# Thread: How to find inverse matrix of a 3*3 or of having order > 3 in a easy way

1. ## How to find inverse matrix of a 3*3 or of having order > 3 in a easy way

I know finding inverse of a matrix in 3 ways 1) if 2*2 1/(ad-bc)[d -b] [-c a] 2) if order is >3 by elementary row operations with identity matrix [ -1 2 -3 | 1 0 0 ] [ 2 1 0 | 0 1 0 ] [ 4 -2 5 | 0 0 1 ] like dis on reducing into In ; B 3) through eigen value process sir all of dese processes are taking bulk of time in xams. I request you to tell me if u r known with any other easiest way

2. Originally Posted by achasiri
I know finding inverse of a matrix in 3 ways 1) if 2*2 1/(ad-bc)[d -b] [-c a] 2) if order is >3 by elementary row operations with identity matrix [ -1 2 -3 | 1 0 0 ] [ 2 1 0 | 0 1 0 ] [ 4 -2 5 | 0 0 1 ] like dis on reducing into In ; B 3) through eigen value process sir all of dese processes are taking bulk of time in xams. I request you to tell me if u r known with any other easiest way
Write out your post properly, pressing the return' key every once in a while and spelling these' correctly, and then I will tell you my thoughts on the matter (your post is pretty much illegible the way it is).

Also, to get the matrix to be displayed correctly, use LaTeX.

Finally, what do you mean by `order 3'? Do you mean a 3x3 matrix, or a matrix $\displaystyle A$ such that $\displaystyle A^3=Id$?

3. Frankly, I believe that row-reduction is the easiest way to find the inverse matrix, even for 3 by 3 and 2 by 2 matrices. However, you can also use the fact that the inverse of the matrix
$\displaystyle A= \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33}\end{bmatrix}$
is given by
$\displaystyle A^{-1}= \frac{1}{det(A)}\begin{bmatrix}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{13} & b_{23} & b_{33}\end{bmatrix}$
where $\displaystyle b_{ij}$ is the "minor" of $\displaystyle a_{ji}$ (note the reversal of i and j). The minor is the determinant of the matrix with the row and column $\displaystyle a_{ji}$ is in removed, multiplied by $\displaystyle (-1)^{i+j}$. This, in fact, will work for any n by n matrix but probably is easier than row-reduction for 2 by 2 and some 3 by 3 matrices.

As a simple 2 by 2 example, if
$\displaystyle A= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}$
then the determinant is $\displaystyle a_{11}a_{22}- a_{12}a_{21}$, the minor of $\displaystyle a_{11}$ is just $\displaystyle a_{22}$, the minor of $\displaystyle a_{12}$ is $\displaystyle -a_{21}$, the minor of $\displaystyle a_{21}$ is $\displaystyle -a_{12}$, and the minor of $\displaystyle a_{22}$ is $\displaystyle a_{11}$. Remembering the reversal of "i" and "j", the inverse matrix is
$\displaystyle A^{-1}= \frac{1}{a_{11}a_{22}- a_{12}a_{21}}\begin{bmatrix}a_{22} & -a_{12} \\ -a_{21} & a_{11}\end{bmatrix}$

### special tips for find inverse of a in matrix 3×3 order

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