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I need tricks or shortcuts to find the inverse of $2 \times 2$ and $3 \times 3$ matrices. I have to take a time-based exam, in which I have to find the inverse of square matrices.

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Tangentially: how sad that people are being timed computing matrix inverses.... I can think of very few less useful abilities than being able to compute the inverse of a $3\times3$ matrix fast! –  Mariano Suárez-Alvarez Feb 11 '11 at 15:11
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Will it be a multiple choice test? In that case you can just multiply by the provided answers and look for which one works. –  Ross Millikan Feb 11 '11 at 15:49
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@Ross: ssh! That's so much faster and easier as to defeat the intended purpose of the question. (Compare: what is the factorization of $N$? Here, let me give you some choices...) :) –  Pete L. Clark Feb 11 '11 at 16:04
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@Pete: I agree, but that is often what is asked on tests. And one should take advantage of the situation. –  Ross Millikan Feb 11 '11 at 16:41
    
@Ross: Sure, I was not being entirely serious. If my warning was to anyone, it was to those (hopeless souls?) who write multiple choice linear algebra tests. –  Pete L. Clark Feb 11 '11 at 19:30

3 Answers 3

up vote 3 down vote accepted

For a 2x2 matrix, the inverse is: $$ \left(\begin{array}{cc} a&b\\ c&d \end{array}\right)^{-1} = {1 \over a d - b c} \left(\begin{array}{rr} d&-b\\ -c&a \end{array}\right)~,~~\text{ where } ad-bc \ne 0. $$

just swap the 'a' and 'd', negate the 'b' and 'c', then divide all by the determinant $a d - b c$.

That's really the most straightforward 'trick', just memorize that pattern.

For 3x3, it's lot more complicated but there is a pattern. As usual compute the determinant first (kind of a pain; but surely you already know the pattern to compute that quickly).

$$ \left(\begin{array}{ccc} a&b&c\\ d&e&f\\ g&h&i \end{array}\right)^{-1} = {1 \over {\rm{det}}} \left(\begin{array}{rrr} e i - f h&-(b i - c h)&b f - c e\\ -(d i - f g)&a i - c g&-(a f -c d)\\ d h - e g&-(a h - b g)&a e - b d \end{array}\right). $$ The pattern is that each entry is

  • the determinant of the submatrix gotten by removing that row and column. I.e. for row 2 column 3 (the $f$ position, the determinant is $a h - b g$: $$ \det\left(\begin{array}{cc} a&b\\ g&h \end{array}\right) = a h - b g $$

  • then multiply in the checkerboard pattern. (i.e.1x1 is positive, 1x2 is negative... mathematically it's multiply by $(-1)^r(-1)^c$.

  • Then transpose.

See? There's a pattern, but I feel it's about the same symbolic complexity as just doing it brute force Gaussian-elimination style.

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Is there any formula for 3x3 , like the one above –  SpongeBob SquarePants Feb 11 '11 at 15:17
    
Yes...but...it involves the determinant of the 3x3 and all the 2x2 submatrices. I thought that that isn't much of a trick or shortcut; it seems about the same complexity as just plodding through row/column operations to convert the 3x3 into an identity matrix and applying those operations to an identity matrix at the same time. Of course, if there's an expectation that the determinant is 1, then maybe it's appropriate. Also, be warned that the row/column operations are 'meaningful' (you see that they are computing the inverse) but the 'trick' is just blind application of a formula. –  Mitch Feb 11 '11 at 15:24
    
Yes there is but is not as nice. There is a formula like this for square matrices of any size. If $A$ is your matrix then the $ij$ entry of $A^{-1}$ is $(-1)^{i+j}/\det(A) \cdot \det(minor(A(j,i))$ where $minor(A(j,i))$ is my nonstandard and poor notation for the $(n-1)\times (n-1)$ minor of $A$ obtained by removing the $j$th row and the $i$th column. In the $3\times 3$ case this says the entry in the first column and first row of $A^{-1}$ is $+1/\det(A)\cdot (a_{22}a_{33} - a_{23}a_{32})$. For sparse matrices Gauss-Jordan is ok. If all entires of A are nonzero this formula is faster. –  solbap Feb 11 '11 at 15:29
    
@Mitch The above mentioned method, does it work for all non - singular 2x 2 matrices ? –  SpongeBob SquarePants Feb 11 '11 at 15:35
    
@abcdefghijklmnopqrstuvwxyz: Well, sorta. if it's nonsingular, the determinant is 0, and so the method will work in that it will also fail when the inverse of a matrix will fail (when it is non-singular). –  Mitch Feb 11 '11 at 15:48

Your best bet is the Gauss-Jordan method.

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any other method ? –  SpongeBob SquarePants Feb 11 '11 at 15:21
    
@SpongeBob: By the way, Gauss-Jordan elimination is much faster than the Laplace expansion (the method Mitch mentioned). It also skips over having to find the determinant, and it works for any square matrix. –  Zach Langley Apr 17 '11 at 14:17

So we want to find out a way to compute $2 \times 2 ~\text{ or }~ 3 \times 3$ matrix systems the most efficient way. Well I think the route that we want to go would be to use Cramer's Rule for the $2 \times 2 \text{ or } 3 \times 3$ case. To state the $2 \times 2$ case we will use the following:

For some coefficient matrix A= $\left[ \begin{array}{rr} a & b \\ c & d \end{array} \right]$

$A^{-1}=\dfrac{1}{ad-bc} \cdot \left[ \begin{array}{rr} d & -b \\ -c & a \end{array} \right]~ \iff ad-bc \ne 0$ $~~~~~~~~~\Big($i.e., Det(A)$~\ne ~ 0\Big)$

For the $3 \times 3$ case, we will denote that as the folllowing:

$x_{1} = \dfrac{|b~~x_{2}~~x_{3}|}{|\bf{A}|}~~,$

$x_{2} = \dfrac{|x_{1}~~b~~x_{3}|}{|\bf{A}|}~~,$

$x_{3} = \dfrac{|x_{1}~~x_{2}~~b|}{|\bf{A}|}.$

This comes from the matrix equation: ${\bf{A\vec{x}}}={\bf{\vec{b}}},~~~$ where $\vec{x}=[x_{1}~~x_{2}~~x_{3}]^{T}$.

For the elements of matrix $A = \left|\begin{array}{rrr} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right|,~~$ it can be extended for the solutions $x_{1},~x_{2},~x_{3}$

as so knowing that ${\bf|{A}| =} ~ |a_{ij}| ~ \not= ~ 0.$

$x_{1} = \dfrac{1}{|{\bf{A}}|} \left|\begin{array}{rrr} b_1 & a_{12} & a_{13} \\ b_2 & a_{22} & a_{23} \\ b_3 & a_{32} & a_{33} \end{array} \right|$,

$x_{2} = \dfrac{1}{|{\bf{A}}|} \left|\begin{array}{rrr} a_{11} & b_1 & a_{13} \\ a_{21} & b_2 & a_{23} \\ a_{31} & b_3 & a_{33} \end{array} \right|$,

$x_{3} = \dfrac{1}{|{\bf{A}}|} \left|\begin{array}{rrr} a_{11} & a_{12} & b_1 \\ a_{21} & a_{22} & b_2 \\ a_{31} & a_{32} & b_3 \end{array} \right|$.

An alternate way of doing this would be using row reducing methods, known as either Gaussian Elimination( ref ) or Gauss-Jordan Elimination( rref ).

I hope this helped out. Let me know if there if anything you do not understand.

Thanks.

Good Luck.

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