Finding the Inverse of a 2x2 Matrix from Scratch
Finding the Inverse of a 2x2 Matrix from Scratch
This post shows a complete, step-by-step derivation of the inverse of a 2x2 matrix. Everything is expressed using stable, browser-safe ASCII formatting so the layout displays correctly on all devices and all templates.
FIRST PART.
Start with the matrix equation:
A = [[a, b], [c, d]] A^(-1) = [[w, x], [y, z]] Goal: A * A^(-1) = I
This produces the column equations:
[aw + by, cw + dy]^T = [1, 0]^T [ax + bz, cx + dz]^T = [0, 1]^T
Which gives the four equations:
aw + by = 1 cw + dy = 0 ax + bz = 0 cx + dz = 1
SECOND PART.
Use the first two equations to find w.
aw + by = 1 cw + dy = 0
Multiply:
(ad)w + (bd)y = d (first eq multiplied by d) (bc)w + (bd)y = 0 (second eq multiplied by b)
Subtract:
(ad - bc)w = d w = d / (ad - bc) (ad - bc != 0)
THIRD PART.
Use the next pair to find x.
ax + bz = 0 cx + dz = 1
Multiply:
(ad)x + (bd)z = 0 (bc)x + (bd)z = b
Subtract:
(ad - bc)x = -b x = -b / (ad - bc) (ad - bc != 0)
FOURTH PART.
Find y from cw + dy = 0.
cw + dy = 0 dy = -cw y = (-cw) / d
Substitute w = d / (ad - bc):
y = (-c / d) * (d / (ad - bc)) y = -c / (ad - bc) (ad - bc != 0)
FIFTH PART.
Find z from ax + bz = 0.
ax + bz = 0 bz = -ax
Substitute x = -b / (ad - bc):
bz = (-a) * (-b / (ad - bc)) bz = ab / (ad - bc) z = a / (ad - bc) (ad - bc != 0)
FINAL PART.
Collect the four entries:
w = d / (ad - bc) x = -b / (ad - bc) y = -c / (ad - bc) z = a / (ad - bc)
So the matrix inverse is:
A = [[a, b], [c, d]] A^(-1) = (1 / (ad - bc)) * [[d, -b], [-c, a]] Condition = ad - bc != 0
Check:
A * A^(-1) = I A^(-1) * A = I
And:
det(A) = ad - bc
FINAL NOTES.
- If det(A) = 0, then A is singular and has no inverse.
- The same idea applies to 3x3 matrices: determinant zero means no inverse.
- The minor of an element in a 3x3 matrix is the determinant of the submatrix left after removing the element’s row and column.
Personalised notes based on FP3, Edexcel, Pearson.
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