International Baccalaureate · DP 2: Applications and Interpretation HL

Paper 3 Workshop

⌚ ~27 min ▢ 3 blocks ✓ 10 exercises ➣ Prerequisites : Revision: Mixed Practice, Eigenvalues and Eigenvectors

I Toolbox: Eigenvalues of a 2 × 2 Matrix

Recappreviously studied — click to reread

Definition — Eigenvalues and Eigenvectors

Let $\mathbf{A}$ be a square matrix. An eigenvector of $\mathbf{A}$ is a non-zero vector $\mathbf{x}$ such that:$$ \mathbf{A}\mathbf{x} = \lambda \mathbf{x} $$ where $\lambda$ is a scalar called the eigenvalue associated with the eigenvector $\mathbf{x}$.

Method — Finding Eigenvalues and Eigenvectors

To find the eigenvalues and eigenvectors of a matrix $\mathbf{A}$:

Find the eigenvalues ($\lambda$): Solve the characteristic equation $$ \det(\mathbf{A} - \lambda \mathbf{I}) = 0, $$ where $\mathbf{I}$ is the identity matrix. The solutions are the eigenvalues.
Find the eigenvectors ($\mathbf{x}$): For each eigenvalue $\lambda$, substitute it into $$ \mathbf{A}\mathbf{x} = \lambda\mathbf{x}\text{ or } (\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = \mathbf{0} $$ and solve for the non-zero vector $\mathbf{x}$.

Example

Find the eigenvalues and corresponding eigenvectors of the matrix$$ \mathbf{A} = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}. $$

Answer

Find eigenvalues: $$\begin{aligned} \det(\mathbf{A} - \lambda \mathbf{I}) &= 0\\ \det\left( \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) &= 0\\ \det\left( \begin{pmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{pmatrix} \right) &= 0\\ (4-\lambda)(3-\lambda) - (2)(1) &= 0\\ \lambda^2 - 7\lambda + 12 - 2 &= 0\\ \lambda^2 - 7\lambda + 10 &= 0\\ (\lambda - 5)(\lambda - 2) &= 0 \end{aligned}$$ The eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = 2$.
Find an eigenvector for $\lambda_1 = 5$: $$\begin{aligned} (\mathbf{A} - 5 \mathbf{I})\mathbf{x} &= \mathbf{0}\\ \left( \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} - 5 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right)\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} 4-5 & 2 \\ 1 & 3-5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} -x + 2y \\ x - 2y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{aligned}$$ This gives $-x + 2y = 0$ and $x - 2y = 0$, so $x = 2y$.
Letting $y = t$, we have $x = 2t$, so $$ \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2t \\ t \end{pmatrix} = t\begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad t \neq 0. $$ Any vector of the form $t\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, $t\neq 0$, is an eigenvector corresponding to the eigenvalue $5$.
Find an eigenvector for $\lambda_2 = 2$: $$\begin{aligned} (\mathbf{A} - 2 \mathbf{I})\mathbf{x} &= \mathbf{0}\\ \left( \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} - 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right)\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} 4-2 & 2 \\ 1 & 3-2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} 2x + 2y \\ x + y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{aligned}$$ This gives $2x + 2y = 0$ and $x + y = 0$, so $y = -x$.
Letting $x = t$, we have $y = -t$, so $$ \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} t \\ -t \end{pmatrix} = t\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \quad t \neq 0. $$ Any vector of the form $t\begin{pmatrix} 1 \\ -1 \end{pmatrix}$, $t\neq 0$, is an eigenvector corresponding to the eigenvalue $2$.

Skills to practice

Finding the Eigenvalues of a 2 × 2 Matrix

II Toolbox: The Laws of Logarithms

Recappreviously studied — click to reread

Proposition — Laws of Logarithms

For $x>0$, $y>0$, $a>0$ with $a\neq 1$, and for any real number $k$:

Product Rule: $\log_a (xy) = \log _a x + \log _a y$
Quotient Rule: $\log _a \left(\dfrac{x}{y}\right) = \log _a x - \log_a y$
Power Rule: $\log_a (x^k) = k \log_a x$

Example

Write as a single logarithm: $\log (5)+\log (3)$.

Answer

$\begin{aligned}[t]\log (5)+\log (3) &= \log (5\times 3) &&\text{(product rule)} \\&= \log 15\\\end{aligned}$

Example

For $a>0$ and $b>0$, write $\log\left(\dfrac{a^2 b}{10}\right)$ in terms of $\log a$ and $\log b$.

Answer

$$\begin{aligned}[t] \log\left(\dfrac{a^2 b}{10}\right) &= \log(a^2 b) - \log 10 &&\text{(quotient rule)}\\ &= \log(a^2) + \log b - \log 10 &&\text{(product rule)}\\ &= 2\log a + \log b - \log 10 &&\text{(power rule)}\\ &= 2\log a + \log b - 1 \end{aligned}$$ The laws also work in this direction: they break a single logarithm into a combination of simpler logarithms.

Skills to practice

Writing as a Single Logarithm

III Exam-Style Questions

Skills to practice

Investigating with Technology: Paper 3