Determinants & Inverses of $2 \times 2$ Matrices
For a $2 \times 2$ matrix $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:
If $\det(\mathbf{A}) = 0$, the matrix is singular and has no inverse. In geometry, $|\det(\mathbf{A})|$ represents the area scale factor of the transformation.
Eigenvalues & Eigenvectors
A non-zero vector $\mathbf{v}$ is an eigenvector of $\mathbf{A}$ with scalar eigenvalue $\lambda$ if:
Geometrically, matrix transformation $\mathbf{A}$ simply stretches or compresses the eigenvector along its own line by a scale factor of $\lambda$ without changing its directional axis!
Matrix Diagonalization & Powers $\mathbf{A}^n$
If an $n \times n$ matrix $\mathbf{A}$ has $n$ linearly independent eigenvectors $\mathbf{v}_1, \dots, \mathbf{v}_n$, it can be diagonalized:
Computing matrix powers becomes straightforward:
Find the determinant and inverse of the matrix $\mathbf{M} = \begin{pmatrix} 5 & 2 \\ 3 & 1 \end{pmatrix}$.
Solution:
$\det(\mathbf{M}) = (5)(1) - (2)(3) = 5 - 6 = -1$. [M1][A1]
$\mathbf{M}^{-1} = \frac{1}{-1}\begin{pmatrix} 1 & -2 \\ -3 & 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 3 & -5 \end{pmatrix}$. [A1]
For what values of $k$ is the matrix $\mathbf{K} = \begin{pmatrix} k & 6 \\ 2 & k - 1 \end{pmatrix}$ singular?
Solution:
A matrix is singular when its determinant is zero: $\det(\mathbf{K}) = 0$. [R1]
$k(k - 1) - (6)(2) = 0 \implies k^2 - k - 12 = 0$. [M1]
$(k - 4)(k + 3) = 0$. [M1]
$k = 4$ or $k = -3$. [A1]
Find the eigenvalues of $\mathbf{A} = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix}$.
Solution:
$\det(\mathbf{A} - \lambda\mathbf{I}) = \det\begin{pmatrix} 1 - \lambda & 4 \\ 2 & 3 - \lambda \end{pmatrix} = 0$. [M1]
$(1 - \lambda)(3 - \lambda) - 8 = 0 \implies 3 - 4\lambda + \lambda^2 - 8 = 0$. [M1]
$\lambda^2 - 4\lambda - 5 = 0 \implies (\lambda - 5)(\lambda + 1) = 0$. [M1]
$\lambda_1 = 5, \quad \lambda_2 = -1$. [A1]
Find the eigenvectors corresponding to $\lambda_1 = 5$ and $\lambda_2 = -1$ for the matrix in Exercise 3.
Solution:
For $\lambda_1 = 5$: $(\mathbf{A} - 5\mathbf{I})\mathbf{v}_1 = \begin{pmatrix} -4 & 4 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x = y \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. [M1][A1]
For $\lambda_2 = -1$: $(\mathbf{A} + \mathbf{I})\mathbf{v}_2 = \begin{pmatrix} 2 & 4 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 2x + 4y = 0 \implies x = -2y \implies \mathbf{v}_2 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$. [M1][A1]
Diagonalize the matrix $\mathbf{A} = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix}$ in the form $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$, writing down matrices $\mathbf{P}$, $\mathbf{D}$, and $\mathbf{P}^{-1}$.
Solution:
$\mathbf{P} = \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & -1 \end{pmatrix}$. [A2]
$\det(\mathbf{P}) = (1)(1) - (-2)(1) = 3$. [M1]
$\mathbf{P}^{-1} = \frac{1}{3}\begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{2}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{pmatrix}$. [A2]
Using the diagonalization from Exercise 5, calculate $\mathbf{A}^4$.
Solution:
$\mathbf{A}^4 = \mathbf{P}\mathbf{D}^4\mathbf{P}^{-1}$. [M1]
$\mathbf{D}^4 = \begin{pmatrix} 5^4 & 0 \\ 0 & (-1)^4 \end{pmatrix} = \begin{pmatrix} 625 & 0 \\ 0 & 1 \end{pmatrix}$. [A1]
$\mathbf{P}\mathbf{D}^4 = \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 625 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 625 & -2 \\ 625 & 1 \end{pmatrix}$. [M1]
$\mathbf{A}^4 = \begin{pmatrix} 625 & -2 \\ 625 & 1 \end{pmatrix} \cdot \frac{1}{3}\begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 627 & 1248 \\ 624 & 1251 \end{pmatrix} = \begin{pmatrix} 209 & 416 \\ 208 & 417 \end{pmatrix}$. [A1]
A transformation in the plane maps triangle $T$ of area $14\text{ cm}^2$ to an image triangle $T'$ under matrix $\mathbf{T} = \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix}$. Find the area of $T'$.
Solution:
The area scale factor is $|\det(\mathbf{T})|$. [R1]
$\det(\mathbf{T}) = (3)(4) - (-1)(2) = 12 + 2 = 14$. [M1]
$\text{Area}(T') = 14 \times \text{Area}(T) = 14 \times 14 = 196\text{ cm}^2$. [A1]
Prove that the sum of the eigenvalues of any $2 \times 2$ matrix $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ equals the trace of the matrix ($\text{tr}(\mathbf{A}) = a + d$), and their product equals $\det(\mathbf{A})$.
Solution:
The characteristic equation is $\det(\mathbf{A} - \lambda\mathbf{I}) = (a - \lambda)(d - \lambda) - bc = 0$. [M1]
$\lambda^2 - (a + d)\lambda + (ad - bc) = 0$. [A1]
By Vieta's formulas for the quadratic equation $\lambda^2 - B\lambda + C = 0$: [M1]
Sum of roots $\lambda_1 + \lambda_2 = a + d = \text{tr}(\mathbf{A})$, and product $\lambda_1 \lambda_2 = ad - bc = \det(\mathbf{A})$. [A1]
Solve the matrix equation $\mathbf{A}\mathbf{X} = \mathbf{B}$ for matrix $\mathbf{X}$, where $\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} 4 & -1 \\ 1 & 2 \end{pmatrix}$.
Solution:
$\mathbf{X} = \mathbf{A}^{-1}\mathbf{B}$ (Left-multiply by $\mathbf{A}^{-1}$). [M1]
$\det(\mathbf{A}) = 6 - 5 = 1 \implies \mathbf{A}^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$. [A1]
$\mathbf{X} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\begin{pmatrix} 4 & -1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 12 - 1 & -3 - 2 \\ -20 + 2 & 5 + 4 \end{pmatrix}$. [M1]
$\mathbf{X} = \begin{pmatrix} 11 & -5 \\ -18 & 9 \end{pmatrix}$. [A1]
Explain why matrix multiplication is not commutative in general ($\mathbf{A}\mathbf{B} \ne \mathbf{B}\mathbf{A}$) using an illustrative geometric example.
Solution:
Matrix transformations represent sequential operations in space. [R1]
For example, reflecting across the $x$-axis and then rotating $90^\circ$ counter-clockwise results in a reflection across the line $y = x$, whereas rotating $90^\circ$ first and then reflecting across the $x$-axis results in reflection across $y = -x$. [A2]
Core Formulas (Topic 1 AHL)
| Matrix Inverse ($2 \times 2$) | $$\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$ |
| Characteristic Equation | $$\det(\mathbf{A} - \lambda\mathbf{I}) = \lambda^2 - \text{tr}(\mathbf{A})\lambda + \det(\mathbf{A}) = 0$$ |
| Diagonalization | $$\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1} \implies \mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}$$ |
| Area Scale Factor | $$\text{New Area} = |\det(\mathbf{A})| \times \text{Original Area}$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Matrix Multiplication Order: $\mathbf{A}\mathbf{B} \ne \mathbf{B}\mathbf{A}$. If $\mathbf{A}\mathbf{X} = \mathbf{B}$, then $\mathbf{X} = \mathbf{A}^{-1}\mathbf{B}$ (Left-multiply). Writing $\mathbf{B}\mathbf{A}^{-1}$ is incorrect!
- Trap 2: Column order mismatch in $\mathbf{P}$ and $\mathbf{D}$: If the first column of $\mathbf{P}$ is eigenvector $\mathbf{v}_1$, the $(1, 1)$ entry of diagonal matrix $\mathbf{D}$ must be its corresponding eigenvalue $\lambda_1$.
- Trap 3: Negative determinant in area scaling: When $\det(\mathbf{A}) = -3$, the area scale factor is $|-3| = +3$. Area cannot be negative (the minus sign indicates an orientation flip / reflection).
- Trap 4: Dividing by matrices: There is no such operation as matrix division; always use the inverse matrix $\mathbf{A}^{-1}$.
- Trap 5: Eigenvector scaling: Any non-zero scalar multiple of an eigenvector is still a valid eigenvector (e.g. $[1, 1]^T$ and $[2, 2]^T$ are equally valid).
5-Question Concept Check
Score: - / 51. What is the determinant of $\mathbf{A} = \begin{pmatrix} 6 & 3 \\ 4 & 2 \end{pmatrix}$?
2. If $\mathbf{A}\mathbf{v} = 4\mathbf{v}$, what does the scalar 4 represent?
3. In the diagonalization $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$, the columns of $\mathbf{P}$ are formed by:
4. If $\mathbf{D} = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$, what is $\mathbf{D}^3$?
5. A triangle with area $10\text{ cm}^2$ is transformed by $\mathbf{M}$ with $\det(\mathbf{M}) = -2.5$. What is the area of the transformed triangle?