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. Also, 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.

1. Introduction to Core Concepts

Counterintuitive, but true Worth keeping that in mind..

Understanding these building blocks allows us to tackle more sophisticated topics like singular value decomposition (SVD), least‑squares fitting, and Markov chains Turns out it matters..

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)).

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) Practical, not theoretical..

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}) Not complicated — just consistent..

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) No workaround needed..

• 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]).

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) Not complicated — just consistent. Took long enough..

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)).

Honestly, this part trips people up more than it should Most people skip this — try not to..

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}0.2\0.3 & 0.1 & 0 Worth keeping that in mind..

[ \mathbf{I}-\mathbf{A}= \begin{bmatrix}0.Which means 7 & -0. 2\-0.1 & 0.6\end{bmatrix},; (\mathbf{I}-\mathbf{A})^{-1}= \frac{1}{0.Consider this: 4}\begin{bmatrix}0. 6 & 0.2\0.This leads to 1 & 0. 7\end{bmatrix} = \begin{bmatrix}1.Even so, 5 & 0. In practice, 5\0. 25 & 1 Took long enough..

[ \mathbf{x}= \begin{bmatrix}1.75\end{bmatrix} \begin{bmatrix}100\80\end{bmatrix} = \begin{bmatrix}1.Even so, 25\cdot100+1. 5\cdot80\0.Also, 5 & 0. Think about it: 25 & 1. In real terms, 5\0. Here's the thing — 5\cdot100+0. 75\cdot80\end{bmatrix} = \begin{bmatrix}190\150\end{bmatrix}.

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 Worth keeping that in mind..

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 Less friction, more output..