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I am relatively new to mathematics, especially matrices. I saw a similar question on here about my question but I could not follow what was being said.

I am trying to work out the inverse of a $2\times 2$ matrix $A$.

I know the inverse matrix $A^{-1}$ reads:

$$A^{-1} = \dfrac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$

My question is: how is the inverse matrix obtained?

I have tried transposing the cofactor matrix

$$\begin{pmatrix}a & -b \\ c & -d\end{pmatrix} \longrightarrow \begin{pmatrix} a & c \\ -b & -d\end{pmatrix}$$

I am using the book Maths for Chemistry Paul Monk Lindsey J Munro as a starting point. Thank you.

Alternatively, I have tried using the determinant $(ad-bc)$ therefore, \begin{pmatrix}a & -b \\ -c & d\end{pmatrix}.

However, I am not sure how the $a$ and $d$ terms swap as I thought this was not allowed during a transpose, as the principal diagonal remains unchanged.

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    $\begingroup$ The cofactor matrix of $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is $\begin{pmatrix}d&-c\\-b&a\end{pmatrix}.$ $\endgroup$ Commented Nov 13, 2022 at 19:45
  • $\begingroup$ Assume $ax+by=A ...(1)$ $cx+dy=B ...(2)$ then $(1) \times d -(2) \times b$ to remove $y$ gives $(ad-bc)x = ...$ and this is the source of the determinant. $\endgroup$
    – Paul
    Commented Nov 13, 2022 at 19:53
  • $\begingroup$ How is the formula obtained? Well, this is the simplest non-trivial case for the usage of the adjugate matrix en.wikipedia.org/wiki/Adjugate_matrix to get a formula for the inverse of a matrix, in case its determinant is invertible. Once we have the formula, we can simply check it, just multiply the given $A$ with its to-be-inverse, this is an other way to "obtain" the inverse, of course not in a constructive manner. A cheap further possibility would be to use unknowns $x,y,z,w$ and the matrix $$B=\begin{bmatrix}x&y\\t&w\end{bmatrix}$$, then solve the system with four equations $AB=I$. $\endgroup$
    – dan_fulea
    Commented Nov 14, 2022 at 0:01

2 Answers 2

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To get the minors of a matrix, you delete a row and column of the matrix and calculate the determinant of the remaining matrix. For a 2x2 matrix, that means that we would take the minors as follows:

$A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$

$\begin{array}{cc}{M_{11} = \det \begin{pmatrix} \Box & \Box \\ \Box & d\end{pmatrix} = d} & {M_{12} = \det \begin{pmatrix} \Box & \Box \\ c & \Box \end{pmatrix} = c} \\ {M_{21} = \det \begin{pmatrix} \Box & b \\ \Box & \Box \end{pmatrix} = b} & {M_{22} = \det \begin{pmatrix} a & \Box \\ \Box & \Box \end{pmatrix} = a} \end{array}$

So the matrix of minors is $\begin{pmatrix}d & c \\ b & a \end{pmatrix}$, and the adjoint matrix (which is the matrix of cofactors) is $\begin{pmatrix}d & -c \\ -b & a\end{pmatrix}$, which comes by multiplying the $(i,j)$th entry by $(-1)^{i+j}$.

The adjoint does not relate directly to the transpose - as you've noted, in the adjoint you swap $a$ and $d$ which a transpose cannot do. In fact, the way you get the adjoint by swapping elements of the original matrix is something that's only true for 2x2 matrices - larger matrix adjoints are much messier.

You can also find the inverse by performing Gaussian elimination - start with the augmented matrix

$\left( \begin{array}{cc|cc} a & b & 1 & 0 \\ c & d & 0 & 1 \end{array} \right)$

and, using the operations of

  1. Multiplying a row by a non-zero constant (i.e. $R_i \rightarrow \lambda R_i$)
  2. Swapping two rows (i.e. $R_i \leftrightarrow R_j$)
  3. Adding a multiple of one row to another (i.e. $R_i \rightarrow R_i + \lambda R_j$)

try to make the left-hand side become the identity matrix $\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ and you should see the right-hand side become the required inverse.

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[![Solved the adjugate of a 2x2 matrix][1]][1]

Hi, i've managed to get my head around this equation. I have poseted my solution below. Thank you for all your help as I could not have viewed the problem differently without your comments! [1]: https://i.sstatic.net/TMPok.jpg

Step 1 - Determine every minor for the 2x2 matrix

Matrix A = a11 a12 a21 a22

Determine the minor for each element. This is done by selecting an element, amn, where m is the row and n the column. Eliminate from the matrix the rows m and columns n as in the selected element.

So, for example, a11 we would eliminate the row 1 and column 1, leaving us with the minor a22. a22 is our minor for the element a11. We will refer to minors as Mij, so a22 = M22

Proceeding forward, we get the following minors in order a22 a21 a12 a11 or M22 M21 M12 M11.

These minors M22 M21 M12 M11 correspond to the matrix elements a11, a12, a21, a22 respectively.

Now we have our minor terms in the correct place in the matrix, we can proceed to step 2.

Step 2 - Determining the cofactor matrix or getting the negative signs

The equation cofactor = (-1)^i-j x Minor ij is required.

So now, we have to multiply each minor by (-1)^ij, where i corresponds to the row and j corresponds to the column of the minor.

So for M22 we have: (-1)^2+2 2+2 = 4 --> (-1)^4 = 1
1 x M22 = M22

So for M21 we have: (-1)^2+1 2+1 = 3 --> (-1)^3 = -1
-1 x M21 = -M21

As we proceed forward, we get, in order M22, -M21, -M12, M11

Therefore, we have obtained the adjunct of a 2x2 matrix. In other words, we have obtained d -c -b a which is the second part of the equation required to determine the inverse of a 2x2 square matrix.

I would highly appreciate it if someone could edit and put the matrices in as I do not know how to do this.

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    $\begingroup$ It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. $\endgroup$ Commented Nov 13, 2022 at 20:34

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