Invertible matrices

What is an invertible matrix?

In plain words: A^⁻¹ undoes A in the way 1/5 undoes 5, since 5 × 1/5 = 1. Because matrix products depend on order, both AB = I and BA = I are asked for. This is the matrix version of an invertible function, where gof and fog must both be identities.

NCERT’s example (page 68):

AB has rows 4 − 3, −6 + 6 and 2 − 2, −3 + 4, that is I, and BA = I as well. So B = A^⁻¹, and also A = B^⁻¹: if B is the inverse of A, then A is the inverse of B (NCERT page 69).

Not every square matrix has an inverse. For example, the matrix with rows 1, 2 and 2, 4 has none: in any product with it on the left, the second row is twice the first, so the product can never be I.

Finding an inverse from AB = I

NCERT Section 3.7 gives no method for finding an inverse. Beyond the NCERT text (reasoned from NCERT Definition 6), one can be found within this chapter by writing B with unknown entries and solving AB = I. For A with rows 3, 1 and 5, 2, let B have rows p, q and r, s:

• row 1 of AB: 3p + r = 1 and 3q + s = 0;
• row 2 of AB: 5p + 2r = 0 and 5q + 2s = 1.

The first and third give p = 2, r = −5; the second and fourth give q = −1, s = 3. A check shows BA = I too, so A^⁻¹ has rows 2, −1 and −5, 3.

A shortcut for 2 × 2 matrices. Beyond this section, Chapter 4 of NCERT proves it with the adjoint: if ad − bc ≠ 0, the matrix with rows a, b and c, d has inverse 1/(ad − bc) times the matrix with rows d, −b and −c, a. For the example above, ad − bc = 6 − 5 = 1, and the formula gives rows 2, −1 and −5, 3 again.

The inverse is unique

Inverse of a square matrix, if it exists, is unique.

NCERT Mathematics Part I (Class 12), Section 3.7, page 69

Proof (NCERT Theorem 3, page 69): if B and C are both inverses of A, then AB = BA = I and AC = CA = I, so

B = BI = B(AC) = (BA)C = IC = C.

So it makes sense to speak of the inverse A^⁻¹.

Inverse of a product

NCERT Theorem 4 (page 69): if A and B are invertible matrices of the same order, then

The order reverses, as for the transpose and as when undoing two steps: to undo “put on socks, then shoes”, take off the shoes first. A direct check: (AB)(B^⁻¹A^⁻¹) = A(BB^⁻¹)A^⁻¹ = A I A^⁻¹ = AA^⁻¹ = I, and in the same way (B^⁻¹A^⁻¹)(AB) = I. NCERT’s proof starts from (AB)(AB)^⁻¹ = I and multiplies on the left by A^⁻¹ and then B^⁻¹.

NCERT slip (page 69): the first line of the proof of Theorem 4 is printed “(AB) (AB)^⁻¹ = 1”; it should be I, the identity matrix, as the next line (“= A^⁻¹I”) uses. In an exam, write I.

NCERT’s Note 2 (page 69), that A is the inverse of B whenever B is the inverse of A, already gives (A^⁻¹)^⁻¹ = A. One more fact, beyond the NCERT text: (A′)^⁻¹ = (A^⁻¹)′, since A′(A^⁻¹)′ = (A^⁻¹A)′ = I and (A^⁻¹)′A′ = (AA^⁻¹)′ = I.

Solved examples

Common mistakes to avoid

Quick revision summary

• A square matrix A is invertible if some B of the same order has AB = BA = I; then B = A^⁻¹ and A = B^⁻¹.
• The inverse, when it exists, is unique; rectangular matrices have none.
• (AB)^⁻¹ = B^⁻¹A^⁻¹, with the order reversed.

Questions students ask

Do invertible matrices have to be square, or can a rectangular matrix have AB = I?

Invertible matrices have to be square, but a rectangular matrix can meet one of the two conditions: beyond the NCERT text, a B with AB = I is called a right inverse of A, and one with BA = I a left inverse (Wikipedia, “Invertible matrix”). For the row matrix A = [3, 2], the columns B with entries 1, −1 and C with entries −1, 2 both give AB = AC = [1], the identity of order 1. But no column X makes XA the identity of order 2, since every row of XA is a multiple of [3, 2]: BA, for one, has rows 3, 2 and −3, −2. So a one-sided inverse need not be unique; it is the two conditions together, AB = I and BA = I, that the proof of uniqueness uses.

Is the sum of two invertible matrices invertible?

No, the sum of two invertible matrices need not be invertible, although their product always is, with (AB)^⁻¹ = B^⁻¹A^⁻¹. The simplest case is I + (−I) = O, and the zero matrix has no inverse. A less obvious one: B with rows 1, 2 and 1, 0 is invertible (its inverse has rows 0, 1 and ½, −½), yet I + B has rows 2, 2 and 1, 1, whose first row is twice its second; then every product (I + B)X also has its first row twice its second, so it is never I. A sum can be invertible too, as A + A = 2A is whenever A is.

What is the difference between the additive inverse and the inverse of a matrix?

The additive inverse of a matrix A is −A, which gives the zero matrix when added, A + (−A) = O (NCERT page 47), while the inverse A^⁻¹ gives the identity matrix when multiplied, AA^⁻¹ = A^⁻¹A = I. Every matrix has an additive inverse, found by changing the sign of every element, as in the laws for addition, but only some square matrices have an inverse. For A with rows 4, 3 and 5, 4, −A has rows −4, −3 and −5, −4, while A^⁻¹ has rows 4, −3 and −5, 4, as AA^⁻¹ = A^⁻¹A = I confirms. The zero matrix O is its own additive inverse but has no inverse at all, since OB = O ≠ I for every B.

Is an invertible matrix the same as an inverse matrix?

An invertible matrix is not the same thing as an inverse matrix: invertible describes a square matrix A that has an inverse, while the inverse A^⁻¹ is the partner B of the same order with AB = BA = I (NCERT Definition 6, page 68). The two come in pairs, since B is then invertible too, with inverse A (NCERT page 69). For example, A with rows 2, 1 and 1, 1 is invertible, and its inverse, with rows 1, −1 and −1, 2, is also invertible: both products are I. A square matrix with no such partner is neither; Chapter 4 shows these are exactly the singular matrices, with |A| = 0.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Show that the matrices with rows 4, 7 and 1, 2, and with rows 2, −7 and −1, 4, are inverses of each other.

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   Call them A and B. AB has rows 8 − 7, −28 + 28 and 2 − 2, −7 + 8, that is I; BA has rows 8 − 7, 14 − 14 and −4 + 4, −7 + 8, also I.

2. Q2. Find the inverse of the matrix with rows 2, 5 and 1, 3 by solving AB = I.

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   With B having rows p, q and r, s: 2p + 5r = 1, p + 3r = 0 give p = 3, r = −1; 2q + 5s = 0, q + 3s = 1 give q = −5, s = 2. So A^⁻¹ has rows 3, −5 and −1, 2, and BA = I as well.

3. Q3. If A and B are invertible matrices of the same order with A^⁻¹ having rows 1, 2 and 0, 1, and B^⁻¹ having rows 3, 0 and 1, 1, find (AB)^⁻¹.

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   (AB)^⁻¹ = B^⁻¹A^⁻¹. Row 1 of B^⁻¹ times the columns of A^⁻¹ gives 3, 6; row 2 gives 1, 3. So (AB)^⁻¹ has rows 3, 6 and 1, 3.

4. Q4. A is a square matrix with A² = I. What is A^⁻¹?

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   AA = I, so A is its own inverse: A^⁻¹ = A. For example, the matrix with rows 0, 1 and 1, 0 squares to I.