Find the inverse of a 3 by 3 matrix

Question

Challenge

Given nine numbers, a, b, c, d, e, f, g, h, i, as input which correspond to the square matrix:

$$\mathbf{M} = \begin{pmatrix}a& b& c\\ d& e& f\\ g& h& i\end{pmatrix}$$

Find the inverse of the matrix, \$\mathbf{M}^{-1}\$ and output its components.

Inverse Matrix

The inverse of a matrix 3 by 3 obeys the following equation:

$$\mathbf{MM}^{-1} = \mathbf{M}^{-1}\mathbf{M} = \mathbf{I} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$

And can be calculated as:

$$\mathbf{M}^{-1} = \frac{1}{\det(\mathbf{M})}\mathbf{C}^T$$

Where \$\mathbf{C}\$ is the matrix of cofactors:

$$\mathbf{C}=\begin{pmatrix}ei-fh&fg-di&dh-eg\\ch-bi&ai-cg&bg-ah\\bf-ce&cd-af&ae-bd\end{pmatrix}$$

And \$\mathbf{C}^T\$ is the transpose of \$\mathbf{C}\$:

$$\mathbf{C}^T = \begin{pmatrix}ei-fh&ch-bi&bf-ce\\fg-di&ai-cg&cd-af\\dh-eg&bg-ah&ae-bd\end{pmatrix}$$

And \$\det(\mathbf{M})\$ is the determinant of \$\mathbf{M}\$:

$$\det(\mathbf{M}) = a(ei-fh)-b(di-fg)+c(dh-eg)$$

Worked Example

For example, let's say the input is 0, -3, -2, 1, -4, -2, -3, 4, 1. This corresponds to the matrix:

$$\mathbf{M} = \begin{pmatrix}0&-3&-2\\1&-4&-2\\-3&4&1\end{pmatrix}$$

Firstly, let's calculate what's known as the determinant using the formula above:

$$\det(\mathbf{M}) = 0(-4\times1-(-2)\times4) - (-3)(1\times1-(-2)\times-3) + (-2)(1\times4-(-4)\times-3) = 1$$

Next let's calculate the matrix of cofactors:

$$\mathbf{C} = \begin{pmatrix}-4\times1-(-2)\times4& -(1\times1-(-2)\times-3)&1\times4-(-4)\times-3\\-(-3\times1-(-2)\times4)&0\times1-(-2)\times-3&-(0\times4-(-3)\times-3)\\-3\times-2-(-2)\times-4&-(0\times-2-(-2)\times1)&0\times-4-(-3)\times1\end{pmatrix}$$

$$= \begin{pmatrix}4&5&-8\\-5&-6&9\\-2&-2&3\end{pmatrix}$$

We then need to transpose \$\mathbf{C}\$ (flip the rows and columns) to get \$\mathbf{C}^T\$:

$$\mathbf{C}^T = \begin{pmatrix}4&-5&2\\5&-6&-2\\-8&9&3\end{pmatrix}$$

Finally, we can find the inverse as:

$$\mathbf{M}^{-1} = \frac{1}{\det(\mathbf{M})}\mathbf{C}^T = \frac{1}{1}\begin{pmatrix}4&-5&2\\5&-6&-2\\-8&9&3\end{pmatrix}=\begin{pmatrix}4&-5&2\\5&-6&-2\\-8&9&3\end{pmatrix}$$

So the output would be 4, -5, -2, 5, -6, -2, -8, 9, 3.

Rules

• The given matrix will always have an inverse (i.e. non-singular). The matrix may be self-inverse

• The given matrix will always be a 3 by 3 matrix with 9 integers

• The numbers in the input will always be integers in the range \$-1000 \leq n \leq 1000\$

• Non-integer components of the matrix may be given as a decimal or a fraction

Examples

Input > Output
1, 0, 0, 0, 1, 0, 0, 0, 1 > 1, 0, 0, 0, 1, 0, 0, 0, 1
0, -3, -2, 1, -4, -2, -3, 4, 1 > 4, -5, -2, 5, -6, -2, -8, 9, 3
1, 2, 3, 3, 1, 2, 2, 1, 3 > -1/6, 1/2, -1/6, 5/6, 1/2, -7/6, -1/6, -1/2, 5/6
7, 9, 4, 2, 7, 9, 3, 4, 5 > -1/94, -29/94, 53/94, 17/94, 23/94, -55/94, -13/94, -1/94, 31/94

Winning

The shortest code in bytes wins.

Answer 1

MATL, 54 bytes

th3LZ)t,3:q&XdpswP]w-lw/GtY*tXdsGXdsUw-IXy*2/+GtXds*-*

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Just to keep it interesting, doesn't use the inbuilt matrix division or determinant functions to do it.

Instead, computes the determinant using the Rule of Sarrus.

And the adjugate (transposed cofactor matrix) using Cayley–Hamilton formula.

$$ {\displaystyle \operatorname {adj} (\mathbf {A} )={\frac {1}{2}}\left((\operatorname {tr} \mathbf {A} )^{2}-\operatorname {tr} \mathbf {A} ^{2}\right)\mathbf {I} _{3}-\mathbf {A} \operatorname {tr} \mathbf {A} +\mathbf {A} ^{2}.} $$

Commented code:

% Finding determinant
th    % concatenate the matrix to itself sideways
3LZ)  % chop off the last column (since the Rule of Sarrus doesn't need it)
t     % duplicate this matrix (say S)
,     % do this twice:
  3:q&Xd  % get the first three diagonals of S
  ps      % multiply each diagonal's values and add the results
  wP      % switch and flip the matrix (to get the popposing diagonals next time)
]w    % close loop, switch to have correct order of sums
-     % subtract - we now have the determinant
lw/   % invert that

% Finding adjugate using Cayley–Hamilton formula
GtY*  % A^2 term (last term of the formula)
tXds  % trace(A^2) for term 1 of formula
GXdsU % (trace(A))^2 for term1 of formula
w-    % (trace(A))^2 - trace(A^2)
IXy*  % multiply that by the identity matrix
2/    % divide that by 2 - term 1 complete
+
GtXds* % A*trA for term 2 of formula
-      % subtract to get adj(A)

*      % multiply by the inverse of determinant we found earlier
       % implicit output

We could go even more insane purer by replacing the matrix multiplication GtY* done for \$ A^2 \$, with something like 3:"Gt!@qYS*!s] 3$v t&v 3:K-&Xd (Try it on MATL Online).

The more direct and obvious way:

4 bytes

-1Y^

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(-1 byte thanks to @Luis Mendo.)

-1 - Push the literal -1

Y^ - Raise input to that power (implicit input, implicit output)

Answer 2

APL(Dyalog Classic),1 byte

⌹

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if a flat lis is required this is 8 bytes

,∘⌹3 3∘⍴

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Answer 3

R, 51 35 27 8 5 bytes

solve

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First go at doing one of these golf challenges. Sorry if my formatting is wrong!

Saved a total additional 11 bytes thanks to Giuseppe! Saved an additional 19 bytes thanks to JAD!

Answer 4

Jelly, 3 bytes

æ*-

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Assuming we can take input and provide as a 2D list of integers. If a flat list of integers is really required for both input and output, then this works for 6 bytes.

Answer 5

JavaScript (ES6), 123 bytes

Saved 2 bytes thanks to @Mr.Xcoder
Saved 1 byte thanks to @ETHproductions

Takes input as 9 distinct values.

(a,b,c,d,e,f,g,h,i)=>[x=e*i-h*f,c*h-b*i,b*f-c*e,y=f*g-d*i,a*i-c*g,d*c-a*f,z=d*h-g*e,g*b-a*h,a*e-d*b].map(v=>v/=a*x+b*y+c*z)

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Answer 6

J, 2 bytes

%.

Just a built-in primitive

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Answer 7

Python 2, 139 bytes

def F(a,b,c,d,e,f,g,h,i):x=e*i-f*h;y=f*g-d*i;z=d*h-e*g;print[j/(a*x+b*y+c*z)for j in x,c*h-b*i,b*f-c*e,y,a*i-c*g,c*d-a*f,z,b*g-a*h,a*e-b*d]

Try it online! (Has return instead of print for ease of testing.)

Answer 8

Python 3, 77 bytes

import numpy
lambda l:(numpy.matrix(l).reshape(-1,3)**-1).ravel().tolist()[0]

Takes input as a flat list.

It's 63 bytes if input is taken as a 2D array:

import numpy
lambda l:(numpy.matrix(l)**-1).ravel().tolist()[0]

Answer 9

Google Sheets, 16 bytes

=MINVERSE(A1:C3)

Input is in the range A1:C3

^{Built-ins are boring}

Answer 10

Clean, 143 bytes

import StdEnv
$a b c d e f g h i#p=e*i-h*f
#q=f*g-d*i
#r=d*h-g*e
=[v/(a*p+b*q+c*r)\\v<-[p,c*h-b*i,b*f-c*e,q,a*i-c*g,d*c-a*f,r,g*b-a*h,a*e-d*b]]

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Answer 11

Perl 5, 179 bytes

sub{($a,$b,$c,$d,$e,$f,$g,$h,$i)=@_;map$_/($a*$x+$b*$y+$c*$z),$x=$e*$i-$f*$h,$c*$h-$b*$i,$b*$f-$c*$e,$y=$f*$g-$d*$i,$a*$i-$c*$g,$c*$d-$a*$f,$z=$d*$h-$e*$g,$b*$g-$a*$h,$a*$e-$b*$d}

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Answer 12

Pari/GP, 6 bytes

m->1/m

Takes multiplicative inverse in the matrix ring \$\mathrm{M}_n\$.

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Answer 13

APL (Dyalog), 7 bytes

,⌹3 3⍴⎕

Takes input as a flat list and outputs as a flat list

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Answer 14

Factor + math.matrices.extras, 5 bytes

recip

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Answer 15

Ruby + -rmatrix, 24 bytes

->a{Matrix[*a].inv.to_a}

Or, if taking in and outputting a Matrix is apart of standard I/O, then this becomes 10 bytes.

->a{a.inv}

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Answer 16

Perl, 226 + 4 (-plF, flag) = 230 bytes

$_=join', ',map$_/($a*$x+$b*$y+$c*$z),$x=($e=$F[4])*($i=$F[8])-($f=$F[5])*($h=$F[7]),($c=$F[2])*$h-($b=$F[1])*$i,$b*$f-$c*$e,$y=$f*($g=$F[6])-($d=$F[3])*$i,($a=$F[0])*$i-$c*$g,$c*$d-$a*$f,$z=$d*$h-$e*$g,$b*$g-$a*$h,$a*$e-$b*$d

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Answer 17

Noether, 168 bytes

I~aI~bI~cI~dI~eI~fI~gI~hI~iei*fh*-a*di*fg*-b*-dh*eg*-c*+~zei*fh*-z/P","~nPch*bi*-z/PnPbf*ce*-z/PnPfg*di*-z/PnPai*cg*-z/PnPcd*af*-z/PnPdh*eg*-z/PnPbg*ah*-z/PnPae*bd*-z/P

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Answer 18

Clojure, 165 bytes

(fn[a b c d e f g h i](let[M map C(M -(M *[e f d c a b b c a][i g h h i g f d e])(M *[f d e b c a c a b][h i g i g h e f d]))](for[i C](/ i(apply +(M *[a b c]C))))))

I'm sorry this outputs C in transpose, and I'm feeling lazy to re-do those long character sequences to fix it at the moment.