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Linear Algebra for Team-Based Inquiry Learning: PREVIEW Edition — Instructor Version

Steven Clontz, Drew Lewis

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Section 4.2 The Inverse of a Matrix (MX2)

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Subsection 4.2.1 Warm Up

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Activity 4.2.1.

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Consider the matrices:
A=[15−1032], B=[72−11032−211−1−3].
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Without using technology, what is the third column of the product AB?
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Subsection 4.2.2 Class Activities

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Activity 4.2.2.

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Let A=[27−103211−1]. Find a 3×3 matrix B such that BA=A, that is,
[?????????][27−103211−1]=[27−103211−1]
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Check your guess using technology.
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Definition 4.2.3.

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The identity matrix In (or just I when n is obvious from context) is the n×n matrix
In=[10⋯001⋱⋮⋮⋱⋱00⋯01].
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It has a 1 on each diagonal element and a 0 in every other position.
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Activity 4.2.5.

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Let T:Rn→Rm be a linear map with standard matrix A. Sort the following items into three groups of statements: a group that means T is injective, a group that means T is surjective, and a group that means T is bijective.
  1. T(x→)=b→ has a solution for all b→∈Rm
  2. T(x→)=b→ has a unique solution for all b→∈Rm
  3. T(x→)=0→ has a unique solution.
  4. The columns of A span Rm
  5. The columns of A are linearly independent
  6. The columns of A are a basis of Rm
  7. Every column of RREF(A) has a pivot
  8. Every row of RREF(A) has a pivot
  9. m=n and RREF(A)=I
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Definition 4.2.6.

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Let T:Rn→Rn be a linear bijection with standard matrix A.
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By item (B) from Activity 4.2.5 we may define an inverse map T−1:Rn→Rn that defines T−1(b→) as the unique solution x→ satisfying T(x→)=b→, that is, T(T−1(b→))=b→.
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Furthermore, let
A−1=[T−1(e→1)⋯T−1(e→n)]
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be the standard matrix for T−1. We call A−1 the inverse matrix of A, and we also say that A is an invertible matrix.
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Activity 4.2.7.

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Let T:R3→R3 be the linear bijection given by the standard matrix A=[2−1−6213114].
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(a)
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To find x→=T−1(e→1), we need to find the unique solution for T(x→)=e→1. Which of these linear systems can be used to find this solution?
  1. 2x1−1x2−6x3=x12x1+1x2+3x3=01x1+1x2+4x3=0
  2. 2x1−1x2−6x3=x12x1+1x2+3x3=x21x1+1x2+4x3=x3
  3. 2x1−1x2−6x3=12x1+1x2+3x3=01x1+1x2+4x3=0
  4. 2x1−1x2−6x3=12x1+1x2+3x3=11x1+1x2+4x3=1
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(b)
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Use that system to find the solution x→=T−1(e→1) for T(x→)=e→1.
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(c)
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Similarly, solve T(x→)=e→2 to find T−1(e→2), and solve T(x→)=e→3 to find T−1(e→3).
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Activity 4.2.8.

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Find the inverse A−1 of the matrix
A=[000−110−1−4110−41−1−12]
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by computing how it transforms each of the standard basis vectors for R4: T−1(e→1), T−1(e→2), T−1(e→3), and T−1(e→4).
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Activity 4.2.9.

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Is the matrix [231−1−420−55] invertible?
  1. Yes, because its transformation is a bijection.
  2. Yes, because its transformation is not a bijection.
  3. No, because its transformation is a bijection.
  4. No, because its transformation is not a bijection.
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Observation 4.2.10.

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An n×n matrix A is invertible if and only if RREF(A)=In.
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Activity 4.2.11.

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Let T:R2→R2 be the bijective linear map defined by T([xy])=[2x−3y−3x+5y], with the inverse map T−1([xy])=[5x+3y3x+2y].
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(b)
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If A is the standard matrix for T and A−1 is the standard matrix for T−1, find the 2×2 matrix
A−1A=[????].
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Observation 4.2.12.

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T−1∘T=T∘T−1 is the identity map for any bijective linear transformation T. Therefore A−1A=AA−1 equals the identity matrix I for any invertible matrix A.
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Subsection 4.2.3 Individual Practice

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Subsection 4.2.4 Videos

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Figure 43. Video: Invertible matrices
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Figure 44. Video: Finding the inverse of a matrix
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Exercises 4.2.5 Exercises

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Subsection 4.2.6 Mathematical Writing Explorations

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Exploration 4.2.14.

Assume A is an n×n matrix. Prove the following are equivalent. Some of these results you have proven previously.
  • A row reduces to the identity matrix.
  • For any choice of b→∈Rn, the system of equations represented by the augmented matrix [A|b→] has a unique solution.
  • The columns of A are a linearly independent set.
  • The columns of A form a basis for Rn.
  • The rank of A is n.
  • The nullity of A is 0.
  • A is invertible.
  • The linear transformation T with standard matrix A is injective and surjective. Such a map is called an isomorphism.
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Exploration 4.2.15.

  • Assume T is a square matrix, and T4 is the zero matrix. Prove that (I−T)−1=I+T+T2+T3. You will need to first prove a lemma that matrix multiplication distributes over matrix addition.
  • Generalize your result to the case where Tn is the zero matrix.
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Subsection 4.2.7 Sample Problem and Solution

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