How to Solve: IB AI HL – Matrices (Eigenvalues, Transition Matrices, Solving Systems)

How to Solve: IB AI HL – Matrices (Eigenvalues, Transition Matrices, Solving Systems)

Introduction

"Mastering eigenvalues and transition matrices doesn’t just get you 10–15% of your IB AI HL exam—it unlocks real-world problems like predicting population growth, Google’s PageRank algorithm, and even how diseases spread. One question on this topic can be worth 6–8 marks, so let’s break it down step by step."

What You Need To Know First

Before diving in, you must already understand: 1. Matrix operations (addition, multiplication, determinants). 2. Solving linear systems (using inverse matrices or row reduction). 3. Basic probability (for transition matrices).

If any of these feel shaky, pause and review them first.

Key Vocabulary

Formulas To Know

Step-by-Step Method

Part 1: Finding Eigenvalues and Eigenvectors

Step 1: Write the characteristic equation. - Subtract ( λI ) from matrix ( A ). - Take the determinant: ( \det(A - λI) = 0 ).

Step 2: Solve for ( λ ). - Expand the determinant to get a polynomial in ( λ ). - Solve the equation (factor or use quadratic formula).

Step 3: Find eigenvectors for each ( λ ). - Plug each ( λ ) back into ( (A - λI)\vec{v} = \vec{0} ). - Solve the system to find ( \vec{v} ) (usually parametric form).

Part 2: Transition Matrices

Step 1: Identify the transition matrix ( T ). - Each column must sum to 1 (probabilities). - Entry ( T_{ij} ) = probability of moving from state ( j ) to state ( i ).

Step 2: Find the steady-state vector ( \vec{v} ). - Solve ( T\vec{v} = \vec{v} ) (or ( (T - I)\vec{v} = \vec{0} )). - Use the condition that probabilities sum to 1 (e.g., ( v_1 + v_2 = 1 )).

Step 3: Predict long-term behavior. - Compute ( T^n \vec{v}_0 ) for large ( n ) (or use steady-state).

Part 3: Solving Systems with Matrices

Step 1: Write the system in matrix form ( A\vec{x} = \vec{b} ). - ( A ) = coefficient matrix, ( \vec{x} ) = variables, ( \vec{b} ) = constants.

Step 2: Solve using inverse matrices (if ( A ) is invertible). - ( \vec{x} = A^{-1}\vec{b} ).

Step 3: If ( A ) is not invertible, use row reduction. - Augment ( A ) with ( \vec{b} ) and reduce to row-echelon form.

Worked Examples

Example 1 – Basic: Find Eigenvalues and Eigenvectors

Matrix: ( A = \begin{pmatrix} 4 & 1 \ 2 & 3 \end{pmatrix} )

Step 1: Characteristic equation. [ \det(A - λI) = \det \begin{pmatrix} 4-λ & 1 \ 2 & 3-λ \end{pmatrix} = (4-λ)(3-λ) - 2 = 0 ] [ λ^2 - 7λ + 10 = 0 ]

Step 2: Solve for ( λ ). [ (λ - 5)(λ - 2) = 0 ] [ λ = 5 \text{ or } 2 ]

Step 3: Find eigenvectors. - For ( λ = 5 ): [ (A - 5I)\vec{v} = \begin{pmatrix} -1 & 1 \ 2 & -2 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} ] [ -x + y = 0 \implies \vec{v} = \begin{pmatrix} 1 \ 1 \end{pmatrix} ]

• For ( λ = 2 ): [ (A - 2I)\vec{v} = \begin{pmatrix} 2 & 1 \ 2 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} ] [ 2x + y = 0 \implies \vec{v} = \begin{pmatrix} 1 \ -2 \end{pmatrix} ]

What we did and why: We used the characteristic equation to find eigenvalues, then solved ( (A - λI)\vec{v} = \vec{0} ) to find eigenvectors. This is the standard method for any 2x2 matrix.

Example 2 – Medium: Transition Matrix Steady State

Transition Matrix: ( T = \begin{pmatrix} 0.7 & 0.2 \ 0.3 & 0.8 \end{pmatrix} )

Step 1: Set up ( T\vec{v} = \vec{v} ). [ \begin{pmatrix} 0.7 & 0.2 \ 0.3 & 0.8 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} v_1 \ v_2 \end{pmatrix} ]

Step 2: Solve the system. [ 0.7v_1 + 0.2v_2 = v_1 ] [ 0.3v_1 + 0.8v_2 = v_2 ] Simplify: [ -0.3v_1 + 0.2v_2 = 0 ] [ 0.3v_1 - 0.2v_2 = 0 ] (Same equation!) Use ( v_1 + v_2 = 1 ): [ v_1 = \frac{2}{5}, v_2 = \frac{3}{5} ]

What we did and why: We found the steady-state vector by solving ( T\vec{v} = \vec{v} ) and using the probability condition ( v_1 + v_2 = 1 ). This tells us the long-term distribution.

Example 3 – Exam-Style: Solving a System with Matrices

System: [ 2x + y = 5 ] [ 4x - 3y = -5 ]

Step 1: Write in matrix form ( A\vec{x} = \vec{b} ). [ \begin{pmatrix} 2 & 1 \ 4 & -3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 5 \ -5 \end{pmatrix} ]

Step 2: Find ( A^{-1} ). [ \det(A) = (2)(-3) - (1)(4) = -10 ] [ A^{-1} = \frac{1}{-10} \begin{pmatrix} -3 & -1 \ -4 & 2 \end{pmatrix} = \begin{pmatrix} 0.3 & 0.1 \ 0.4 & -0.2 \end{pmatrix} ]

Step 3: Multiply ( A^{-1}\vec{b} ). [ \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0.3 & 0.1 \ 0.4 & -0.2 \end{pmatrix} \begin{pmatrix} 5 \ -5 \end{pmatrix} = \begin{pmatrix} 1 \ 2 \end{pmatrix} ]

What we did and why: We used the inverse matrix method to solve the system. This is efficient for small matrices, but row reduction is better for larger ones.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in. For eigenvalues: write ( \det(A - λI) = 0 ), solve for ( λ ), then find eigenvectors by plugging ( λ ) back in. For transition matrices: set ( T\vec{v} = \vec{v} ), solve, and use ( v_1 + v_2 = 1 ). For systems: write ( A\vec{x} = \vec{b} ), then use ( A^{-1} ) or row reduction. Watch out for non-invertible matrices, probability constraints, and disguised steady-state questions. You’ve got this—go ace that exam!