Linear Algebra and Its Applications – Answer Key
Linear algebra is the branch of mathematics that studies vectors, matrices, and linear transformations. On top of that, it provides the language and tools for modeling and solving problems in science, engineering, economics, computer science, and many other fields. This answer key walks through the core concepts, demonstrates how they are applied, and supplies clear solutions to typical exercises that illustrate the power of linear algebra in real‑world contexts Turns out it matters..
1. Introduction to Core Concepts
| Concept | Definition | Typical Notation |
|---|---|---|
| Vector | An ordered list of numbers representing a point or direction in space. Also, | v = ([v_1, v_2, \dots , v_n]^T) |
| Matrix | A rectangular array of numbers that encodes a linear transformation. | A = ([a_{ij}]) |
| Linear Transformation | A function (T: \mathbb{R}^n \rightarrow \mathbb{R}^m) that preserves vector addition and scalar multiplication. | (T(\mathbf{x}) = \mathbf{A}\mathbf{x}) |
| Determinant | A scalar value that indicates whether a matrix is invertible and measures volume scaling. | (\det(\mathbf{A})) |
| Eigenvalue / Eigenvector | For a matrix A, a non‑zero vector v such that (\mathbf{A}\mathbf{v} = \lambda \mathbf{v}). | (\lambda) = eigenvalue, v = eigenvector |
| Rank | The dimension of the column space; the number of linearly independent columns. |
Understanding these building blocks allows us to tackle more sophisticated topics like singular value decomposition (SVD), least‑squares fitting, and Markov chains.
2. Solving Linear Systems
Problem 1: Solve the system
[ \begin{cases} 2x + 3y - z = 5 \ -4x + y + 2z = -2 \ 3x - 2y + 4z = 7 \end{cases} ]
Solution (Answer Key):
- Write the augmented matrix
[ \left[\begin{array}{ccc|c} 2 & 3 & -1 & 5\ -4 & 1 & 2 & -2\ 3 & -2 & 4 & 7 \end{array}\right] ]
-
Perform Gaussian elimination
- R2 ← R2 + 2R1
[ \left[\begin{array}{ccc|c} 2 & 3 & -1 & 5\ 0 & 7 & 0 & 8\ 3 & -2 & 4 & 7 \end{array}\right] ]
- R3 ← R3 – (3/2)R1
[ \left[\begin{array}{ccc|c} 2 & 3 & -1 & 5\ 0 & 7 & 0 & 8\ 0 & -\frac{13}{2} & \frac{11}{2} & \frac{-1}{2} \end{array}\right] ]
- R3 ← R3 + (\frac{13}{14})R2
[ \left[\begin{array}{ccc|c} 2 & 3 & -1 & 5\ 0 & 7 & 0 & 8\ 0 & 0 & \frac{11}{2} & \frac{99}{14} \end{array}\right] ]
-
Back‑substitution
- (z = \frac{99/14}{11/2} = \frac{9}{7})
- (y = \frac{8 - 0\cdot z}{7} = \frac{8}{7})
- (x = \frac{5 - 3y + z}{2} = \frac{5 - 24/7 + 9/7}{2}= \frac{5 - 15/7}{2}= \frac{20/7}{2}= \frac{10}{7})
Answer: ((x, y, z) = \left(\frac{10}{7}, \frac{8}{7}, \frac{9}{7}\right)).
3. Matrix Inverses and Determinants
Problem 2: Find the inverse of
[ \mathbf{B}= \begin{bmatrix} 1 & 2\ 3 & 4 \end{bmatrix} ]
Solution:
The determinant (\det(\mathbf{B}) = 1\cdot4 - 2\cdot3 = -2).
The adjugate matrix is
[ \operatorname{adj}(\mathbf{B}) = \begin{bmatrix} 4 & -2\ -3 & 1 \end{bmatrix} ]
Thus
[ \mathbf{B}^{-1}= \frac{1}{\det(\mathbf{B})}\operatorname{adj}(\mathbf{B}) = -\frac{1}{2} \begin{bmatrix} 4 & -2\ -3 & 1 \end{bmatrix}
\begin{bmatrix} -2 & 1\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} ]
Answer: (\displaystyle \mathbf{B}^{-1}= \begin{bmatrix}-2 & 1\[2pt] 1.5 & -0.5\end{bmatrix}) Small thing, real impact..
4. Eigenvalues and Eigenvectors
Problem 3: Compute eigenvalues of
[ \mathbf{C}= \begin{bmatrix} 4 & 1\ 2 & 3 \end{bmatrix} ]
Solution:
Characteristic polynomial
[ \det(\mathbf{C}-\lambda\mathbf{I}) = \begin{vmatrix} 4-\lambda & 1\ 2 & 3-\lambda \end{vmatrix} = (4-\lambda)(3-\lambda)-2 = \lambda^{2}-7\lambda+10=0 ]
Solve: (\lambda^{2}-7\lambda+10=0 \Rightarrow (\lambda-5)(\lambda-2)=0).
Eigenvalues: (\lambda_{1}=5,; \lambda_{2}=2) And that's really what it comes down to..
Corresponding eigenvectors:
-
For (\lambda=5): ((\mathbf{C}-5\mathbf{I})\mathbf{v}=0) → (\begin{bmatrix}-1 & 1\2 & -2\end{bmatrix}\mathbf{v}=0) → (v_{1}=v_{2}). Choose (\mathbf{v}_{1}=[1,1]^T).
-
For (\lambda=2): (\begin{bmatrix}2 & 1\2 & 1\end{bmatrix}\mathbf{v}=0) → (2v_{1}+v_{2}=0) → (\mathbf{v}_{2}=[-1,2]^T) It's one of those things that adds up. Took long enough..
Answer: Eigenvalues 5 and 2 with eigenvectors ([1,1]^T) and ([-1,2]^T) respectively That's the part that actually makes a difference. Still holds up..
5. Real‑World Applications
5.1 Computer Graphics – Transformations
A 3‑D point (\mathbf{p} = [x, y, z, 1]^T) is transformed by a homogeneous transformation matrix (\mathbf{M}). For a rotation of ( \theta ) about the (z)-axis:
[ \mathbf{M}_{\text{rot}}= \begin{bmatrix} \cos\theta & -\sin\theta & 0 & 0\ \sin\theta & \cos\theta & 0 & 0\ 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1 \end{bmatrix} ]
Multiplying (\mathbf{M}_{\text{rot}}\mathbf{p}) yields the rotated coordinates. The answer key for a typical exercise (rotate ([1,0,0]) by (90^\circ) about (z)) is ([0,1,0]).
5.2 Data Science – Least Squares Regression
Given data matrix (\mathbf{X}\in\mathbb{R}^{m\times n}) and response vector (\mathbf{y}), the normal equations provide the best‑fit coefficients (\boldsymbol{\beta}):
[ \boldsymbol{\beta}= (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{y} ]
Answer key example: For (\mathbf{X} = \begin{bmatrix}1&1\1&2\1&3\end{bmatrix}) and (\mathbf{y} = [2,3,5]^T),
[ \mathbf{X}^T\mathbf{X}= \begin{bmatrix}3&6\6&14\end{bmatrix},; (\mathbf{X}^T\mathbf{X})^{-1}= \frac{1}{6}\begin{bmatrix}14&-6\-6&3\end{bmatrix}, ] [ \boldsymbol{\beta}= \frac{1}{6}\begin{bmatrix}14&-6\-6&3\end{bmatrix} \begin{bmatrix}10\22\end{bmatrix}= \begin{bmatrix}1\1.5\end{bmatrix}. ]
Thus the fitted line is (y = 1 + 1.5x).
5.3 Quantum Mechanics – State Evolution
The state vector (|\psi\rangle) evolves via a unitary matrix (\mathbf{U}=e^{-i\mathbf{H}t/\hbar}), where (\mathbf{H}) is the Hamiltonian. If (\mathbf{H}) has eigenvalues (\lambda_k) and eigenvectors (|\phi_k\rangle), then
[ |\psi(t)\rangle = \sum_k e^{-i\lambda_k t/\hbar}c_k |\phi_k\rangle, ]
with (c_k = \langle\phi_k|\psi(0)\rangle). The answer key often asks to compute probabilities (|c_k|^2); for a two‑level system with (\mathbf{H}= \begin{bmatrix}0 & \Delta\\Delta & 0\end{bmatrix}) and initial state ([1,0]^T), the probability of being in the second level after time (t) is (\sin^2(\Delta t/\hbar)).
5.4 Economics – Input‑Output Models
Leontief’s model uses matrix A (technical coefficients) and final demand vector d to determine total output x:
[ \mathbf{x}= (\mathbf{I}-\mathbf{A})^{-1}\mathbf{d}. ]
Answer key example: With
[ \mathbf{A}= \begin{bmatrix}0.Consider this: 3 & 0. Also, 2\0. 1 & 0.
[ \mathbf{I}-\mathbf{A}= \begin{bmatrix}0.In practice, 7 & -0. 5\0.5 & 0.6\end{bmatrix},; (\mathbf{I}-\mathbf{A})^{-1}= \frac{1}{0.2\-0.4}\begin{bmatrix}0.7\end{bmatrix} = \begin{bmatrix}1.In real terms, 1 & 0. 1 & 0.2\0.6 & 0.25 & 1 Turns out it matters..
[ \mathbf{x}= \begin{bmatrix}1.But 5 & 0. So naturally, 5\0. That's why 25 & 1. 75\end{bmatrix} \begin{bmatrix}100\80\end{bmatrix} = \begin{bmatrix}1.5\cdot100+0.5\cdot80\0.Day to day, 25\cdot100+1. 75\cdot80\end{bmatrix} = \begin{bmatrix}190\150\end{bmatrix} Worth keeping that in mind..
So the economy must produce 190 units of sector 1 and 150 units of sector 2.
5.5 Machine Learning – Principal Component Analysis (PCA)
PCA finds orthogonal directions (principal components) that maximize variance. Steps:
- Center data matrix X (subtract column means).
- Compute covariance matrix (\mathbf{C}= \frac{1}{m-1}\mathbf{X}^T\mathbf{X}).
- Perform eigen‑decomposition (\mathbf{C}= \mathbf{Q}\Lambda\mathbf{Q}^T).
- Project data: (\mathbf{Z}= \mathbf{X}\mathbf{Q}{k}) where (\mathbf{Q}{k}) contains the first k eigenvectors.
Answer key for a 2‑D example: Data points ((2,0), (0,2), (-2,0), (0,-2)). After centering, (\mathbf{C}= \begin{bmatrix}2 & 0\0 & 2\end{bmatrix}). Eigenvalues are both 2, eigenvectors can be chosen as the standard basis. The principal components coincide with the original axes, confirming that variance is equal in both directions.
6. Frequently Asked Questions
Q1. Why are eigenvalues important in engineering?
Eigenvalues describe natural frequencies of mechanical systems, stability of control loops, and convergence rates of iterative algorithms. A system is stable when all eigenvalues of its state‑matrix have negative real parts.
Q2. Can every matrix be inverted?
Only nonsingular (full‑rank) square matrices have inverses. If (\det(\mathbf{A}) = 0) or (\operatorname{rank}(\mathbf{A}) < n), the matrix is singular and no inverse exists.
Q3. How does linear algebra enable modern recommendation systems?
Techniques such as matrix factorization decompose a user‑item rating matrix into low‑rank latent factors, allowing prediction of missing entries. The factorization relies on singular value decomposition, a cornerstone of linear algebra.
Q4. What is the geometric meaning of the determinant?
For a 2‑D matrix, (|\det(\mathbf{A})|) equals the area scaling factor applied to a unit square. In 3‑D, it represents volume scaling. A negative sign indicates orientation reversal.
Q5. Is linear algebra useful for solving nonlinear problems?
Yes. Linearization (e.g., using Jacobian matrices) approximates nonlinear systems near equilibrium points, enabling analysis with linear algebra tools.
7. Conclusion
Linear algebra is more than an abstract mathematical discipline; it is the engine behind countless applications that shape technology, science, and economics. By mastering vectors, matrices, determinants, eigen‑structures, and related algorithms, you gain a versatile toolkit for:
- Modeling physical phenomena (rotations, quantum states, structural vibrations).
- Analyzing data (regression, PCA, collaborative filtering).
- Optimizing systems (least‑squares, linear programming).
The answer key provided above demonstrates step‑by‑step reasoning for typical problems, reinforcing conceptual understanding while showcasing practical relevance. Continual practice with these methods will deepen intuition and prepare you to tackle increasingly complex challenges where linear algebra serves as the foundational language.
