# Inverse of 2x2 matrix

• Jun 21st 2012, 08:30 AM
Soroban
Inverse of 2x2 matrix

In case you didn't know . . .

Find the inverse of: . $A \;=\;\begin{bmatrix}a&b\\c&d\end{bmatrix}$

[1] Switch the two on the main diagonal $(a\text{ and }d): \;\begin{bmatrix}d & . \\ . & a\end{bmatrix}$

[2] Change the signs of the other two: . $\begin{bmatrix}. & -b \\ -c & . \end{bmatrix}$

[3] Divide everything by the determinant: . $\begin{vmatrix}a&b\\c&d\end{vmatrix} \:=\:ad - bc$

$\text{Therefore: }\:A^{-1} \;=\;\begin{bmatrix}\dfrac{d}{ad-bc} & \dfrac{-b}{ad-bc} \\ \dfrac{-c}{ad-bc} & \dfrac{a}{ad-bc}\end{bmatrix}$

• Jun 26th 2012, 05:04 PM
Deveno
Re: Inverse of 2x2 matrix
being somewhat bad at "memorizing things" i often have to "re-derive" this formula. here is how i do it:

suppose A =

[a b]
[c d].

i want to find B =

[x y]
[z w], so that AB = I. so:

ax + bz = 1
ay + bw = 0
cx + dz = 0
cy + dw = 1

that's 4 linear equations in 4 unknowns (x,y,z,w). if it has a solution, there will be only one.

from the first equation, and the third, we have:

-bcx - bdz = 0

thus (ad - bc)x = d. if ad - bc ≠ 0, this gives:

from the second, and fourth equation, we get:

-bcy - bdw = -b

thus (ad - bc)y = -b, and again if ad - bc ≠ 0, this gives us:

to solve for z, we use equations one and three again:

acx + bcz = c

thus (bc - ad)z = c, so

z = c/(bd - ac) = -c/(ad - bc)

finally, from equations two and four:

acy + bcw = 0