10 min readConcept sheet1-mark MCQ or assertion-reason; 2-mark 'show that B is the inverse of A'; the uniqueness proof as a 2 or 3-mark answer

Invertible matrices and the uniqueness of the inverse

A square matrix A is invertible if some B gives AB = BA = I; then B = A⁻¹. The inverse, when it exists, is unique.

Invertible matrixA square matrix A of order n is invertible if there is a matrix B of order n with AB = BA = I. Then B = A⁻¹.

UniquenessIf B and C are both inverses of A, then B = C.

Inverse of a product(AB)⁻¹ = B⁻¹A⁻¹ for invertible A and B of the same order

Not every matrix has oneO has no inverse, and neither has any A with AB = O for some B ≠ O, such as [1, 2; 2, 4].

What an inverse is

For numbers, 5 × (1/5) = 1, so 1/5 undoes multiplication by 5. For a square matrix A, an inverse is a matrix B of the same order with AB = BA = I. If such a B exists, A is called invertible and B is written A⁻¹. Only square matrices can be invertible, and there is no division of matrices: A⁻¹ is not '1/A'. Since the condition AB = BA = I reads the same with A and B swapped, B is invertible too, with B⁻¹ = A. (Matrices are written row by row here, with rows separated by semicolons.)

Proof: the inverse of a square matrix, if it exists, is unique

1
Let A be a square matrix, and suppose B and C are both inverses of A.
2
Then AB = BA = I and AC = CA = I.
3
B = BI, because I is the identity matrix.
4
= B(AC), because AC = I.
5★
= (BA)C, because matrix multiplication is associative.
6
= IC = C, because BA = I.
7
So, B = C: an invertible matrix has exactly one inverse.

★ key step

Proof: (AB)⁻¹ = B⁻¹A⁻¹

1
(AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AIA⁻¹ = AA⁻¹ = I
2
(B⁻¹A⁻¹)(AB) = B⁻¹(A⁻¹A)B = B⁻¹IB = B⁻¹B = I
3★
So B⁻¹A⁻¹ is an inverse of AB. By uniqueness it is the inverse: (AB)⁻¹ = B⁻¹A⁻¹.

★ key step

Example 1: check an inverse

Problem

Show that B = [2, −5; −1, 3] is the inverse of A = [3, 5; 1, 2].

Working

Rule: B = A⁻¹ if AB = I and BA = I.
AB = [(3)(2) + (5)(−1), (3)(−5) + (5)(3); (1)(2) + (2)(−1), (1)(−5) + (2)(3)] = [1, 0; 0, 1]
BA = [(2)(3) + (−5)(1), (2)(5) + (−5)(2); (−1)(3) + (3)(1), (−1)(5) + (3)(2)] = [1, 0; 0, 1]
Both products equal I.

Answer

AB = BA = I, so B = A⁻¹.

Example 2: the inverse from a matrix equation (board style)

Problem

If A = [1, 2; −1, 3], show that A² − 4A + 5I = O. Hence find A⁻¹.

Working

Rule: multiply row by column; then rearrange the equation into the form A × (a matrix) = I.
A² = [(1)(1) + (2)(−1), (1)(2) + (2)(3); (−1)(1) + (3)(−1), (−1)(2) + (3)(3)] = [−1, 8; −4, 7]
4A = [4, 8; −4, 12] and 5I = [5, 0; 0, 5]
A² − 4A + 5I = [−1 − 4 + 5, 8 − 8 + 0; −4 + 4 + 0, 7 − 12 + 5] = O
So 4A − A² = 5I, that is A(4I − A) = 5I. In the same way, (4I − A)A = 5I.
Divide by 5: A × (1/5)(4I − A) = I and (1/5)(4I − A) × A = I.
So A⁻¹ = (1/5)(4I − A) = (1/5)[4 − 1, 0 − 2; 0 + 1, 4 − 3] = (1/5)[3, −2; 1, 1].
Check: A × [3, −2; 1, 1] = [3 + 2, −2 + 2; −3 + 3, 2 + 3] = [5, 0; 0, 5] = 5I.

Answer

A⁻¹ = (1/5)[3, −2; 1, 1]

Which matrices have no inverse?

A matrix that is not square: the definition needs AB and BA to be the same identity matrix.
The zero matrix O: OB = O ≠ I for every B.
Any A with AB = O for some B ≠ O. If A⁻¹ existed, then B = A⁻¹(AB) = A⁻¹O = O, which is false.
So [1, 2; 2, 4] has no inverse, because [1, 2; 2, 4][2, −6; −1, 3] = O.
In the next chapter you will find inverses using determinants.

In real life

A secret message can be coded by multiplying blocks of numbers by a matrix A. The receiver decodes it by multiplying by A⁻¹. This works only because the inverse exists and is unique.

Exam angle

1 marks: “If A is invertible, what is (A⁻¹)⁻¹?” A, because A⁻¹A = AA⁻¹ = I shows that A is the inverse of A⁻¹.

2 marks: “Show that B = [3, −1; −5, 2] is the inverse of A = [2, 1; 5, 3].” AB = [6 − 5, −2 + 2; 15 − 15, −5 + 6] = I and BA = [6 − 5, 3 − 3; −10 + 10, −5 + 6] = I. So B = A⁻¹.

3 marks: “Prove that the inverse of a square matrix, if it exists, is unique.” Suppose B and C are both inverses of A. Then B = BI = B(AC) = (BA)C = IC = C, using AC = I, the associative law and BA = I. So B = C.

Remember it

One lock, one key: an invertible matrix has exactly one inverse. The proof chain: B = BI = B(AC) = (BA)C = IC = C.