inverse
The inverse of a matrix A is the matrix when multiplied to A gives the identity matrix.
Or from our understanding of matrix as transformations, The inverse of a matrix A (a transformation) is the transformation when applied to A brings back all the vectors to the initial position.
If i and j was initially at $\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$and later transformed to$\begin{bmatrix} a & c\\ b& d \end{bmatrix}$ then we can say that the inverse of A is the transformation which brings the i and j back to their initial coordinates.
Currently the coordinate is like this:
Now first we will bring it back so that it aligns with the i and j.Remember that we have to do this from the perspective of the transformed coordinates.
From initial perspective
From transformed perspective.
Now we will convert this rectangle to a square.
For that we will interchange the a and d so that the transformed version of a will be scaled down by a factor of d and transformed vector of d will be scaled up by a factor of a making this a square.
$$\begin{bmatrix} d & -c\\ -b& a \end{bmatrix}$$
This square has an area of determinant of A . So to transform it to the initial i and j we have to make the area 1. So for that we divide this by determinant of A.
This matrix when applied to A gives the identity matrix so this is the inverse of A.
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