2x2 Matrices, Determinants, and Inverses Study Guide

Cell Phone Population Transition Analysis

• Historical Ownership Calculation via Matrix Inversion:

  • Historical population state vectors are computed by multiplying the inverse of the population transition matrix by the current population state vector:     $\begin{pmatrix} 0.57 & 0.13 \cr 0.43 & 0.87 \end{pmatrix}^{-1} \begin{pmatrix} 6073 \cr 9927 \end{pmatrix} = \begin{pmatrix} 9075 \cr 6925 \end{pmatrix}$
  • The resulting historical ownership numbers for last year are:
  • Non-cell phone owners last year: $9075$ people.
  • Cell phone owners last year: $6925$ people.
  • Total population represented in the system: $6073 + 9927 = 16000$ people ($9075 + 6925 = 16000$ people).

• Multi-Year Population Ownership Percentage Analysis:

  • Last Year Ownership Percentage:     $\frac{6925}{16000} \times 100\% = 43.28125\%$
  • Current Year Ownership Percentage:     $\frac{9927}{16000} \times 100\% = 62.04375\%$
  • Next Year Ownership Projection:
  • Calculated by applying the transition matrix forward by one period to find the updated ownership count and corresponding population percentage.
  • Two-Year Future Estimation:
  • Computed by applying the transition matrix twice ($T^2$) to the current population state vector to estimate ownership percentage two years from now.
  • Two-Year Past Estimation:
  • Computed by applying the inverse transition matrix twice ($(T^{-1})^2$) to the current state vector to estimate cell phone ownership count two years ago.

Matrix Multiplicative Inverses

• Definition and Properties:

  • Two square matrices $A$ and $B$ are multiplicative inverses of each other if and only if their product in both multiplication orders yields the identity matrix $I$:     $A \cdot B = I \quad \text{and} \quad B \cdot A = I$
  • For $2 \times 2$ matrices, the multiplicative identity matrix is:     $I = \begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix}$

• Verification Procedure for Inverse Matrices:

  • To verify that two matrices are multiplicative inverses, evaluate both products $A \cdot B$ and $B \cdot A$. Both operations must evaluate to $I$.
  • Example Pair 1:     $\begin{pmatrix} 1 & 4 \cr 3 & 4 \end{pmatrix}$
  • Multiplied with candidate inverse matrix to confirm product equals $I$
  • Example Pair 2:     $\begin{pmatrix} 3 & 2 \cr 4 & 3 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 3 & -2 \cr -4 & 3 \end{pmatrix}$
  • Multiplication verification:       $\begin{pmatrix} 3 & 2 \cr 4 & 3 \end{pmatrix} \begin{pmatrix} 3 & -2 \cr -4 & 3 \end{pmatrix} = \begin{pmatrix} (3)(3)+(2)(-4) & (3)(-2)+(2)(3) \cr (4)(3)+(3)(-4) & (4)(-2)+(3)(3) \end{pmatrix} = \begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix}$

Determinants of 2x2 Matrices

• Formula for Determinant Evaluation:

  • For a general $2 \times 2$ matrix $A = \begin{pmatrix} a & b \cr c & d \end{pmatrix}$, the determinant scalar is defined as:     $\det(A) = ad - bc$

• Step-by-Step Determinant Calculations:

  • Exercise 6:     $\det\begin{pmatrix} 4 & 0 \cr 3 & 6 \end{pmatrix} = (4)(6) - (0)(3) = 24 - 0 = 24$
  • Exercise 7:     $\det\begin{pmatrix} 3 & 2 \cr 5 & 4 \end{pmatrix} = (3)(4) - (2)(5) = 12 - 10 = 2$
  • Exercise 8:     $\det\begin{pmatrix} 1 & 4 \cr 2 & 8 \end{pmatrix} = (1)(8) - (4)(2) = 8 - 8 = 0$
  • Exercise 9:     $\det\begin{pmatrix} 2 & 1 \cr 0 & 3 \end{pmatrix} = (2)(3) - (1)(0) = 6 - 0 = 6$
  • Exercise 10:     $\det\begin{pmatrix} 2 & 3 \cr 1 & 4 \end{pmatrix} = (2)(4) - (3)(1) = 8 - 3 = 5$
  • Exercise 11:     $\det\begin{pmatrix} -2 & 3 \cr 4 & -1 \end{pmatrix} = (-2)(-1) - (3)(4) = 2 - 12 = -10$
  • Exercise 12:     $\det\begin{pmatrix} 3 & 4 \cr 1 & 2 \end{pmatrix} = (3)(2) - (4)(1) = 6 - 4 = 2$
  • Exercise 13:     $\det\begin{pmatrix} 2 & 3 \cr 5 & 7 \end{pmatrix} = (2)(7) - (3)(5) = 14 - 15 = -1$

Matrix Inverses and Invertibility

• Existence Condition for Inverse Matrices:

  • A square matrix $A$ is invertible (non-singular) if and only if its determinant is non-zero:     $\det(A) \neq 0$
  • If $\det(A) = 0$, the matrix is singular and has no multiplicative inverse.

• Explicit Inverse Formula for a 2x2 Matrix:

  • Given $A = \begin{pmatrix} a & b \cr c & d \end{pmatrix}$ with $\det(A) \neq 0$:     $A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \cr -c & a \end{pmatrix} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \cr -c & a \end{pmatrix}$

• Determination and Calculation for Specific Matrices:

  • Exercise 14:     $\begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix}$
  • Determinant: $\det = (1)(1) - (0)(0) = 1$
  • Inverse matrix: $\begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix}$
  • Exercise 18:     $\begin{pmatrix} 6 & 8 \cr 3 & 4 \end{pmatrix}$
  • Determinant: $\det = (6)(4) - (8)(3) = 24 - 24 = 0$
  • Conclusion: Determinant is zero; matrix has no inverse (singular).
  • Exercise 19:     $\begin{pmatrix} 1 & 2 \cr 3 & 4 \end{pmatrix}$
  • Determinant: $\det = (1)(4) - (2)(3) = 4 - 6 = -2$
  • Inverse calculation:       $A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \cr -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \cr 1.5 & -0.5 \end{pmatrix}$
  • Exercise 20:     $\begin{pmatrix} 15 & 0.5 \cr \dots & \dots \end{pmatrix}$
  • Evaluate determinant $\det(A)$ and apply inverse formula if $\det(A) \neq 0$
  • Exercise 21:     $\begin{pmatrix} 3 & 2 \cr \dots & \dots \end{pmatrix}$
  • Evaluate determinant $\det(A)$ and apply inverse formula if $\det(A) \neq 0$

Matrix Equations

• Solution Procedure for Systems of Matrix Equations:

  • Given a matrix equation of the form $A X = B$ where $A$ is an $n \times n$ matrix and $X, B$ are vectors or matrices:
  1. Calculate $\det(A)$ to check for invertibility.
  2. If $\det(A) \neq 0$, multiply both sides from the left by $A^{-1}$:        $A^{-1} (A X) = A^{-1} B$$I X = A^{-1} B$$X = A^{-1} B$
  3. If $\det(A) = 0$, the equation cannot be solved uniquely using matrix inversion.

• Application Problems:

  • Exercise 22: Solve linear matrix equation $A X = B$ or demonstrate non-existence of unique inverse.
  • Exercise 24: Matrix equation involving vector $X = \begin{pmatrix} 1 & 5 \end{pmatrix}$ or coefficient multiplication.

Cable Television Transition Data Analysis

• Problem Parameters and Transition Rates:

  • Annual shift rate from No Cable TV to Cable TV: $1.5\% = 0.015$
  • Annual retention rate for No Cable TV: $100\% - 1.5\% = 98.5\% = 0.985$
  • Annual shift rate from Cable TV to No Cable TV: $2\% = 0.02$
  • Annual retention rate for Cable TV: $100\% - 2\% = 98\% = 0.98$
  • Population parameters:
  • Total stable population: $30{,}000$ people.
  • Current cable television subscribers: $20{,}000$ people.
  • Current non-subscribers: $10{,}000$ people.

• Transition Matrix Construction (Part a):

  • Define the annual transition matrix $T$ where column 1 represents current Cable TV subscribers and column 2 represents current non-subscribers:     $T = \begin{pmatrix} 0.98 & 0.015 \cr 0.02 & 0.985 \end{pmatrix}$

• Next Year Subscriber Prediction (Part b):

  • Matrix multiplication equation for next year's subscriber states:     $\begin{pmatrix} \text{Cable TV subscribers next year} \cr \text{Non-subscribers next year} \end{pmatrix} = \begin{pmatrix} 0.98 & 0.015 \cr 0.02 & 0.985 \end{pmatrix} \begin{pmatrix} 20000 \cr 10000 \end{pmatrix}$
  • Exact calculations:
  • Predicted Cable TV subscribers next year:       $(0.98)(20000) + (0.015)(10000) = 19600 + 150 = 19750$
  • Predicted non-subscribers next year:       $(0.02)(20000) + (0.985)(10000) = 400 + 9850 = 10250$
  • Verification of total population:       $19750 + 10250 = 30000$ people.

• Previous Year Subscriber Estimation (Part c):

  • Determinant of transition matrix $T$:     $\det(T) = (0.98)(0.985) - (0.015)(0.02) = 0.9653 - 0.0003 = 0.965$
  • Inverse transition matrix $T^{-1}$:     $T^{-1} = \frac{1}{0.965} \begin{pmatrix} 0.985 & -0.015 \cr -0.02 & 0.98 \end{pmatrix}$
  • Matrix equation for last year's subscriber counts:     $\begin{pmatrix} \text{Cable TV subscribers last year} \cr \text{Non-subscribers last year} \end{pmatrix} = T^{-1} \begin{pmatrix} 20000 \cr 10000 \end{pmatrix}$
  • Exact calculation for last year's cable subscribers:     $\text{Cable TV subscribers last year} = \frac{(0.985)(20000) + (-0.015)(10000)}{0.965} = \frac{19700 - 150}{0.965} = \frac{19550}{0.965} \approx 20259$$\text{Non-subscribers last year} = \frac{(-0.02)(20000) + (0.98)(10000)}{0.965} = \frac{-400 + 9800}{0.965} = \frac{9400}{0.965} \approx 9741$