Theory
Matrices have no division key
If 3x = 12, you divide by 3 and get x = 4.
Now the matrix version: A·X = B, where A and B are known matrices and X is unknown. You want to "divide by A", but matrix division does not exist. There is no ÷ for grids.
So mathematicians built the next best thing: a matrix that undoes A. Multiply by it, and A's effect is cancelled.
Theory
The undo button
Dividing by 3 is really multiplying by ⅓, because 3 × ⅓ = 1. The inverse matrix plays the role of ⅓: multiplying A by A⁻¹ gives the matrix version of 1, called the identity matrix I, with 1s on the diagonal and 0s elsewhere. Multiplying anything by I changes nothing, just like multiplying a number by 1.
Theory
The inverse, formally
For a square matrix A, the inverse A⁻¹ is the matrix with:
A·A⁻¹ = A⁻¹·A = I
It exists only when A's determinant is not zero. For a 2×2:
A = ⎡ a b ⎤ |A| = ad - bc
⎣ c d ⎦
If |A| ≠ 0, A is non-singular and has an inverse. If |A| = 0, A is singular: no inverse, full stop.
Quiz
What is the determinant of the matrix with rows (3 1) and (5 2)?
- 1
- 11
- -1
- 6
Show the answer
1
ad - bc = 3·2 - 1·5 = 6 - 5 = 1. If you got 11 you added ad + bc, the single most common determinant slip. It is a difference, not a sum.
Follow along
The 2×2 inverse recipe
- Compute |A| = ad - bc One subtraction. Do it first, always.
- If |A| = 0, stop A is singular: no inverse exists. Writing anything else loses marks.
- Swap the main diagonal a and d trade places.
- Flip the signs of the other two b becomes -b, c becomes -c. They stay in their seats, only the signs change.
- Divide every entry by |A| The step everyone forgets. The answer is that whole matrix over the determinant.
Theory
Worked example, with a professor's check
A = ⎡ 3 1 ⎤ and |A| = 3·2 - 1·5 = 1.
⎣ 5 2 ⎦
Swap 3 and 2, negate 1 and 5, divide by 1:
A⁻¹ = ⎡ 2 -1 ⎤
⎣ -5 3 ⎦
Now verify, row into column: 3·2 + 1·(-5) = 1, 3·(-1) + 1·3 = 0, 5·2 + 2·(-5) = 0, 5·(-1) + 2·3 = 1. That is exactly I. Thirty seconds of multiplication turns "I hope it is right" into "it is right".
Think first
Your turn
B = ⎡ 2 1 ⎤
⎣ 7 4 ⎦
Find B⁻¹ on paper: determinant first, then swap, sign, divide.
Show the answer
|B| = 2·4 - 1·7 = 1. Swap 2 and 4, negate 1 and 7, divide by 1:
B⁻¹ = ⎡ 4 -1 ⎤
⎣ -7 2 ⎦
If your answer had 2 and 4 negated instead, you flipped the wrong diagonal: the main diagonal swaps, the other one changes sign.
Quiz
Which of these matrices has NO inverse?
- Rows (2 4) and (1 2)
- Rows (1 0) and (0 1)
- Rows (3 1) and (5 2)
- Rows (0 1) and (1 0)
Show the answer
Rows (2 4) and (1 2)
Its determinant is 2·2 - 4·1 = 0, so it is singular: no inverse. The identity (option B) is its own inverse, and option D has determinant -1, which is fine: negative determinants still give inverses, only zero kills it.
Watch out
The three mark-eaters
Forgetting to divide by |A|: your matrix is right but every entry is scaled wrong. Flipping the wrong diagonal: a and d swap seats, b and c swap signs, never both. Ignoring |A| = 0: if the determinant is zero, the only correct answer is "A is singular, the inverse does not exist", and that sentence earns the marks.
Formula
Exam recipe
For "find the inverse of A": write |A| = ad - bc first. If zero, state "singular, no inverse". Otherwise apply swap, sign, divide, then multiply A·A⁻¹ in the margin to confirm it gives I. This layout is exactly what examiners award full marks to.
Summary
Key takeaways
- I (1s on the diagonal, 0s elsewhere) is the matrix version of 1.
- A⁻¹ undoes A: AA⁻¹ = A⁻¹A = I.
- The inverse exists only when |A| = ad - bc is not zero (non-singular).
- 2×2 recipe: swap a and d, negate b and c, divide everything by |A|.
- Always verify: multiply back and check you get I.
- Memory hook: swap, sign, divide.