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Lesson 33 — Transpose, Identity, Inverse Matrix

The transpose mirrors the matrix. The inverse undoes multiplication — exists only when the determinant is non-zero.

Used in: 1.º ano do EM (16 anos) · Math I japonês cap. matrizes · Klasse 11 alemã Lineare Algebra

AA1=A1A=I,(AT)ij=ajiA A^{-1} = A^{-1} A = I, \qquad (A^T)_{ij} = a_{ji}
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Rigorous notation, full derivation, hypotheses

Transpose and inverse

Transpose

(AT)ij=aji(A^T)_{ij} = a_{ji}. Rows are swapped for columns. Properties:

PropertyFormula
Involution(AT)T=A(A^T)^T = A
Sum(A+B)T=AT+BT(A + B)^T = A^T + B^T
Scalar(αA)T=αAT(\alpha A)^T = \alpha A^T
Product(AB)T=BTAT(AB)^T = B^T A^T (reverses order)
Inverse-transpose(AT)1=(A1)T(A^T)^{-1} = (A^{-1})^T

Symmetric matrix: AT=AA^T = A. Orthogonal matrix: ATA=IA^T A = IA1=ATA^{-1} = A^T.

Identity

InI_n: n×nn \times n square matrix with 1's on the diagonal and 0 off. For every An×nA_{n \times n}: AI=IA=AAI = IA = A

Inverse

An×nA_{n \times n} is invertible (non-singular) if there exists A1A^{-1} such that AA1=A1A=IAA^{-1} = A^{-1}A = I. Equivalence theorem — for square AA, the following are equivalent:

  1. AA is invertible.
  2. detA0\det A \neq 0.
  3. Ax=0A\mathbf{x} = \mathbf{0} has only x=0\mathbf{x} = \mathbf{0}.
  4. The columns of AA are linearly independent.
  5. The rows of AA are linearly independent.
  6. AA has full rank: rank(A)=n\text{rank}(A) = n.
  7. AA is a product of elementary matrices.

2x2 inverse

A=(abcd),A1=1adbc(dbca)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

(Valid if adbc0ad - bc \neq 0.)

Inverse properties

  • (A1)1=A(A^{-1})^{-1} = A
  • (AB)1=B1A1(AB)^{-1} = B^{-1} A^{-1} (reverses order!)
  • (AT)1=(A1)T(A^T)^{-1} = (A^{-1})^T
  • (αA)1=(1/α)A1(\alpha A)^{-1} = (1/\alpha) A^{-1}
  • det(A1)=1/det(A)\det(A^{-1}) = 1/\det(A)

Computation via Gauss-Jordan

Form [AI][A | I] → row-reduce until [IA1][I | A^{-1}]. Cost O(n3)O(n^3).

Exercise list

48 exercises · 12 with worked solution (25%)

Application 30Understanding 8Modeling 6Challenge 3Proof 1
  1. Ex. 33.1ApplicationAnswer key
    Transpose of (1234)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}. (Ans: (1324)\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}.)
  2. Ex. 33.2Application
    Transpose of (123456)\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}.
  3. Ex. 33.3ApplicationAnswer key
    Inverse of (1234)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}. (Ans: (213/21/2)\begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}.)
  4. Ex. 33.4Application
    Inverse of (2003)\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}.
  5. Ex. 33.5Application
    Inverse of (1001)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.
  6. Ex. 33.6Application
    Does (1224)\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} have an inverse? Justify. (Ans: no, det=0\det = 0.)
  7. Ex. 33.7ApplicationAnswer key
    Verify that AA1=IA \cdot A^{-1} = I for A=(3152)A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}.
  8. Ex. 33.8Application
    Inverse of (cosθsinθsinθcosθ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}. (Ans: rotation by θ-\theta.)
  9. Ex. 33.9Application
    Solve Ax=bA\mathbf{x} = \mathbf{b} via the inverse: A=(2113)A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}, b=(5,7)T\mathbf{b} = (5, 7)^T.
  10. Ex. 33.10ApplicationAnswer key
    Verify whether (1234)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} is symmetric. (Ans: no — a12a21a_{12} \neq a_{21}.)
  11. Ex. 33.11Application
    Verify (AB)T=BTAT(AB)^T = B^T A^T for A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} and B=(0110)B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.
  12. Ex. 33.12Application
    For which kk does the matrix (1k24)\begin{pmatrix} 1 & k \\ 2 & 4 \end{pmatrix} fail to have an inverse? (Ans: k=2k = 2.)
  13. Ex. 33.13Application
    Inverse of (1111)\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.
  14. Ex. 33.14Application
    Numerically show that A+ATA + A^T is symmetric for A=(1425)A = \begin{pmatrix} 1 & 4 \\ 2 & 5 \end{pmatrix}.
  15. Ex. 33.15Application
    Numerically show that AATA - A^T is skew-symmetric.
  16. Ex. 33.16ApplicationAnswer key
    (A1)1=A(A^{-1})^{-1} = A — verify for A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}.
  17. Ex. 33.17Application
    For which diagonal (a00b)\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} is invertible? (Ans: ab0ab \neq 0.)
  18. Ex. 33.18Application
    Inverse of (ab0d)\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} (triangular, ad0ad \neq 0).
  19. Ex. 33.19Application
    A=(0110)A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. Compute A4A^4 and A1A^{-1}.
  20. Ex. 33.20ApplicationAnswer key
    Decompose (1425)\begin{pmatrix} 1 & 4 \\ 2 & 5 \end{pmatrix} as symmetric + skew-symmetric.
  21. Ex. 33.21Application
    Compute A1A^{-1} of A=(100020005)A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{pmatrix} (diagonal).
  22. Ex. 33.22Application
    Compute A1A^{-1} of A=(111011001)A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} (triangular).
  23. Ex. 33.23Application
    Apply Gauss-Jordan to [AI][A|I] for A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}.
  24. Ex. 33.24Application
    Verify that (123014560)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} has an inverse (compute det\det).
  25. Ex. 33.25Application
    Solve Ax=bA\mathbf x = \mathbf b with A=(1123)A = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}, b=(3,8)T\mathbf b = (3, 8)^T, via A1A^{-1}.
  26. Ex. 33.26Application
    Inverse of the permutation matrix P=(010100001)P = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}. (Ans: PT=P1P^T = P^{-1}.)
  27. Ex. 33.27Application
    Compute A1A^{-1} for A=(1211)A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}.
  28. Ex. 33.28ApplicationAnswer key
    Use A1A^{-1} to solve {x+2y=5x+y=1\begin{cases} x + 2y = 5 \\ -x + y = 1 \end{cases}.
  29. Ex. 33.29ApplicationAnswer key
    Verify that if AT=A1A^T = A^{-1}, then ATA=IA^TA = I (orthogonal matrix).
  30. Ex. 33.30Application
    Verify whether (1/21/21/21/2)\begin{pmatrix} 1/\sqrt 2 & 1/\sqrt 2 \\ -1/\sqrt 2 & 1/\sqrt 2 \end{pmatrix} is orthogonal.
  31. Ex. 33.31Understanding
    Show that if AA is symmetric and invertible, A1A^{-1} is also symmetric.
  32. Ex. 33.32UnderstandingAnswer key
    Show that if A2=IA^2 = I, then A=A1A = A^{-1}.
  33. Ex. 33.33Understanding
    Show that an orthogonal matrix (ATA=IA^T A = I) has A1=ATA^{-1} = A^T.
  34. Ex. 33.34Understanding
    Show that if A,BA, B invertible, ABAB is invertible.
  35. Ex. 33.35Understanding
    Show that (AT)1=(A1)T(A^T)^{-1} = (A^{-1})^T.
  36. Ex. 33.36Understanding
    Show that if AA is invertible triangular, A1A^{-1} is also triangular of the same type.
  37. Ex. 33.37UnderstandingAnswer key
    Show that the product of orthogonal matrices is orthogonal.
  38. Ex. 33.38Understanding
    Show that det(A1)=1/det(A)\det(A^{-1}) = 1/\det(A).
  39. Ex. 33.39Modeling
    Use the inverse to solve: {2x+y=7x3y=2\begin{cases} 2x + y = 7 \\ x - 3y = -2 \end{cases}.
  40. Ex. 33.40Modeling
    In matrix cryptography (Hill cipher), encrypt the message as a vector m\mathbf{m} via AmA\mathbf{m}. Decrypt = A1(Am)A^{-1}(A\mathbf{m}). Model with A2×2A_{2\times 2}.
  41. Ex. 33.41ModelingAnswer key
    In CG, the inverse transformation is fundamental: applying a transformation to the camera is applying the inverse to objects. Explain why.
  42. Ex. 33.42Modeling
    In economics, the Leontief matrix LL relates production to demand. Solution: x=(IL)1d\mathbf{x} = (I - L)^{-1} \mathbf{d}. For L=(0.20.10.30.4)L = \begin{pmatrix} 0.2 & 0.1 \\ 0.3 & 0.4 \end{pmatrix}, d=(10,20)T\mathbf d = (10, 20)^T, compute x\mathbf x.
  43. Ex. 33.43Modeling
    Identify whether (123014001)\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix} is upper-triangular. Is the inverse also upper-triangular?
  44. Ex. 33.44Modeling
    In statistics, regression β^=(XTX)1XTy\hat\beta = (X^TX)^{-1}X^T\mathbf y. Why is row reduction preferable to direct inversion?
  45. Ex. 33.45Challenge
    Find a matrix AA with A3=IA^3 = I but AIA \neq I. (Hint: rotation by 120°120°.)
  46. Ex. 33.46ChallengeAnswer key
    Show that if AB=IAB = I for A,BMnA, B \in M_n, then BA=IBA = I (non-trivial).
  47. Ex. 33.47Challenge
    Compute A1A^{-1} for A=I+uvTA = I + \mathbf u \mathbf v^T with u,vRn\mathbf u, \mathbf v \in \mathbb R^n and 1+vTu01 + \mathbf v^T\mathbf u \neq 0 (Sherman-Morrison).
  48. Ex. 33.48Proof
    Prove (AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1} via (AB)(B1A1)=I(AB)(B^{-1}A^{-1}) = I.

Sources

Updated on 2026-04-30 · Author(s): Clube da Matemática

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