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Math Help - Inverse of 2x2 matrix

  1. #1
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    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}

    Thanks from emakarov
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  2. #2
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    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:

    adx + bdz = d
    -bcx - bdz = 0

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

    x = d/(ad - bc).

    from the second, and fourth equation, we get:

    ady + bdw = 0
    -bcy - bdw = -b

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

    y = -b/(ad - bc).

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

    acx + bcz = c
    -acx - adz = 0

    thus (bc - ad)z = c, so

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

    finally, from equations two and four:

    acy + bcw = 0
    -acy - adw = -a

    (bc - ad)w = -a

    w = -a/(bc - ad) = a/(ad - bc).

    it should come as no surprise that is is 1/(det(A) (= 1/(ad - bc)) times:

    [d -b]
    [-c a], as Soroban points out above.
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