Matrices · Lesson 7 of 8
Inverse Matrices and Product Equations
“Recognise two-sided inverses, prove their laws, and solve product equations without changing factor order.”
• Verify that two square matrices are inverses by identity products. • Distinguish an inverse under multiplication from a negative under addition. • Prove uniqueness of an inverse and the reverse-order law for product inverses. • Use transpose or a matrix identity to identify an inverse when appropriate. • Solve product equations entry by entry and check every original condition.
Undoing a matrix product
For a nonzero real number, multiplication by its reciprocal undoes multiplication by the number. An inverse matrix plays a related role, with the identity matrix taking the place of the number 1. Since multiplication order matters, the defining requirement includes both orders of multiplication.
A square matrix A is invertible if a square matrix B of the same order satisfies AB = BA = I. B is the inverse of A, denoted A⁻¹. If no such matrix exists, A is not invertible.
Problem
Show that A = [2,3;1,2] and B = [2,−3;−1,2] are inverses.
- 1.Both matrices have order 2 × 2. Calculate AB = [2×2+3(−1),2(−3)+3×2;1×2+2(−1),1(−3)+2×2].
- 2.This simplifies to [1,0;0,1] = I₂.
- 3.Check the other order: BA = [2×2+(−3)×1,2×3+(−3)×2;(−1)×2+2×1,(−1)×3+2×2] = [1,0;0,1].
- 4.Therefore B = A⁻¹. The same pair of equations also says A = B⁻¹; the inverse relationship is mutual.
- 5.Neither negating A nor taking reciprocals of its entries produces B. Inverses are defined by a full row-column product.
The zero matrix has no inverse because multiplying it by a compatible matrix still gives a zero matrix, not I. Being square is necessary but not sufficient for invertibility: some nonzero square matrices also cannot produce I by multiplication. We will use direct checks and equations here rather than introduce a new general inverse-finding formula.
Problem
Explain why A = [1,2;2,4] is not invertible.
- 1.Suppose a candidate B = [p,q;r,s] existed. The first column of AB would be [p+2r;2p+4r].
- 2.To make this the first column [1;0] of I₂, we would need p+2r = 1 and 2p+4r = 0.
- 3.The second left-hand side is twice the first, so the first equation forces it to be 2, contradicting the required 0.
- 4.Thus no B can even make AB = I₂. A is square and nonzero but has no two-sided inverse.
A rectangular matrix has no two-sided inverse under this definition. If A is m × n and B is n × m, then AB is m × m while BA is n × n. For m ≠ n they cannot both be the one common identity required here. One-sided identity relationships for rectangular arrays are different statements; they do not make the rectangular matrix invertible in the chapter’s sense.
The negative −A satisfies A+(−A) = O and exists for every matrix. The inverse A⁻¹ satisfies AA⁻¹ = A⁻¹A = I and exists only for suitable square matrices. Even when an inverse exists, A⁻¹ is generally not −A.
Why an inverse cannot have two different values
An inverse, when it exists, is unique. The proof uses only the defining equations, associativity, and identity products. It provides a model for matrix reasoning that does not cancel or move factors without justification.
Suppose B and C are both inverses of A. We then know BA = AB = I and AC = CA = I. Start with B, insert the identity AC on its right, and regroup without reversing any factor. This connects B to C directly.
Notice the work done by the assumptions. We did not assert that B and C commute with each other, and we did not divide by A. Each replacement uses a known identity product. This style of proof is safer whenever matrix factors appear on different sides of an equation.
The inverse of a product reverses the order
If two operations are performed in sequence, undoing them requires undoing the later one first. For matrices this intuition is expressed by a precise reverse-order law. We establish existence by checking a candidate in both orders, rather than assume the product already has an inverse.
Let C = B⁻¹A⁻¹. Then (AB)C = A(BB⁻¹)A⁻¹ = AIA⁻¹ = I. In the reverse order, C(AB) = B⁻¹(A⁻¹A)B = B⁻¹IB = I. Thus C is a two-sided inverse of AB, proving both that AB is invertible and that the stated expression is its inverse. Uniqueness ensures there cannot be a different answer.
Problem
Take A = [1,1;0,1] with A⁻¹ = [1,−1;0,1], and B = [2,0;0,1] with B⁻¹ = [1/2,0;0,1]. Verify the inverse of AB.
- 1.First AB = [2,1;0,1]. The reverse-order candidate is B⁻¹A⁻¹ = [1/2,−1/2;0,1].
- 2.Multiply (AB)(B⁻¹A⁻¹): its entries are [1,−1+1;0,1], giving I₂. Multiplication in the reverse order also gives [1,1/2−1/2;0,1] = I₂.
- 3.The unreversed expression A⁻¹B⁻¹ is [1/2,−1;0,1]. Multiplying AB by this gives [1,−1;0,1], which is not I₂.
- 4.The order is therefore part of the formula, not a choice of notation. The inverse factors can be interchanged only if an additional commutativity condition makes that valid.
Recognising an inverse from an identity
An inverse can sometimes be identified without solving for arbitrary entries. If a known expression produces the identity in both orders, it is the inverse. The transpose exercises and identities of the form A² = I offer useful cases.
Problem
For A = [cos α,sin α;−sin α,cos α], show that Aᵀ is its inverse.
- 1.Transpose the array: Aᵀ = [cos α,−sin α;sin α,cos α].
- 2.In AᵀA, both diagonal entries become cos²α+sin²α = 1. The off-diagonal entries are cos α sin α−sin α cos α = 0. Thus AᵀA = I₂.
- 3.Calculate AAᵀ as well. Its diagonal entries again become 1 and its off-diagonal entries again become 0.
- 4.Therefore AAᵀ = AᵀA = I₂, so A⁻¹ = Aᵀ. This holds for this family because of the verified equations, not for arbitrary matrices.
If A² = I for a square matrix, A multiplied by itself produces I in both orders, so A⁻¹ = A. For A = [α,β;γ,−α], multiplication gives A² = (α²+βγ)I: the off-diagonal terms cancel. Thus A² = I exactly when α²+βγ = 1. This is a condition on the entries, not a claim that every matrix of that displayed shape is its own inverse.
Solving product equations with entries
A product equation can always be translated into equations for entries when its dimensions are fixed. When a suitable inverse is known, multiplication on the appropriate side gives a shorter route. Both approaches must preserve the original factor order and be checked in the original equation.
Problem
Let A = [2,−1;3,4], B = [5,2;7,4], C = [2,5;3,8]. Find D satisfying CD−AB = O.
- 1.The product AB is [10−7,4−4;15+28,6+16] = [3,0;43,22]. The equation therefore requires CD = [3,0;43,22].
- 2.Since CD has two rows and two columns and C is 2 × 2, D must be 2 × 2. Write D = [a,b;c,d]. Then CD = [2a+5c,2b+5d;3a+8c,3b+8d].
- 3.The first column gives 2a+5c = 3 and 3a+8c = 43. Multiply the first equation by 3 and the second by 2: 6a+15c = 9 and 6a+16c = 86. Subtraction gives c = 77, then a = −191.
- 4.The second column gives 2b+5d = 0 and 3b+8d = 22. The same elimination gives d = 44 and then b = −110.
- 5.Thus D = [−191,−110;77,44]. Checking CD gives [−382+385,−220+220;−573+616,−330+352] = [3,0;43,22] = AB.
- 6.As a connection to inverses, C⁻¹ = [8,−5;−3,2] can be verified directly by two identity products. Pre-multiplying CD = AB by C⁻¹ then gives D = C⁻¹AB, the same matrix. Post-multiplying by C⁻¹ would not isolate D.
An equation XA = B with a rectangular A can still have a solution even though A has no inverse in this chapter’s sense. Write the unknown X with the order forced by the product, compare entries, and check every resulting column. Solvability of one particular product equation and invertibility of its factor are different questions.
Quiz
Which condition defines B as the inverse of a same-order square A?
If B is the inverse of A, what is A in relation to B?
For invertible A and B of the same order, which is (AB)⁻¹?
Which statement about an inverse is correct?
If A² = I for a square A, which follows?
If C is invertible and CD = E, how do we isolate D?
Which statement about transposes is justified?
Practice Problems
- Verify both products for A = [2,3;1,2] and B = [2,−3;−1,2]. Explain why AB = BA alone would not establish that they are inverses.
- For A = [sin α,cos α;−cos α,sin α], calculate AᵀA and AAᵀ. Identify A⁻¹ and compare with the cosine-sine family in the lesson.
- Find all real x, y, z for which A = [0,2y,z;x,y,−z;x,−y,z] satisfies AᵀA = I. Calculate the three diagonal entries and the off-diagonal entries before solving the conditions.
- Find X such that X[1,2,3;4,5,6] = [−7,−8,−9;2,4,6]. Determine X’s order, solve from the first two columns, and check the third. Explain why no rectangular inverse is needed.
- For A = [α,β;γ,−α], calculate A² and give the condition on α, β, γ that makes A its own inverse. Give one numerical example.
- Suppose B and C are both claimed inverses of A. Reconstruct B = BI = B(AC) = (BA)C = IC = C, giving the reason for every equality.
- For invertible A and B, verify both identity products for B⁻¹A⁻¹. Explain why this proves the product AB has an inverse rather than merely guessing its value.
- For C = [2,5;3,8], verify [8,−5;−3,2] is its inverse, then solve CD = [3,0;43,22]. Explain why left multiplication isolates D.
Key Takeaways
• A⁻¹ is defined by two identity products, AA⁻¹ = A⁻¹A = I. • A matrix must be square for this two-sided inverse, but being square does not guarantee an inverse. • An inverse is unique and the inverse relationship is mutual. • The product inverse reverses factor order: (AB)⁻¹ = B⁻¹A⁻¹. • A transpose or the matrix itself can be an inverse only when the required identity products are established. • Use entries or a verified inverse to solve product equations; multiplication must occur on the correct side.