Linear Algebra And Its Applications Answer Key

Linear Algebra and Its Applications – Answer Key

Linear algebra is the branch of mathematics that studies vectors, matrices, and linear transformations. 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.

Easier said than done, but still worth knowing.

1. Introduction to Core Concepts

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):

1. Write the augmented matrix

[ \left[\begin{array}{ccc|c} 2 & 3 & -1 & 5\ -4 & 1 & 2 & -2\ 3 & -2 & 4 & 7 \end{array}\right] ]

2. 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] ]

3. 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)) And that's really what it comes down to..

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) Not complicated — just consistent..

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}) It's one of those things that adds up..

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).

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) Worth knowing..

• 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).

Answer: Eigenvalues 5 and 2 with eigenvectors ([1,1]^T) and ([-1,2]^T) respectively.

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]) Not complicated — just consistent..

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) It's one of those things that adds up. Less friction, more output..

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)) Small thing, real impact..

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}0.3 & 0.2\0.1 & 0 Simple, but easy to overlook..

[ \mathbf{I}-\mathbf{A}= \begin{bmatrix}0.Consider this: 2\0. Here's the thing — 1 & 0. 7\end{bmatrix} = \begin{bmatrix}1.Day to day, 5 & 0. 5\0.Plus, 6\end{bmatrix},; (\mathbf{I}-\mathbf{A})^{-1}= \frac{1}{0. 7 & -0.1 & 0.6 & 0.2\-0.4}\begin{bmatrix}0.25 & 1.

[ \mathbf{x}= \begin{bmatrix}1.5\cdot100+0.Day to day, 5\0. 75\end{bmatrix} \begin{bmatrix}100\80\end{bmatrix} = \begin{bmatrix}1.5 & 0.5\cdot80\0.Plus, 25 & 1. Still, 25\cdot100+1. 75\cdot80\end{bmatrix} = \begin{bmatrix}190\150\end{bmatrix} Worth keeping that in mind. And it works..

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:

1. Center data matrix X (subtract column means).
2. Compute covariance matrix (\mathbf{C}= \frac{1}{m-1}\mathbf{X}^T\mathbf{X}).
3. Perform eigen‑decomposition (\mathbf{C}= \mathbf{Q}\Lambda\mathbf{Q}^T).
4. 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 It's one of those things that adds up..

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.