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[2020-02-05 Wed 21:19]
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Session 11: Matrix Inverses | Part B: Matrices and Systems of Equations | 1. Vectors and Matrices | Multivariable Calculus | Mathematics | MIT OpenCourseWare

1 Chalkboard

Figure 1: Inverse Matrix

Figure 1: Inverse Matrix

Figure 2: Inverse Matrix formula

Figure 2: Inverse Matrix formula

Figure 3: Inverse Matrix Formula: Minors and cofactors

Figure 3: Inverse Matrix Formula: Minors and cofactors

Figure 4: Inverse Matrix Formula: Transpose and Divide by \(\det(A)\)

Figure 4: Inverse Matrix Formula: Transpose and Divide by det⁡(A)\det(A)

2 How can we solve a small squared linear system?

2.1 Front

How can we solve a small squared linear system?

  • Low dimensional
  • 2×22 \times 2 system, or 3×33 \times 3 system
  • x1=a11y1+a12y2+a13y3x_1 = a_{11}y_1 + a_{12}y_2 + a_{13}y_{3}
  • x2=a21y1+a22y2+a23y3x_2 = a_{21}y_1 + a_{22}y_2 + a_{23}y_{3}
  • x3=a31y1+a32y2+a33y3x_3 = a_{31}y_1 + a_{32}y_2 + a_{33}y_{3}

2.2 Back

  • Using inverse matrices
  • \(A \vb{x} = \vb{b}\)
  • AA matrix with coefficients (aij)(a_{ij})
  • \({\displaystyle \vb{x} = \begin{pmatrix} x_1 \ x_2 \x_3 \end{pmatrix}}\)
  • \({\displaystyle \vb{b} = \begin{pmatrix} b_1 \ b_2 \ b_3 \end{pmatrix}}\)

Solving (multiplying by inverse matrix MM at the left)

\begin{align*} A \vb{x} &= \vb{b} \\\ M (A \vb{x}) &= M \vb{b}\\\ \vb{x} = M \vb{b} \end{align*}

3 When can we say that the inverse matrix of AA exits?

3.1 Front

When can we say that the inverse matrix of $A$ exits?

3.2 Back

  • MM is the inverse matrix of AA
  • M exists ⇔det⁡(A)≠0M \text{ exists } \Leftrightarrow \det(A) \neq 0
  • That’s because of determinant law of matrix multiplication
    • det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A) \det(B)
    • MA=I  ⟹  det⁡(M)det⁡(A)=det⁡(I)=1  ⟹  det⁡(A)≠0MA = I \implies \det(M)\det(A) = \det(I) = 1 \implies \det(A) \neq 0

4 Which step do we need to calculate a matrix inverse?

4.1 Front

Which step do we need to calculate a matrix inverse?

Set AA matrix, and A−1A^{-1} as inverse matrix

4.2 Back

  1. AA is a n×nn \times n matrix
  2. det⁡(A)≠0\det(A) \neq 0
  3. Unique solution
    • A−1A=AA−1=IA^{-1}A = AA^{-1} = I, Identity matrix
    • \(A \vb{x} = \vb{b}\)
    • \(M(A \vb{x}) = M \vb{b}\)
    • \(\vb{x} = M\vb{b}\)
  4. A−1=1det⁡(A)adj(A)=1det⁡(A)(A11A12A13 A21A22A23 A31A32A33)T{\displaystyle A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{\det(A)} \begin{pmatrix} A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33} \end{pmatrix}^{T}}
    • adj(A)\text{adj}(A), is the adjoint or adjugate of AA
    • \(A_{ij} = (-1)^{i+j} \abs{A_{ij}}\) is the ij-cofactors
    • \(\abs{A_{ij}}\) is the ij-minor, the determinant after remove the row ii and column jj

Formula steps

  1. Calculate the matrix of minors
  2. Change the signs of the entries according to the checkerboard rule
  3. Transpose the resulting matrix, this gives adj(A)\text{adj}(A)
  4. Divide every entry by \(\abs{A}\)

5 What is the quick formula for inverse of 2×22 \times 2 matrix AA?

5.1 Front

What is the quick formula for inverse of $2 \times 2$ matrix $A$?

A=(ab cd){\displaystyle A = \begin{pmatrix}a & b \ c & d\end{pmatrix}}

5.2 Back

  • Cofactors (d−c −ba){\displaystyle \begin{pmatrix}d & -c \ -b & a \end{pmatrix}}
  • adj AA: (d−b −ca){\displaystyle \begin{pmatrix}d & -b \ -c & a \end{pmatrix}}
  • inverse of AA: \({\displaystyle \frac{1}{\abs{A}} \begin{pmatrix}d & -b \ -c & a \end{pmatrix}}\)

Steps:

  1. Switch aa and dd
  2. Change the signs on bb and cc
  3. Divide by the determinant