Determinants · Lesson 4 of 6
Adjoints and Inverse Matrices
“Build an adjoint from cofactors, establish when an inverse exists, and connect inverse calculations to matrix identities.”
• Construct the adjoint by transposing the cofactor matrix. • Explain the diagonal and off-diagonal entries of A(adj A) = |A|I. • Distinguish singular and nonsingular matrices using their determinants. • Justify the condition for invertibility and calculate inverses by adjoints. • Use determinant product relationships and reverse the order when inverting a product. • Derive an inverse from an appropriate matrix polynomial identity.
From a cofactor grid to an adjoint
Cofactors do more than shorten determinant expansion. If we arrange them carefully in a matrix, they produce a tool for finding an inverse. The crucial arrangement is a transpose: a row of cofactors must become a column before multiplication with the original matrix.
The adjoint of a square matrix A, written adj A, is the transpose of its cofactor matrix. If C = [Aᵢⱼ] contains cofactors at their original positions, then adj A = Cᵀ, so the (i,j) entry of adj A is Aⱼᵢ.
The adjoint is still a matrix of the same order as A. It exists even when A has no inverse, since minors and cofactors can always be calculated. Do not confuse the word adjoint with either determinant or transpose alone: a transpose rearranges the original entries, while an adjoint transposes the calculated cofactors.
Problem
Find adj A for A = [2,3;1,4].
- 1.Calculate the minors: M₁₁=4, M₁₂=1, M₂₁=3, M₂₂=2.
- 2.Attach signs to get the cofactor matrix C = [4,−1;−3,2].
- 3.Transpose C by exchanging its rows and columns: adj A = [4,−3;−1,2].
- 4.For a general order-two matrix [a,b;c,d], this procedure gives adj A = [d,−b;−c,a]. It swaps the main-diagonal entries and changes the signs of the other two entries.
Why multiplication produces a scalar times the identity
Multiply A by its adjoint using the row-column rule. Because the adjoint contains transposed cofactors, each resulting entry is a cofactor sum of the kind we studied in the previous lesson. Matching row indices give the determinant; different row indices give zero.
The (1,1) entry of A(adj A) is a₁₁A₁₁+a₁₂A₁₂+a₁₃A₁₃, which is det A. Its (1,2) entry is a₁₁A₂₁+a₁₂A₂₂+a₁₃A₂₃, which is zero. The same argument gives det A in every diagonal position and zero in every off-diagonal position. Reversing the multiplication uses the matching and different-column cofactor sums and gives the same result.
This identity is useful even when the determinant is zero. In that case the product is the zero matrix O, although A and adj A may both be nonzero matrices. It is a familiar warning from matrix multiplication: a zero product does not let us conclude that one factor is the zero matrix.
Problem
For A = [2,3;−4,−6], find adj A and verify both products in the adjoint identity.
- 1.The determinant is 2(−6)−3(−4)=−12+12=0. The order-two adjoint rule gives adj A = [−6,−3;4,2].
- 2.Multiplying A(adj A) gives [−12+12,−6+6;24−24,12−12] = [0,0;0,0].
- 3.The reverse product is [−12+12,−18+18;8−8,12−12] = O as well.
- 4.Thus both products equal (det A)I = 0I. Neither matrix is O, and no division by det A is possible.
The determinant decides whether an inverse exists
An inverse of A is a matrix B of the same order satisfying AB = BA = I. The adjoint identity nearly has that form. When its scalar det A is nonzero, we can divide by that scalar; when it is zero, that operation is not defined.
A square matrix is singular if det A = 0 and nonsingular if det A ≠ 0. These definitions concern its determinant, not whether some individual entries are zero.
For example, [1,2;4,8] is singular because 1·8−2·4=0, even though none of its entries is zero. The identity matrix is nonsingular despite many zero entries. The terms singular and zero matrix therefore do not mean the same thing. A whole-matrix determinant test is needed.
We use a product relationship for square matrices A and B of the same order: the determinant of their matrix product equals the product of their numerical determinants. We state this general result here and apply it to invertibility. It does not say that AB = BA; equality of these scalar products is much weaker than equality of the matrices.
First assume A has an inverse B. Taking determinants in AB = I gives (det A)(det B)=det I=1. Neither factor can be zero, so A must be nonsingular. For the reverse direction, assume det A ≠ 0. Divide the adjoint identity by det A. The matrix (adj A)/(det A) then multiplies with A to give I in both orders. It is therefore an inverse. These two directions together establish the equivalence, rather than merely a convenient sufficient condition.
The adjoint exists for a singular matrix, but its inverse does not. A nonzero adjoint cannot rescue a zero denominator in A⁻¹ = (adj A)/det A. Calculate the determinant first, and say that no inverse exists when it is zero.
Problem
For A = [1,3,3;1,4,3;1,3,4], verify A(adj A) = |A|I and find A⁻¹.
- 1.Expanding row 1 gives det A = 1(16−9)−3(4−3)+3(3−4)=7−3−3=1. Therefore A is invertible.
- 2.The first cofactor row is [7,−1,−1], the second [−3,1,0], and the third [−3,0,1]. For example, A₁₂ = −(4−3)=−1 and A₂₁ = −(12−9)=−3.
- 3.Transpose the cofactor matrix to obtain adj A = [7,−3,−3;−1,1,0;−1,0,1].
- 4.The first product row is [7−3−3, −3+3+0, −3+0+3] = [1,0,0]. The second is [7−4−3,−3+4+0,−3+0+3] = [0,1,0].
- 5.The third product row is [7−3−4,−3+3+0,−3+0+4] = [0,0,1]. Hence A(adj A)=I=(det A)I.
- 6.Since det A = 1, the inverse equals the adjoint itself: A⁻¹ = [7,−3,−3;−1,1,0;−1,0,1]. The reverse product is I by the same identity and can be checked directly.
Determinants of adjoints and inverses
The adjoint identity also relates the numerical determinants of these matrices. We must keep track of order when taking the determinant of a scalar multiple of I. This brings together the scaling rule and the determinant product rule.
For a nonsingular order-three matrix, take determinants in (adj A)A = (det A)I. On the left, obtain det(adj A)·det A. On the right, multiplying every entry of I by det A multiplies its determinant by (det A)³. Since det I=1, the right side is (det A)³. Dividing by the nonzero det A gives det(adj A)=(det A)². For order n the same reasoning gives exponent n−1.
The inverse relationship follows from AA⁻¹=I: (det A)det(A⁻¹)=1. For instance, if an order-three matrix has determinant −3, its adjoint has determinant 9 and its inverse has determinant −1/3. These are scalar checks on calculations; they are not formulas for the entries of the inverse.
Inverting a product reverses the factors
An inverse must undo a product in a particular order. To undo AB, B⁻¹ is placed nearest B and A⁻¹ nearest A when the products are associated. This is why B⁻¹A⁻¹ is the candidate inverse, even though ordinary scalar fractions may make the reversed order seem unnecessary.
Check both products. (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AIA⁻¹ = I. In the reverse order, (B⁻¹A⁻¹)(AB)=B⁻¹(A⁻¹A)B = B⁻¹IB = I. These steps use associativity without swapping any two matrices. Since a two-sided inverse is unique, the candidate is the required inverse.
Problem
Let A = [2,3;1,−4] and B = [1,−2;−1,3]. Verify (AB)⁻¹ = B⁻¹A⁻¹.
- 1.Compute AB = [−1,5;5,−14]. Its determinant is 14−25=−11, so (AB)⁻¹ = (1/11)[14,5;5,1].
- 2.Separately, det A = −11 gives A⁻¹ = (1/11)[4,3;1,−2]. Since det B = 1, B⁻¹ = [3,2;1,1].
- 3.Now B⁻¹A⁻¹ = (1/11)[3,2;1,1][4,3;1,−2] = (1/11)[14,5;5,1].
- 4.The entries match the direct inverse. Calculating A⁻¹B⁻¹ instead would give (1/11)[15,11;1,0], so maintaining order is essential.
An identity can reveal an inverse
Sometimes a matrix satisfies a polynomial equation involving A, its powers, and I. We can rearrange that equation into a product equal to I. This gives an inverse without repeating a full cofactor calculation, provided the algebra really produces a two-sided inverse or invertibility has already been established.
Problem
For A = [2,3;1,2], verify A²−4A+I=O and use it to obtain A⁻¹.
- 1.Calculate A² = [7,12;4,7]. Then A²−4A+I = [7−8+1,12−12;4−4,7−8+1] = O.
- 2.Rearrange to 4A−A²=I, or A(4I−A)=I. The same factors also satisfy (4I−A)A=4A−A²=I.
- 3.Therefore A⁻¹ = 4I−A = [4,0;0,4]−[2,3;1,2] = [2,−3;−1,2].
- 4.As a check, det A = 4−3=1 and its order-two adjoint is [2,−3;−1,2], in agreement. If multiplying a polynomial identity by A⁻¹ instead, first justify that det A is nonzero.
For a higher-degree relation such as A³+pA²+qA+rI=O with r≠0, isolate rI and factor A from the other terms. This yields A·[−(A²+pA+qI)/r]=I; the reverse product is also I because these powers of the same matrix associate in the same way. If the constant coefficient is zero, this particular rearrangement cannot divide by it. Each given identity must be used with its actual coefficients and signs.
Quiz
The cofactor matrix is [4,−1;−3,2]. What is the adjoint?
Which condition makes a square matrix invertible?
If det A = 0, what is A(adj A)?
A and B are invertible. Which is (AB)⁻¹?
For an order-three matrix with det A = −3, what is det(adj A)?
If det A = 5, what is det(A⁻¹)?
A²−4A+I=O. Which matrix multiplies with A to give I?
If det A = −2 and det B = 3, what is det(AB)?
Practice Problems
- Find adj A for [1,2;3,4] and for [1,−1,2;2,3,5;−2,0,1]. Show the cofactor matrix before transposing it.
- Verify A(adj A)=(adj A)A=(det A)I for [2,3;−4,−6] and for [1,−1,2;3,0,−2;1,0,3]. Explain the difference between the two determinant cases.
- Find inverses, if they exist, for [2,−2;4,3] and [−1,5;−3,2]. Verify both products for one answer.
- Find inverses, if they exist, for [1,2,3;0,2,4;0,0,5] and [1,0,0;3,3,0;5,2,−1]. Do not assume that every matrix containing zeros is singular.
- Find inverses, if they exist, for [2,1,3;4,−1,0;−7,2,1] and [1,−1,2;0,2,−3;3,−2,4]. Check each determinant first.
- Find the inverse of [1,0,0;0,cos α,sin α;0,sin α,−cos α] for real α. Verify your answer using sin²α+cos²α=1.
- For A=[3,7;2,5] and B=[6,8;7,9], verify (AB)⁻¹=B⁻¹A⁻¹ by calculating the two sides.
- For A=[3,1;−1,2], verify A²−5A+7I=O and hence find A⁻¹.
- For A=[3,2;1,1], find real numbers a and b such that A²+aA+bI=O, and check every entry of the equation.
- For A=[1,1,1;1,2,−3;2,−1,3], verify A³−6A²+5A+11I=O and hence express A⁻¹ as a polynomial in A. Evaluate that expression.
- For A=[2,−1,1;−1,2,−1;1,−1,2], verify A³−6A²+9A−4I=O and hence find A⁻¹.
- For a nonsingular order-three matrix, choose and justify the value of det(adj A) from det A, (det A)², (det A)³, and 3det A.
- For an invertible order-two matrix, choose and justify the value of det(A⁻¹) from det A, 1/det A, 1, and 0.
- A calculation claims det A=0 but A⁻¹=(1/2)adj A. Explain why these two claims cannot both be correct.
Key Takeaways
• The adjoint is the transpose of the cofactor matrix, with no further sign changes during the transpose. • A(adj A)=(adj A)A=(det A)I, including when A is singular. • A square matrix has an inverse exactly when its determinant is nonzero. • For a nonsingular matrix, A⁻¹=(adj A)/det A; the determinant product rule also shows that products of nonsingular matrices are nonsingular. • For order n≥2, det(adj A)=(det A)^(n−1), and det(A⁻¹)=1/det A when the inverse exists. • Inverting a product reverses its factors, and a suitable matrix identity can provide an alternative route to an inverse.