Determinants and Linear Equations

Overview

Linear algebra forms the bedrock of modern data science, providing the mathematical framework to understand and manipulate complex datasets. This chapter delves into two fundamental concepts within linear algebra: determinants and systems of linear equations. Mastering these topics is not merely an academic exercise; it's a critical prerequisite for tackling advanced data science techniques, from understanding the behavior of machine learning models to optimizing algorithms and performing robust statistical analysis. Determinants, in particular, offer profound insights into the properties of a matrix, indicating whether a system has a unique solution, how transformations scale space, and playing a crucial role in concepts like eigenvalues and eigenvectors – foundational elements for techniques such as Principal Component Analysis (PCA) and dimensionality reduction. Simultaneously, systems of linear equations are ubiquitous in data science, appearing in regression models, network analysis, optimization problems, and the core mechanics of many machine learning algorithms where you often need to solve for unknown parameters. For your CMI examinations, a solid conceptual understanding combined with computational proficiency in these areas is indispensable. Questions often test your ability to apply these principles to solve practical problems involving data matrices, analyze model behavior, or interpret the solvability of systems that model real-world scenarios. This chapter is designed to equip you with the essential tools and intuition necessary to excel in these foundational mathematical methods, paving the way for your success in advanced data science topics. ---

Chapter Contents

| # | Topic | What You'll Learn | |---|-------|-------------------| | 1 | Calculating Determinants | Compute determinants for various matrix orders. | | 2 | Properties of Determinants | Utilize determinant properties for efficient computation. | | 3 | Solving Systems of Linear Equations | Apply matrix methods to solve linear systems. | ---

Learning Objectives

--- Now let's begin with Calculating Determinants... ## Part 1: Calculating Determinants

Introduction

In the realm of linear algebra, the determinant of a square matrix is a fundamental scalar value that encapsulates critical properties of the matrix. For a Masters in Data Science, understanding determinants is not merely an academic exercise; it is crucial for grasping concepts in various domains, including solving systems of linear equations, analyzing the invertibility of matrices, computing eigenvalues, and understanding the scaling effect of linear transformations. These concepts are foundational for advanced topics like Principal Component Analysis (PCA), Singular Value Decomposition (SVD), optimization algorithms, and the theoretical underpinnings of many statistical and machine learning models. This section will guide you through the methods for calculating determinants and their essential properties, focusing on techniques relevant to the CMI exam. ---

Key Concepts

# ## 1. Determinant of $2 \times 2$ and $3 \times 3$ Matrices For small matrices, determinants can be calculated using straightforward formulas. These serve as building blocks for understanding more complex methods. # ### $2 \times 2$ Matrix For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is given by: # ### $3 \times 3$ Matrix (Sarrus' Rule) For a $3 \times 3$ matrix $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$, Sarrus' Rule provides a convenient way to calculate the determinant: Step 1: Rewrite the first two columns of the matrix to the right of the third column. Step 2: Sum the products of the elements along the three main diagonals from top-left to bottom-right. Step 3: Subtract the sum of the products of the elements along the three anti-diagonals from top-right to bottom-left. Worked Example: Problem: Calculate the determinant of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$. Solution: Step 1: Apply Sarrus' Rule by extending the matrix. Step 2: Calculate the sum of products along the main diagonals. Step 3: Calculate the sum of products along the anti-diagonals. Step 4: Subtract the second sum from the first. Answer: $0$ --- # ## 2. Cofactor Expansion (Laplace Expansion) Cofactor expansion is a general method for calculating the determinant of any $n \times n$ matrix. It involves breaking down the determinant of a larger matrix into a sum of determinants of smaller submatrices. The determinant of an $n \times n$ matrix $A$ can be calculated by expanding along any row $i$ or any column $j$: Worked Example: Problem: Calculate the determinant of $A = \begin{bmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & 2 & 4 \end{bmatrix}$ using cofactor expansion. Solution: Step 1: Choose a row or column for expansion. The first row has a zero, so let's expand along the first row ($i=1$). Step 2: Calculate the cofactors for the first row elements. Step 3: Substitute the elements and cofactors into the expansion formula. Answer: $16$ --- # ## 3. Properties of Determinants Understanding determinant properties is crucial for efficient calculation, especially for larger matrices, and for theoretical understanding.

Row/Column Interchange: If a matrix $B$ is obtained from $A$ by interchanging two rows or two columns, then $|B| = -|A|$.

Scalar Multiplication of a Row/Column: If a matrix $B$ is obtained from $A$ by multiplying a single row or column by a scalar $k$, then $|B| = k|A|$.

Scalar Multiplication of a Matrix: If $A$ is an $n \times n$ matrix and $k$ is a scalar, then $|kA| = k^n|A|$.

Row/Column Addition: If a matrix $B$ is obtained from $A$ by adding a multiple of one row (or column) to another row (or column), then $|B| = |A|$. This property is fundamental for row reduction.

Zero Row/Column: If a matrix $A$ has a row or a column consisting entirely of zeros, then $|A| = 0$.

Identical/Proportional Rows/Columns: If a matrix $A$ has two identical rows or two identical columns, or if one row/column is a scalar multiple of another row/column (i.e., linearly dependent), then $|A| = 0$.

Determinant of Triangular Matrices: If $A$ is a triangular matrix (upper triangular, lower triangular, or diagonal), its determinant is the product of its diagonal elements.

Determinant of Transpose: The determinant of a matrix is equal to the determinant of its transpose: $|A^T| = |A|$.

Determinant of Product: The determinant of a product of matrices is the product of their determinants:

Determinant of Inverse: If $A$ is an invertible matrix, then $|A^{-1}| = \frac{1}{|A|}$. This property is directly linked to matrix invertibility.

--- # ## 4. Determinant Calculation using Row Reduction (Gaussian Elimination) For larger matrices (e.g., $4 \times 4$ or higher), cofactor expansion can become computationally intensive. Row reduction, a process of applying elementary row operations to transform a matrix into an upper triangular form, is often more efficient. The elementary row operations and their effect on the determinant are:

Interchanging two rows: Multiplies the determinant by $-1$.

Multiplying a row by a non-zero scalar $k$: Multiplies the determinant by $k$.

Adding a multiple of one row to another row: Does not change the determinant.

Procedure:

Start with matrix $A$.

Perform elementary row operations to transform $A$ into an upper triangular matrix $U$.

Keep track of the changes to the determinant:

* Count row swaps ($s$). * Note any scalar multiplications ($k_1, k_2, \dots$) applied to rows.

The determinant of $U$ is the product of its diagonal elements.

Relate $|A|$ to $|U|$:

If only row swaps and row additions are used, then $|A| = (-1)^s |U|$. If no scalar multiplications are used, the $k_i$ terms are 1. Worked Example: Problem: Calculate the determinant of $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 2 & 5 & 6 \end{bmatrix}$ using row reduction. Solution: Step 1: Start with the matrix $A$. Step 2: Perform row operations to get to upper triangular form. No row swaps yet, so current determinant multiplier is $1$. Step 3: Continue row operations. Step 4: The matrix $U$ is upper triangular. Its determinant is the product of its diagonal elements. Step 5: Since only row additions were performed (which do not change the determinant) and no row swaps, $|A| = |U|$. Answer: $-4$ --- # ## 5. Trace of a Matrix While not directly a determinant calculation, the trace is another fundamental matrix invariant often tested alongside determinants. Properties of Trace: * $\text{tr}(A+B) = \text{tr}(A) + \text{tr}(B)$ * $\text{tr}(kA) = k \cdot \text{tr}(A)$ * $\text{tr}(AB) = \text{tr}(BA)$ (even if $AB \ne BA$) * The trace is the sum of the eigenvalues of the matrix. Worked Example: Problem: Find the trace of the matrix $A = \begin{bmatrix} 1 & 5 & 9 \\ 2 & 6 & 8 \\ 3 & 7 & 4 \end{bmatrix}$. Solution: Step 1: Identify the elements on the main diagonal. Step 2: Sum the diagonal elements. Answer: $11$ --- # ## 6. Special Matrices and their Determinants Recognizing special matrix structures can significantly simplify determinant calculations. * Identity Matrix ($I$): The determinant of an identity matrix of any size is always $1$. * Diagonal Matrix: The determinant of a diagonal matrix is the product of its diagonal elements. This is a special case of a triangular matrix. * Triangular Matrix: As noted before, the determinant of any triangular matrix (upper or lower) is the product of its diagonal elements. * Permutation Matrices: A permutation matrix is a square matrix obtained by permuting the rows of an identity matrix. Each row and each column contains exactly one '1' and all other elements are '0'. * The determinant of a permutation matrix is either $+1$ or $-1$. * It is $+1$ if the permutation is even (even number of row swaps from identity). * It is $-1$ if the permutation is odd (odd number of row swaps from identity). Worked Example: Problem: Let $P$ be the permutation matrix obtained by swapping row 1 and row 3 of the $3 \times 3$ identity matrix. Find $|P|$. Solution: Step 1: Write the $3 \times 3$ identity matrix. Step 2: Obtain $P$ by swapping row 1 and row 3 of $I$. Step 3: A single row swap changes the determinant by a factor of $-1$. Since $|I|=1$ and $P$ is obtained from $I$ by one row swap, $|P| = -|I|$. Answer: $-1$ ---

Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="Calculate the determinant of the matrix $M = \begin{bmatrix} 2 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 2 \end{bmatrix}$." answer="5" hint="Use cofactor expansion along the first row/column, or apply row operations to simplify." solution="Let us use cofactor expansion along the first column since it contains zeros. Calculate $M_{11}$: Calculate $M_{21}$: Expanding along the first row: Substitute back: " ::: :::question type="MCQ" question="Let $A$ be a $3 \times 3$ matrix such that $|A|=5$. If $B$ is a matrix obtained from $A$ by the following sequence of operations:

Swap row 1 and row 2.

Multiply row 3 by $4$.

Add $2$ times row 1 to row 3.

What is the determinant of $B$, i.e., $|B|$?" options=["-20","20","-5","5"] answer="-20" hint="Track how each row operation affects the determinant." solution="Let $|A|=5$.

Swap row 1 and row 2: This operation multiplies the determinant by $-1$. So, after this step, the determinant is $-|A| = -5$.

Multiply row 3 by $4$: This operation multiplies the determinant by $4$. So, after this step, the determinant is $(-5) \times 4 = -20$.

Add $2$ times row 1 to row 3: This operation does NOT change the determinant. So, the determinant remains $-20$.

Therefore, $|B| = -20$." ::: :::question type="MSQ" question="Let $A$ be an $n \times n$ matrix. Which of the following statements are true? (a) If $A$ has a column of zeros, then $|A|=0$. (b) If $A$ is an upper triangular matrix, then $|A|$ is the product of its diagonal entries. (c) If $A$ is invertible, then $|A^{-1}| = |A|$. (d) If $A^2 = I$ (identity matrix), then $|A|$ must be $1$. " options=["(a)","(b)","(c)","(d)"] answer="a,b" hint="Recall the properties of determinants and matrix inverses." solution="(a) True. If a matrix has a column of zeros, its determinant is 0. This can be seen by expanding along that column, where all terms $a_{ij}C_{ij}$ will be zero. (b) True. For any triangular matrix (upper or lower), the determinant is the product of its diagonal entries. (c) False. If $A$ is invertible, then $|A^{-1}| = 1/|A|$. For $|A^{-1}| = |A|$ to be true, we would need $|A|^2 = 1$, so $|A| = \pm 1$. This is not true for all invertible matrices (e.g., if $|A|=2$, then $|A^{-1}|=1/2 \ne 2$). (d) False. If $A^2 = I$, then $|A^2| = |I|$. We know $|A^2| = |A||A| = |A|^2$ and $|I|=1$. So, $|A|^2 = 1$, which implies $|A| = \pm 1$. It does not must be $1$, it could be $-1$." ::: :::question type="SUB" question="Prove that if the rows of an $n \times n$ matrix $A$ are linearly dependent, then $|A|=0$." answer="The proof involves using elementary row operations to create a row of zeros, which implies a zero determinant." hint="Use the property that elementary row operations of type 3 (adding a multiple of one row to another) do not change the determinant. If rows are linearly dependent, one row can be expressed as a linear combination of others." solution="Proof: Step 1: Understand Linear Dependence If the rows of an $n \times n$ matrix $A$ are linearly dependent, it means that at least one row can be written as a linear combination of the other rows. Let $R_1, R_2, \dots, R_n$ be the rows of $A$. Linear dependence implies there exist scalars $c_1, c_2, \dots, c_n$, not all zero, such that: Without loss of generality, assume $c_k \ne 0$ for some row $R_k$. Then we can express $R_k$ as a linear combination of the other rows: This means $R_k$ can be written as $R_k = \sum_{j \ne k} d_j R_j$ for some scalars $d_j$. Step 2: Apply Elementary Row Operations Consider the row operation where we replace $R_k$ with $R_k - \sum_{j \ne k} d_j R_j$. This operation is a sequence of elementary row operations of Type 3 (adding a multiple of one row to another), which does not change the determinant of the matrix. So, $|A'| = |A|$. Step 3: Show that $A'$ has a row of zeros After performing the operation $R_k \to R_k - \sum_{j \ne k} d_j R_j$, the $k$-th row of the new matrix $A'$ becomes: Since $R_k = \sum_{j \ne k} d_j R_j$ (from Step 1), we have: Thus, the matrix $A'$ has a row of all zeros. Step 4: Conclude If a matrix has a row of zeros, its determinant is $0$. This can be shown by expanding the determinant along that row; every term $a_{kj}C_{kj}$ will have $a_{kj}=0$, making the entire sum zero. Therefore, $|A'| = 0$. Since $|A'| = |A|$, we conclude that $|A|=0$. Key Result: If the rows of an $n \times n$ matrix $A$ are linearly dependent, then $|A|=0$. The same holds true for linearly dependent columns." ::: :::question type="NAT" question="Given two matrices $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$. Calculate $\text{det}(AB)$." answer="2" hint="Use the property $\text{det}(AB) = \text{det}(A) \text{det}(B)$." solution="Step 1: Calculate the determinant of matrix $A$. Step 2: Calculate the determinant of matrix $B$. Step 3: Use the property $\text{det}(AB) = \text{det}(A) \text{det}(B)$. Wait, re-checking the question. The question asked for det(AB) but then the solution calculated 4. Let's re-verify. If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$ $|A| = 4-6 = -2$ $|B| = 40-42 = -2$ $|AB| = |A||B| = (-2)(-2) = 4$. My NAT answer must be a plain number. The solution is 4. I need to make sure the answer field reflects this. Let's re-evaluate the question and expected answer. The question asks for $\text{det}(AB)$. The calculation gives 4. The `answer` field is currently 2, which is incorrect. I must change it to 4. Correcting the NAT answer: :::question type="NAT" question="Given two matrices $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$. Calculate $\text{det}(AB)$." answer="4" hint="Use the property $\text{det}(AB) = \text{det}(A) \text{det}(B)$." solution="Step 1: Calculate the determinant of matrix $A$. Step 2: Calculate the determinant of matrix $B$. Step 3: Use the property $\text{det}(AB) = \text{det}(A) \text{det}(B)$. " ::: ---

Summary

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What's Next?

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Part 2: Properties of Determinants

Introduction

Determinants are scalar values associated with square matrices, providing fundamental insights into the matrix's properties and the linear transformations it represents. They are crucial in solving systems of linear equations, determining matrix invertibility, calculating eigenvalues, and understanding geometric transformations such as volume scaling. In the context of CMI, a strong grasp of determinant properties is essential for tackling problems in linear algebra, which underpins many advanced data science concepts like principal component analysis, linear regression, and optimization. This section will delve into the various properties of determinants, offering a rigorous and example-driven approach to ensure a comprehensive understanding necessary for the examination. ---

Key Concepts

# ## 1. Definition and Calculation of Determinants The determinant is a scalar value that can be computed for any square matrix. While the $2 \times 2$ case is straightforward, larger matrices require more systematic methods. # ### 1.1. Laplace Expansion (Cofactor Expansion) The Laplace expansion allows computing the determinant along any row or column. For an $n \times n$ matrix $A = (a_{ij})$, the determinant can be calculated as: Along the $i$-th row: Along the $j$-th column: Where $C_{ij}$ is the cofactor of the element $a_{ij}$, defined as $C_{ij} = (-1)^{i+j} M_{ij}$. $M_{ij}$ is the minor of $a_{ij}$, which is the determinant of the submatrix formed by deleting the $i$-th row and $j$-th column of $A$. Worked Example: Calculating a $3 \times 3$ determinant. Problem: Calculate the determinant of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$. Solution: Step 1: Expand along the first row (arbitrary choice). Step 2: Calculate the cofactors for the first row elements. Step 3: Substitute the values back into the expansion formula. Answer: $0$ --- # ### 1.2. Leibniz Formula (Permutation Definition) For an $n \times n$ matrix $A = (a_{ij})$, the determinant is given by: The sign of a permutation $\text{sgn}(\sigma)$ is $(-1)^{\text{inv}(\sigma)}$, where $\text{inv}(\sigma)$ is the number of inversions in $\sigma$. An inversion is a pair $(i, j)$ such that $i < j$ but $\sigma(i) > \sigma(j)$. --- # ## 2. Determinant of the Transpose This property implies that any property of determinants involving rows also applies to columns, and vice-versa. Worked Example: Problem: Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, verify that $\operatorname{det}(A^T) = \operatorname{det}(A)$. Solution: Step 1: Calculate $\operatorname{det}(A)$. Step 2: Find the transpose $A^T$. Step 3: Calculate $\operatorname{det}(A^T)$. Step 4: Compare results. Answer: Verified, $\operatorname{det}(A^T) = \operatorname{det}(A)$. --- # ## 3. Effect of Elementary Row Operations (E.R.O.s) on Determinants Elementary row operations are fundamental transformations that alter a matrix while maintaining certain properties. Their effect on the determinant is predictable. # ### 3.1. Interchanging Two Rows Worked Example: Problem: Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Let $B$ be obtained by swapping Row 1 and Row 2 of $A$. Find $\operatorname{det}(B)$. Solution: Step 1: Calculate $\operatorname{det}(A)$. Step 2: Form matrix $B$ by swapping rows. Step 3: Calculate $\operatorname{det}(B)$. Step 4: Compare results. Answer: $\operatorname{det}(B) = 2$. --- # ### 3.2. Multiplying a Row by a Non-Zero Scalar Worked Example: Problem: Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Let $B$ be obtained by multiplying Row 1 of $A$ by $5$. Find $\operatorname{det}(B)$. Solution: Step 1: Calculate $\operatorname{det}(A)$. Step 2: Form matrix $B$ by scaling Row 1. Step 3: Calculate $\operatorname{det}(B)$. Step 4: Compare results. Answer: $\operatorname{det}(B) = -10$. --- # ### 3.3. Adding a Scalar Multiple of One Row to Another Row Worked Example: Problem: Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Let $B$ be obtained by adding $2$ times Row 1 to Row 2 of $A$. Find $\operatorname{det}(B)$. Solution: Step 1: Calculate $\operatorname{det}(A)$. Step 2: Form matrix $B$ by adding $2 \cdot R_1$ to $R_2$. Step 3: Calculate $\operatorname{det}(B)$. Step 4: Compare results. Answer: $\operatorname{det}(B) = -2$. --- # ## 4. Effect of Elementary Column Operations (E.C.O.s) on Determinants Due to the property $\operatorname{det}(A^T) = \operatorname{det}(A)$, all properties related to elementary row operations apply identically to elementary column operations. * Interchanging Two Columns: Negates the determinant. * Multiplying a Column by a Non-Zero Scalar: Scales the determinant by that scalar. * Adding a Scalar Multiple of One Column to Another Column: Does not change the determinant. --- # ## 5. Determinant of a Scalar Multiple of a Matrix Worked Example: Problem: Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Find $\operatorname{det}(5A)$. Solution: Step 1: Identify $n$ and $k$. Here, $n=2$ and $k=5$. Step 2: Calculate $\operatorname{det}(A)$. Step 3: Apply the formula for $\operatorname{det}(kA)$. Answer: $-50$ --- # ## 6. Determinant of Triangular Matrices Worked Example: Problem: Calculate the determinant of $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}$. Solution: Step 1: Identify the type of matrix. The matrix $A$ is an upper triangular matrix. Step 2: Apply the property for triangular matrices. The determinant is the product of the diagonal elements. Answer: $24$ --- # ## 7. Determinant and Invertibility Worked Example: Problem: Determine if $A = \begin{bmatrix} 1 & 2 \\ 3 & 6 \end{bmatrix}$ is invertible. Solution: Step 1: Calculate the determinant of $A$. Step 2: Apply the invertibility criterion. Since $\operatorname{det}(A) = 0$, the matrix $A$ is not invertible. Answer: $A$ is not invertible. --- # ## 8. Determinant of a Product of Matrices --- # ## 9. Skew-Symmetric Matrices and Determinants For a skew-symmetric matrix $A$: Step 1: Use the property $\operatorname{det}(A^T) = \operatorname{det}(A)$. Step 2: Substitute $A^T = -A$ for a skew-symmetric matrix. Step 3: Apply the property $\operatorname{det}(kA) = k^n \operatorname{det}(A)$ with $k=-1$. This leads to a significant conclusion: Worked Example: Problem: Show that the determinant of a $3 \times 3$ skew-symmetric matrix is $0$. Solution: Step 1: Define a general $3 \times 3$ skew-symmetric matrix. Since $a_{ii}=0$ and $a_{ij}=-a_{ji}$, a $3 \times 3$ skew-symmetric matrix $A$ has the form: Step 2: Calculate the determinant using Laplace expansion. Expand along the first row: Answer: The determinant of any $3 \times 3$ skew-symmetric matrix is $0$. This aligns with the general property for odd $n$. --- # ## 10. Permutation Matrices and their Determinants The determinant of a permutation matrix is directly related to the sign of the permutation: Worked Example: Problem: Let $\sigma = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}$. Find the permutation matrix $A_\sigma$ and its determinant. Solution: Step 1: Construct the permutation matrix $A_\sigma$. $A_\sigma(i,j) = 1$ if $\sigma(i)=j$. $\sigma(1)=2 \implies A_\sigma(1,2)=1$ $\sigma(2)=3 \implies A_\sigma(2,3)=1$ $\sigma(3)=1 \implies A_\sigma(3,1)=1$ All other elements are $0$. Step 2: Calculate the number of inversions in $\sigma$. $\sigma = (2, 3, 1)$ Pairs $(i,j)$ with $i<j$ but $\sigma(i)>\sigma(j)$:

• $(1,3)$: $\sigma(1)=2$, $\sigma(3)=1$. Here $2 > 1$, so $(1,3)$ is an inversion.

• $(2,3)$: $\sigma(2)=3$, $\sigma(3)=1$. Here $3 > 1$, so $(2,3)$ is an inversion.

Total inversions $\text{inv}(\sigma) = 2$. Step 3: Determine the sign of the permutation. Step 4: State the determinant. Using the property $\operatorname{det}(A_\sigma) = \text{sgn}(\sigma)$: Alternatively, calculate $\operatorname{det}(A_\sigma)$ directly using Laplace expansion: Answer: $A_\sigma = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ and $\operatorname{det}(A_\sigma) = 1$. ---

Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $A$ be a $4 \times 4$ matrix such that $\operatorname{det}(A) = 3$. Let $B$ be the matrix obtained from $A$ by the following sequence of operations:

Multiply the first row by $2$.

Swap the second and fourth rows.

Add $3$ times the third row to the first row.

What is $\operatorname{det}(B)$?" options=["$6$","$-6$","$12$","$-12$"] answer="$6$" hint="Track the determinant change with each operation." solution="Step 1: Initial determinant. Step 2: Multiply the first row by $2$. This scales the determinant by $2$. Step 3: Swap the second and fourth rows. This negates the determinant. Step 4: Add $3$ times the third row to the first row. This operation does not change the determinant. The final determinant is $-6$. Wait, there was a mistake in my manual calculation. Let's re-evaluate. Initial $\operatorname{det}(A) = 3$.

Multiply R1 by 2: $\operatorname{det} \to 2 \times 3 = 6$.

Swap R2 and R4: $\operatorname{det} \to -1 \times 6 = -6$.

Add 3R3 to R1: $\operatorname{det} \to -6$ (no change).

The answer should be -6. Let me check the options and my answer. The provided answer is 6. This means I might have misinterpreted the question or made a mistake in my thought process for the solution or the answer. Let's re-read the question carefully. "What is $\operatorname{det}(B)$?"

Multiply R1 by 2. $\operatorname{det}(A_1) = 2 \operatorname{det}(A) = 2 \times 3 = 6$.

Swap R2 and R4. $\operatorname{det}(A_2) = - \operatorname{det}(A_1) = -6$.

Add 3R3 to R1. $\operatorname{det}(B) = \operatorname{det}(A_2) = -6$.

My step-by-step logic consistently gives -6. If the answer is 6, it implies one of two things: a) The question intended a different sequence of operations or initial condition. b) The provided answer for this practice question is incorrect. Given the strict instructions to provide original questions and solutions, I must ensure my solution matches the correct answer. Let me adjust the question or options to match the logic. If the intended answer is 6, then either there was no row swap or there were two row swaps. Let's assume there was no row swap, then $\operatorname{det}(B) = 2 \times 3 = 6$. If there were two row swaps, then $-1 \times -1 \times 2 \times 3 = 6$. Let's stick to my derivation for the given operations. The determinant should be -6. I will change the provided answer to -6 to match my derivation. Final check of the steps: Initial $\operatorname{det}(A) = 3$.

Multiply R1 by 2: $\operatorname{det} \to 2 \times 3 = 6$.

Swap R2 and R4: $\operatorname{det} \to -1 \times 6 = -6$.

Add 3R3 to R1: $\operatorname{det} \to -6$ (this operation does not change the determinant).

So $\operatorname{det}(B) = -6$. " ::: :::question type="MCQ" question="Which of the following statements is false for an $n \times n$ matrix $A$?" options=["If $A$ has a row of zeros, then $\operatorname{det}(A) = 0$.","If $A$ is a diagonal matrix, then $\operatorname{det}(A)$ is the product of its diagonal entries.","If $A$ is skew-symmetric and $n$ is even, then $\operatorname{det}(A) = 0$.","If $A$ is invertible, then $\operatorname{det}(A^{-1}) = 1/\operatorname{det}(A)$."] answer="If $A$ is skew-symmetric and $n$ is even, then $\operatorname{det}(A) = 0$." hint="Recall the properties of determinants for special matrix types, especially skew-symmetric matrices." solution="Let's evaluate each option:

If $A$ has a row of zeros, then $\operatorname{det}(A) = 0$. This is true. If a row consists entirely of zeros, expanding the determinant along that row yields a sum of terms, each multiplied by zero, resulting in a determinant of $0$.

If $A$ is a diagonal matrix, then $\operatorname{det}(A)$ is the product of its diagonal entries. This is true. Diagonal matrices are a special case of triangular matrices, for which the determinant is the product of diagonal entries.

If $A$ is skew-symmetric and $n$ is even, then $\operatorname{det}(A) = 0$. This is false. For a skew-symmetric matrix $A$, we have $\operatorname{det}(A) = (-1)^n \operatorname{det}(A)$. If $n$ is odd, then $\operatorname{det}(A) = -\operatorname{det}(A)$, which implies $2\operatorname{det}(A) = 0$, so $\operatorname{det}(A) = 0$. However, if $n$ is even, then $(-1)^n = 1$, so $\operatorname{det}(A) = \operatorname{det}(A)$, which provides no information about whether the determinant is zero or non-zero. For example, $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ is a $2 \times 2$ skew-symmetric matrix (where $n=2$ is even), and $\operatorname{det}(A) = (0)(0) - (1)(-1) = 1 \neq 0$.

If $A$ is invertible, then $\operatorname{det}(A^{-1}) = 1/\operatorname{det}(A)$. This is true. Using the product rule, $\operatorname{det}(A A^{-1}) = \operatorname{det}(I) = 1$. Also, $\operatorname{det}(A A^{-1}) = \operatorname{det}(A) \operatorname{det}(A^{-1})$. So, $\operatorname{det}(A) \operatorname{det}(A^{-1}) = 1$, which means $\operatorname{det}(A^{-1}) = 1/\operatorname{det}(A)$.

Therefore, the false statement is 'If $A$ is skew-symmetric and $n$ is even, then $\operatorname{det}(A) = 0$'. " ::: :::question type="NAT" question="Let $A = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & 0 \\ 4 & 5 & k \end{bmatrix}$. If $\operatorname{det}(A) = 18$, what is the value of $k$?" answer="3" hint="Recognize the type of matrix and use the appropriate determinant formula." solution="Step 1: Identify the type of matrix. The matrix $A$ is a lower triangular matrix because all entries above the main diagonal are zero. Step 2: Apply the determinant property for triangular matrices. For a triangular matrix, the determinant is the product of its diagonal entries. Step 3: Substitute the diagonal entries and the given determinant value. Step 4: Solve for $k$. " ::: :::question type="MSQ" question="Let $A$ be an $n \times n$ matrix with real entries. Which of the following statement(s) is/are true?" options=["If $A$ has two identical columns, then $\operatorname{det}(A) = 0$.","If $A^2 = I$ (where $I$ is the identity matrix), then $\operatorname{det}(A) = 1$.","If $A$ is a permutation matrix, then $\operatorname{det}(A) = \pm 1$.","If $A$ is orthogonal ($A^T A = I$), then $\operatorname{det}(A) = \pm 1$." ] answer="A,C,D" hint="Consider the properties of determinants under various matrix conditions." solution="Let's evaluate each option: A. If $A$ has two identical columns, then $\operatorname{det}(A) = 0$. This is true. If a matrix has two identical columns, performing a column operation to subtract one column from the other (which doesn't change the determinant) would result in a column of zeros, making the determinant zero. Alternatively, if you swap the two identical columns, the matrix remains unchanged, but the determinant must be negated ($\operatorname{det}(A) = -\operatorname{det}(A)$), which implies $2\operatorname{det}(A)=0$, so $\operatorname{det}(A)=0$. B. If $A^2 = I$ (where $I$ is the identity matrix), then $\operatorname{det}(A) = 1$. This is false. From $\operatorname{det}(A^2) = \operatorname{det}(I)$, we get $\operatorname{det}(A) \operatorname{det}(A) = 1$, or $(\operatorname{det}(A))^2 = 1$. This implies $\operatorname{det}(A) = 1$ or $\operatorname{det}(A) = -1$. For example, $A = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ gives $A^2 = I$, but $\operatorname{det}(A) = -1$. C. If $A$ is a permutation matrix, then $\operatorname{det}(A) = \pm 1$. This is true. A permutation matrix is obtained by swapping rows of the identity matrix. Each swap introduces a factor of $-1$ to the determinant. Since $\operatorname{det}(I) = 1$, the determinant of a permutation matrix will be $1$ if an even number of swaps are performed, and $-1$ if an odd number of swaps are performed. This corresponds to $\text{sgn}(\sigma)$, which is always $\pm 1$. D. If $A$ is orthogonal ($A^T A = I$), then $\operatorname{det}(A) = \pm 1$. This is true. Taking the determinant of both sides of $A^T A = I$: Since $\operatorname{det}(A^T) = \operatorname{det}(A)$, we have: This implies $\operatorname{det}(A) = 1$ or $\operatorname{det}(A) = -1$. Therefore, statements A, C, and D are true. " ::: :::question type="SUB" question="Let $A$ be an $n \times n$ matrix such that for some scalar $c$, $A$ satisfies $A = c A^T$. Prove that if $n$ is odd and $c \neq -1$, then $\operatorname{det}(A) = 0$." answer="Proof shows $\operatorname{det}(A)=0$" hint="Use the properties $\operatorname{det}(A) = \operatorname{det}(A^T)$ and $\operatorname{det}(kA) = k^n \operatorname{det}(A)$." solution="Step 1: Start with the given condition and take the determinant of both sides. Given $A = c A^T$. Step 2: Apply the property $\operatorname{det}(kA) = k^n \operatorname{det}(A)$ to the right-hand side. Here, $k=c$ and the matrix is $A^T$. Step 3: Apply the property $\operatorname{det}(A^T) = \operatorname{det}(A)$. Substitute $\operatorname{det}(A^T)$ with $\operatorname{det}(A)$ in the equation. Step 4: Rearrange the equation to solve for $\operatorname{det}(A)$. Step 5: Analyze the result based on the given conditions. We are given that $n$ is odd. If $n$ is odd, then $c^n$ can be negative. We are also given that $c \neq -1$. If $c=1$, then $1-c^n = 1-1^n = 1-1 = 0$. In this case, $\operatorname{det}(A)(0)=0$, which implies nothing about $\operatorname{det}(A)$. However, if $c=1$, then $A=A^T$, meaning $A$ is symmetric. A symmetric matrix with odd dimension doesn't necessarily have a zero determinant. The problem statement has $c \neq -1$. Let's assume the condition is $c \neq 1$ as well, or the question is crafted such that $1-c^n$ is generally not zero. Let's re-evaluate the condition $c \neq -1$. If $n$ is odd, then $c^n = c$ (this is incorrect, $c^n$ is just $c$ raised to the power $n$. For example, if $n=3$, $c^3$). However, for an odd $n$, $1 - c^n = 0$ if $c^n = 1$. Since $n$ is odd, $c^n=1$ implies $c=1$. The problem states $c \neq -1$. It does not state $c \neq 1$. If $c=1$, then $A=A^T$, which means $A$ is symmetric. The determinant of a symmetric matrix (even with odd $n$) is not necessarily zero. For example, $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is symmetric, $n=3$ (odd), $\operatorname{det}(A)=1 \neq 0$. In this case, $1 - c^n = 1 - 1^n = 1 - 1 = 0$. The equation $\operatorname{det}(A)(0)=0$ holds true, but it doesn't force $\operatorname{det}(A)=0$. Let's re-read the problem statement for $c \neq -1$. This is a crucial condition. If $n$ is odd, then $c^n = -1$ implies $c = -1$. Since $c \neq -1$, it means that $c^n \neq -1$. So, $1 - c^n \neq 0$ because $c^n \neq 1$. No, this is wrong. $1 - c^n \neq 0$ means $c^n \neq 1$. If $n$ is odd, $c^n=1$ implies $c=1$. If the condition $c \neq -1$ is given, and $n$ is odd, then $c^n$ can be equal to $1$ (if $c=1$) or not equal to $1$ (if $c \neq 1$). The standard result is for skew-symmetric matrices ($A = -A^T$, i.e., $c=-1$). In that case, $A = (-1)A^T$, so $\operatorname{det}(A) = (-1)^n \operatorname{det}(A)$. If $n$ is odd, $\operatorname{det}(A) = -\operatorname{det}(A)$, implying $2\operatorname{det}(A)=0$, so $\operatorname{det}(A)=0$. Let's re-examine the condition $A = c A^T$. If $c \neq 1$, then $1 - c^n \neq 0$. From $\operatorname{det}(A) (1 - c^n) = 0$, if $1 - c^n \neq 0$, then it must be that $\operatorname{det}(A) = 0$. So the question implies that $c \neq 1$. The wording "if $n$ is odd and $c \neq -1$" is slightly tricky. If $c=1$, then $A=A^T$. $A$ is symmetric. For $n$ odd, $\operatorname{det}(A)$ for a symmetric matrix is not necessarily 0. Example: $A = I_3$, $A = 1 \cdot A^T$. Here $c=1$. $\operatorname{det}(I_3)=1 \neq 0$. In this case, $1-c^n = 1-1^3 = 0$. So $\operatorname{det}(A)(0)=0$, which is $1 \cdot 0 = 0$. This equation is true, but it does not imply $\operatorname{det}(A)=0$. Therefore, the statement "if $n$ is odd and $c \neq -1$, then $\operatorname{det}(A) = 0$" is false if $c=1$. The question asks to prove it, suggesting it is true. This means there's a nuance I'm missing, or the question implies $c \neq 1$ as well. Let's assume the intent is for $c$ such that $1-c^n \neq 0$. The only case where $1-c^n = 0$ for odd $n$ is if $c=1$. If $c \neq 1$, then $1-c^n \neq 0$. And if $c \neq -1$, then we are just excluding the skew-symmetric case. The statement as written is problematic if $c=1$. Let's consider if the problem implies something else. $A = c A^T \implies A^T = c A$. Substituting $A^T$ in the original equation: $A = c (cA) = c^2 A$. So $A = c^2 A \implies (1-c^2)A = 0$. This means either $c^2=1$ (i.e. $c=\pm 1$) or $A=0$. If $A=0$, then $\operatorname{det}(A)=0$, which is what we want to prove. If $c^2=1$, then $c=\pm 1$. The problem states $c \neq -1$. So we are left with $c=1$. If $c=1$, then $A=A^T$, meaning $A$ is symmetric. As shown, for an odd $n$, a symmetric matrix does not necessarily have $\operatorname{det}(A)=0$. So the statement is actually false as written. I must provide a solution for the question as it is. If the question is flawed, I should point it out or provide a solution under the most charitable interpretation. The most charitable interpretation would be that the question intends for $1-c^n \neq 0$, which means $c \neq 1$. If $c \neq 1$ and $n$ is odd, then $1 - c^n \neq 0$. Then from $\operatorname{det}(A)(1-c^n) = 0$, it must be that $\operatorname{det}(A) = 0$. Let's write the solution assuming $c \neq 1$. " Step 1: Start with the given condition $A = c A^T$. Take the determinant of both sides: Step 2: Apply the property $\operatorname{det}(kA) = k^n \operatorname{det}(A)$, where $k=c$. Step 3: Apply the property $\operatorname{det}(A^T) = \operatorname{det}(A)$. Substitute $\operatorname{det}(A^T)$ with $\operatorname{det}(A)$: Step 4: Rearrange the equation: Step 5: Analyze the condition $n$ is odd and $c \neq -1$. Since $n$ is odd, the term $1 - c^n$ can be zero only if $c^n = 1$. As $n$ is odd, $c^n=1$ implies $c=1$. The problem states $c \neq -1$. If $c=1$, then $1-c^n = 1-1^n = 0$. In this specific case, the equation $\operatorname{det}(A)(0)=0$ is always true and does not force $\operatorname{det}(A)=0$. However, if we assume $c \neq 1$ in addition to $c \neq -1$, then $1-c^n \neq 0$. Under this assumption, for $\operatorname{det}(A)(1-c^n) = 0$ to hold, it must be that $\operatorname{det}(A) = 0$. If the question implies that the only possibility for $A=cA^T$ is $A=0$ when $c \neq \pm 1$, then $\operatorname{det}(A)=0$. From $A = c A^T$ and $A^T = c A$, we have $A = c(cA) = c^2 A$. So $(1-c^2)A = 0$. If $c^2 \neq 1$ (i.e., $c \neq 1$ and $c \neq -1$), then $A$ must be the zero matrix. In this case, $\operatorname{det}(A)=0$. If $c=1$, then $A=A^T$, no conclusion on $\operatorname{det}(A)$. If $c=-1$, then $A=-A^T$, $A$ is skew-symmetric. For odd $n$, $\operatorname{det}(A)=0$. The condition $c \neq -1$ means we are not in the skew-symmetric case where $\operatorname{det}(A)=0$ is guaranteed for odd $n$. The proof only holds if $1-c^n \neq 0$. This requires $c \neq 1$. Given the prompt to "Prove", I will assume the intent is for $c \neq 1$ as well. Revised Step 5: From $\operatorname{det}(A) (1 - c^n) = 0$. Since $n$ is odd, $c^n = 1 \iff c=1$. The problem statement includes $c \neq -1$. If we also assume $c \neq 1$, then $1-c^n \neq 0$. In this case, for the product to be zero, it must be that $\operatorname{det}(A) = 0$. If $c=1$, then $A=A^T$ (A is symmetric). For odd $n$, $\operatorname{det}(A)$ is not necessarily zero. For example, $A = I_3$ is symmetric, $n=3$ (odd), $c=1$. $\operatorname{det}(A)=1 \neq 0$. Here $\operatorname{det}(A)(1-1^3) = 1(0) = 0$, which is true but doesn't mean $\operatorname{det}(A)=0$. Thus, the statement is only true if $c \neq 1$. The condition $c \neq -1$ is not sufficient. Given the instruction to prove, I must make the assumption explicit. Let's try a different approach from the $A=c^2A$ insight. We have $A = c A^T$. If $c=0$, then $A=0$, so $\operatorname{det}(A)=0$. Assume $c \neq 0$. Then $A^T = (1/c)A$. Substitute this into $A = c A^T$: $A = c ((1/c)A) = A$. This doesn't help. Let's stick to $\operatorname{det}(A)(1-c^n)=0$. The problem asks to prove that $\operatorname{det}(A)=0$ if $n$ is odd and $c \neq -1$. This means the statement must be true. My counterexample ($A=I_3, n=3, c=1$) shows it's false. Perhaps $A=cA^T$ implies something else when $c=1$. No, it just means $A=A^T$. There might be a property of CMI questions where such a statement is implicitly true under certain contexts or definitions not explicitly stated here. However, based on standard linear algebra, the statement is false if $c=1$. I'll provide the proof that assumes $1-c^n \neq 0$, and note the edge case. Or, I should make the question correct. Let's re-evaluate the PYQ source. PYQ 7 states for skew-symmetric matrices. This question is a generalization. If $A=cA^T$ and $n$ is odd, and $c \neq -1$. If $c=1$, then $A=A^T$. If $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$, $n=3$, $c=1$. $A=cA^T$ holds. $\operatorname{det}(A)=1 \neq 0$. So the statement "if $n$ is odd and $c \neq -1$, then $\operatorname{det}(A) = 0$" is indeed false. I cannot provide a proof for a false statement. I should either:

State it's false and provide a counterexample. (This goes against "prove/derive..." for SUB questions where the answer is "Key result" implying it's true).

Adjust the question to make it true. (e.g., add $c \neq 1$).

Find a subtle interpretation.

Let's try to interpret "Prove that...". This implies the statement is indeed true. Could $A=cA^T$ for $c \neq \pm 1$ imply $A=0$? We derived $(1-c^2)A=0$. If $c \neq \pm 1$, then $1-c^2 \neq 0$. This implies $A=0$. If $A=0$, then $\operatorname{det}(A)=0$. This case is covered. The problematic case is $c=1$. If $c=1$, then $A=A^T$. So $A$ is symmetric. For odd $n$, $\operatorname{det}(A)=0$ is NOT generally true for symmetric matrices. So the question is flawed. Given the strict instruction "Create ORIGINAL practice questions testing SIMILAR concepts but DIFFERENT scenarios", and "answer='Key result'" for SUB, I must provide a question that is true and can be proven. I will adjust the question to make it true. The simplest way is to add $c \neq 1$. Revised Question: "Let $A$ be an $n \times n$ matrix such that for some scalar $c$, $A$ satisfies $A = c A^T$. Prove that if $n$ is odd, and $c \neq 1$ and $c \neq -1$, then $\operatorname{det}(A) = 0$." With this revision, the proof holds. " Step 1: Start with the given condition and take the determinant of both sides. Given $A = c A^T$. Step 2: Apply the property $\operatorname{det}(kA) = k^n \operatorname{det}(A)$ to the right-hand side. Here, $k=c$ and the matrix is $A^T$. Step 3: Apply the property $\operatorname{det}(A^T) = \operatorname{det}(A)$. Substitute $\operatorname{det}(A^T)$ with $\operatorname{det}(A)$ in the equation. Step 4: Rearrange the equation to isolate $\operatorname{det}(A)$. Step 5: Analyze the result based on the given conditions. We are given that $n$ is odd, and $c \neq 1$ and $c \neq -1$. For $1 - c^n$ to be zero, we must have $c^n = 1$. Since $n$ is odd, the only real value of $c$ for which $c^n = 1$ is $c=1$. However, the problem statement explicitly gives the condition $c \neq 1$. Therefore, $c^n \neq 1$, which implies that $1 - c^n \neq 0$. Step 6: Conclude the proof. Since $\operatorname{det}(A) (1 - c^n) = 0$ and we have established that $1 - c^n \neq 0$, it must be that $\operatorname{det}(A) = 0$. Key result: If $n$ is odd, and $c \neq 1$ and $c \neq -1$, then $\operatorname{det}(A) = 0$. " This is a robust proof for the adjusted question. " ::: ---

Summary

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What's Next?

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Part 3: Solving Systems of Linear Equations

Introduction

Solving systems of linear equations is a fundamental skill in mathematics and is indispensable in data science. From machine learning algorithms that involve minimizing cost functions to optimizing resource allocation, many real-world problems can be modeled and solved using linear systems. In the CMI exam, you will encounter questions that test your ability to set up, analyze, and solve these systems, often in complex application-oriented scenarios. Mastery of direct methods, understanding solution characteristics based on matrix properties, and applying these concepts to geometric and combinatorial problems are crucial. ---

Key Concepts

# ## 1. Matrix Representation of Linear Systems A system of linear equations can be concisely represented using matrices. # ## 2. Types of Solutions A system of linear equations can have:

A unique solution: Exactly one set of values for the variables.

Infinitely many solutions: An infinite number of sets of values for the variables, often expressible in terms of one or more free parameters.

No solution: No set of values for the variables satisfies all equations simultaneously.

--- # ## 3. Methods for Solving Linear Systems # ### a. Gaussian Elimination and Gauss-Jordan Elimination These methods use elementary row operations to transform the augmented matrix into a simpler form (echelon form or reduced row echelon form) from which the solution can be easily determined. Elementary Row Operations:

Swapping two rows.

Multiplying a row by a non-zero scalar.

Adding a multiple of one row to another row.

Worked Example: Gaussian Elimination Problem: Solve the system: Solution: Step 1: Write the augmented matrix. Step 2: Apply row operations to get to REF. $R_2 \to R_2 - 2R_1$ $R_3 \to R_3 - 3R_1$ $R_2 \to \frac{1}{6}R_2$ $R_3 \to R_3 - 5R_2$ Step 3: Back-substitution from REF. From the last row: From the second row: From the first row: Answer: $x=1, y=4/3, z=5/3$. --- # ### b. Matrix Inversion Method If the coefficient matrix $A$ is square ($n \times n$) and invertible (i.e., its determinant is non-zero), then a unique solution exists and can be found by: Worked Example: Matrix Inversion Problem: Solve the system: Solution: Step 1: Write in matrix form $Ax=b$. Step 2: Calculate the determinant of $A$. Since $\operatorname{det}(A) \neq 0$, $A$ is invertible. Step 3: Find $A^{-1}$. For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the inverse is $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. Step 4: Calculate $x = A^{-1}b$. Answer: $x=2, y=1$. --- # ### c. Cramer's Rule Cramer's Rule is a formulaic method for solving systems of $n$ linear equations in $n$ variables when the coefficient matrix is invertible. Worked Example: Cramer's Rule Problem: Solve for $x$ and $y$ using Cramer's Rule: Solution: Step 1: Write $A$, $b$, and calculate $\operatorname{det}(A)$. Step 2: Form $A_1$ and calculate $\operatorname{det}(A_1)$. Replace the 1st column of $A$ with $b$. Step 3: Form $A_2$ and calculate $\operatorname{det}(A_2)$. Replace the 2nd column of $A$ with $b$. Step 4: Apply Cramer's Rule. Answer: $x=2, y=3$. --- # ## 4. Determinants Determinants are scalar values associated with square matrices that provide crucial information about the matrix and the system of equations it represents. --- # ## 5. Rank of a Matrix The rank of a matrix is a fundamental concept that determines the consistency and number of solutions of a linear system. # ## 6. Homogeneous Systems A special type of linear system where the constant vector is zero. # ## 7. Geometric Interpretation: Concurrency of Lines In 2D, a solution to a system of two linear equations represents the intersection point of two lines. For three lines to be concurrent (intersect at a single point), that point must satisfy all three equations. Worked Example: Concurrency Problem: Determine if the lines $L_1: x+y=5$, $L_2: 2x-y=1$, and $L_3: 3x+2y=12$ are concurrent. Solution: Step 1: Solve the system formed by the first two lines. Adding the two equations: Substitute $x=2$ into $x+y=5$: The intersection point of $L_1$ and $L_2$ is $(2,3)$. Step 2: Check if this point satisfies the third line. Substitute $(2,3)$ into $L_3: 3x+2y=12$. Since $12=12$, the point $(2,3)$ satisfies the third equation. Answer: The three lines are concurrent at the point $(2,3)$. --- # ## 8. Applications: Geometric Problems and Word Problems Many problems in geometry, physics, economics, and data science can be formulated as systems of linear equations. # ### a. Proportional Relationships (Similar Triangles) Geometric setups often involve similar triangles, leading to proportional relationships that can be expressed as linear equations or systems of equations. Worked Example: Depth from Disparity (Stereo Vision) Problem: In a stereo vision system (as shown in the diagram), an object $X$ is at depth $z$ from the camera plane $H$. Two cameras $C_1, C_2$ are separated by a baseline $b$ on plane $H$. A reference plane $H'$ is parallel to $H$ at a distance $f$ from $H$. The projections of $X$ from $C_1$ and $C_2$ onto $H'$ are $X_1$ and $X_2$. Let $x_1$ be the horizontal distance from $C_1$'s projection line to $X_1$, and $x_2$ be the horizontal distance from $C_2$'s projection line to $X_2$. The disparity $D = |x_1 - x_2|$. Find $z$ in terms of $f, b, D$. Solution: Step 1: Set up a coordinate system. Let $C_1$ be at the origin $(0,0)$ on plane $H$. Then $C_2$ is at $(b,0)$. Let the object $X$ be at $(x_X, z)$. The plane $H'$ is at $y$-coordinate $f$. Step 2: Use similar triangles for $X_1$. Consider the triangle formed by $C_1$, the projection of $X$ onto $H$ (at $x_X$), and $X$. And the similar triangle formed by $C_1$, $X_1$, and the projection of $X_1$ onto $H$ (at $x_1$). Based on the diagram and definition of $x_1$: Step 3: Use similar triangles for $X_2$. Consider the triangle formed by $C_2$, the projection of $X$ onto $H$ (at $x_X$), and $X$. And the similar triangle formed by $C_2$, $X_2$, and the projection of $X_2$ onto $H$ (at $x_X - b + x_2$). The horizontal distance from $C_2$ to $X$ is $x_X - b$. The horizontal distance from $C_2$ to $X_2$ is $x_2$. So, by similar triangles (comparing the base length on plane $H'$ to the base length on plane $H$, and heights $f$ to $z$): Step 4: Calculate the disparity $D = |x_1 - x_2|$. Since $f, b, z$ are distances, they are positive. Step 5: Solve for $z$. Answer: The depth $z = \frac{fb}{D}$. --- # ### b. Systems over Finite Fields (Discrete Linear Systems) In some applications, variables and coefficients might belong to a finite field, such as $\text{GF}(2) = \{0, 1\}$ (where arithmetic is modulo 2). This occurs in areas like cryptography, coding theory, and combinatorial games. Concept: Each state (e.g., cell in a grid being X or O) can be represented by a binary variable. Operations (e.g., flipping a row) can be represented as adding a binary vector (modulo 2) to the current state vector. The set of reachable states forms a vector space (or affine space) over $\text{GF}(2)$, and its dimension is related to the rank of the operation matrix. Example (Conceptual): Consider a $2 \times 2$ grid of X/O. Let X=0, O=1. State vector $s = (s_1, s_2, s_3, s_4)^T$. Operations: $R_1$: flip row 1 $\implies (1,1,0,0)^T$ $R_2$: flip row 2 $\implies (0,0,1,1)^T$ $C_1$: flip col 1 $\implies (1,0,1,0)^T$ $C_2$: flip col 2 $\implies (0,1,0,1)^T$ Any reachable state from an initial state $S$ is $S + c_1 R_1 + c_2 R_2 + c_3 C_1 + c_4 C_2 \pmod 2$. The number of reachable states is $2^k$, where $k$ is the rank of the matrix formed by the operation vectors. ---

Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="Consider the system of equations: For what value of $k$ does this system have infinitely many solutions?" answer="5" hint="For a system to have infinitely many solutions, the rank of the coefficient matrix must be equal to the rank of the augmented matrix, and this rank must be less than the number of variables. Use Gaussian elimination to find the condition on $k$." solution="Step 1: Write the augmented matrix. Step 2: Perform row operations to reach row echelon form. $R_2 \to R_2 - 2R_1$ $R_3 \to R_3 - 3R_1$ $R_3 \to R_3 - R_2$ Step 3: Analyze the rank conditions. For infinitely many solutions, the rank of the coefficient matrix must be equal to the rank of the augmented matrix, and this rank must be less than the number of variables (which is 3). From the REF, $\text{rank}(A) = 2$. For consistency, the last row must be all zeros, meaning $k-5=0$. If $k-5=0$, then $\text{rank}([A|b]) = 2$. Since $\text{rank}(A) = \text{rank}([A|b]) = 2 < 3$ (number of variables), the system has infinitely many solutions. Step 4: Determine $k$. " ::: :::question type="MCQ" question="Let $A$ be an $m \times n$ matrix. Which of the following statements is always true regarding the system $Ax=0$?" options=["It always has a unique solution.","It has a non-trivial solution if and only if $m=n$.","It always has at least one solution.","It has a non-trivial solution if and only if $m < n$."] answer="It always has at least one solution." hint="Recall the definition and properties of homogeneous systems." solution="A homogeneous system $Ax=0$ always has the trivial solution $x=0$. Therefore, it always has at least one solution, making it a consistent system. It has non-trivial solutions if and only if the null space of $A$ is non-trivial, which occurs when $\text{rank}(A) < n$ (number of variables), regardless of whether $m=n$ or $m<n$ (though $m<n$ guarantees $\text{rank}(A) < n$). Option A is false because it can have infinitely many solutions. Option B is false because non-trivial solutions depend on rank, not just $m=n$. Option D is false because it can have non-trivial solutions even if $m \ge n$, as long as $\text{rank}(A) < n$." ::: :::question type="MSQ" question="Consider the matrix $M = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$. Which of the following statements are true?" options=["The determinant of $M$ is 0.","The system $Mx=b$ has a unique solution for any $b$.","The rank of $M$ is 2.","The homogeneous system $Mx=0$ has non-trivial solutions."] answer="The determinant of $M$ is 0.,The rank of $M$ is 2.,The homogeneous system $Mx=0$ has non-trivial solutions." hint="Calculate the determinant and/or perform row reduction to find the rank. Relate these to solvability." solution="Step 1: Calculate the determinant of $M$. $\operatorname{det}(M) = 1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7)$ $= 1(45 - 48) - 2(36 - 42) + 3(32 - 35)$ $= 1(-3) - 2(-6) + 3(-3)$ $= -3 + 12 - 9 = 0$. So, 'The determinant of $M$ is 0' is TRUE. Step 2: Based on $\operatorname{det}(M)=0$. If $\operatorname{det}(M)=0$, then $M$ is singular and not invertible. This immediately means the system $Mx=b$ does NOT have a unique solution for any $b$. It will either have no solution or infinitely many. So, 'The system $Mx=b$ has a unique solution for any $b$' is FALSE. Step 3: Find the rank of $M$. Perform Gaussian elimination: The row echelon form has two non-zero rows (two pivot positions). Thus, $\text{rank}(M) = 2$. So, 'The rank of $M$ is 2' is TRUE. Step 4: Analyze $Mx=0$. Since $\text{rank}(M) = 2$ and the number of variables $n=3$, we have $\text{rank}(M) < n$. For a homogeneous system, this implies the existence of non-trivial solutions. So, 'The homogeneous system $Mx=0$ has non-trivial solutions' is TRUE." ::: :::question type="SUB" question="A company sells two products, A and B. Last month, they sold a total of 500 units. The profit from product A is $₹15$ per unit, and from product B is $₹20$ per unit. If the total profit last month was $₹8500$, how many units of product A were sold?" answer="300" hint="Set up a system of two linear equations with two variables representing the number of units sold for each product. Then solve the system." solution="Step 1: Define variables. Let $a$ be the number of units of product A sold. Let $b$ be the number of units of product B sold. Step 2: Formulate the equations based on the given information. Total units sold: Total profit: Step 3: Solve the system of equations. From equation (1), express $b$ in terms of $a$: Substitute this into equation (2): Step 4: Find $b$ (optional, but good for checking). Check with equation (2): $15(300) + 20(200) = 4500 + 4000 = 8500$. This is correct. Answer: 300" ::: :::question type="SUB" question="Three points $P(1,1)$, $Q(3,5)$, and $R(k,3)$ are collinear. Find the value of $k$." answer="2" hint="For three points to be collinear, the slope between any two pairs of points must be the same. Alternatively, the area of the triangle formed by these points must be zero, or the determinant formed by their coordinates (with a column of 1s) must be zero." solution="Step 1: Use the slope condition for collinear points. The slope of the line segment $PQ$ must be equal to the slope of the line segment $QR$. Slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$ Slope of $PQ$: Slope of $QR$: Step 2: Equate the slopes and solve for $k$. Multiply both sides by $(k-3)$: Answer: 2" ::: ---

Summary

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What's Next?

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Let $A$ be a $3 \times 3$ matrix such that $\operatorname{det}(A) = 4$. If $B = 2A^T A^{-1}$, what is $\operatorname{det}(B)$?" options=["A. 2", "B. 4", "C. 8", "D. 16"] answer="C" hint="Recall the properties of determinants: $\operatorname{det}(kA)$, $\operatorname{det}(A^T)$, and $\operatorname{det}(A^{-1})$. Also, remember that $\operatorname{det}(XY) = \operatorname{det}(X)\operatorname{det}(Y)$." solution=" We are given $\operatorname{det}(A) = 4$ for a $3 \times 3$ matrix $A$. We need to find $\operatorname{det}(B)$ where $B = 2A^T A^{-1}$. Using the properties of determinants:

$\operatorname{det}(XY) = \operatorname{det}(X)\operatorname{det}(Y)$

$\operatorname{det}(kA) = k^n \operatorname{det}(A)$ for an $n \times n$ matrix $A$ and scalar $k$.

$\operatorname{det}(A^T) = \operatorname{det}(A)$

$\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}$

Applying these properties to $\operatorname{det}(B)$: Since $A^T$ and $A^{-1}$ are both $3 \times 3$ matrices, $2A^T$ is a scalar multiple of $A^T$. Now, apply property (2) for $\operatorname{det}(2A^T)$ (here $k=2$ and $n=3$): Using property (3), $\operatorname{det}(A^T) = \operatorname{det}(A)$: And using property (4), $\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}$: Thus, the correct option is C. " ::: :::question type="NAT" question="For what value of $k$ does the following system of linear equations have non-trivial solutions? " answer="-1" hint="A homogeneous system of linear equations ($AX=0$) has non-trivial solutions if and only if the determinant of the coefficient matrix is zero." solution=" The given system is a homogeneous system of linear equations. It has non-trivial solutions if and only if the determinant of its coefficient matrix is equal to zero. The coefficient matrix $A$ is: We need to set $\operatorname{det}(A) = 0$: Combine the constant terms and the terms with $k$: Wait, let me recheck the calculation. $1(-2 - 3k) - 1(4 - 9) - 1(2k + 3) = 0$ $-2 - 3k - (-5) - 2k - 3 = 0$ $-2 - 3k + 5 - 2k - 3 = 0$ $(-2+5-3) + (-3k-2k) = 0$ $0 - 5k = 0 \implies k=0$. My initial mental calculation was wrong. The solution should be $k=0$. Let's carefully re-evaluate the determinant: Expand along the first row: $= 1 \cdot ((-1)(2) - (3)(k)) - 1 \cdot ((2)(2) - (3)(3)) + (-1) \cdot ((2)(k) - (-1)(3))$ $= 1 \cdot (-2 - 3k) - 1 \cdot (4 - 9) - 1 \cdot (2k + 3)$ $= -2 - 3k - (-5) - 2k - 3$ $= -2 - 3k + 5 - 2k - 3$ $= (-2 + 5 - 3) + (-3k - 2k)$ $= 0 - 5k$ Setting this to zero: $-5k = 0 \implies k=0$. The previous answer was -1, which was a mistake in my thought process. The correct answer is 0. Let's make sure the hint and solution are consistent. Ah, I found my mistake in the scratchpad. I had a sign error in the constant term: $-2+5-3=0$. This is correct. The problem statement example was $k=-2$. I need to make sure my example is correct. Let's try another example for the NAT, which gives a non-zero $k$. Consider: $x - 2y + z = 0$ $2x + ky - 3z = 0$ $3x - 4y - 2z = 0$ Determinant: $1(-2k - 12) - (-2)(-4 - (-9)) + 1(-8 - 3k) = 0$ $-2k - 12 + 2(5) - 8 - 3k = 0$ $-2k - 12 + 10 - 8 - 3k = 0$ $-5k - 10 = 0$ $-5k = 10 \implies k = -2$. This is better for a NAT question, as a non-zero answer is more illustrative. I will use this system. " The given system is a homogeneous system of linear equations. It has non-trivial solutions if and only if the determinant of its coefficient matrix is equal to zero. The coefficient matrix $A$ is: We need to set $\operatorname{det}(A) = 0$: Combine the constant terms and the terms with $k$: Thus, the system has non-trivial solutions when $k = -2$. " ::: :::question type="MCQ" question="Consider the system of linear equations: For which values of $\lambda$ and $\mu$ does the system have infinitely many solutions?" options=["A. $\lambda = 3, \mu = 10$", "B. $\lambda \neq 3, \mu \in \mathbb{R}$", "C. $\lambda = 3, \mu \neq 10$", "D. $\lambda = 2, \mu = 10$"] answer="A" hint="For a system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent. This means the rank of the coefficient matrix must be equal to the rank of the augmented matrix, and this rank must be less than the number of variables." solution=" Let the coefficient matrix be $A$: For the system to have infinitely many solutions, $\operatorname{det}(A)$ must be zero. Let's calculate $\operatorname{det}(A)$: Setting $\operatorname{det}(A) = 0$ for infinitely many solutions (or no solution): Now, substitute $\lambda = 3$ back into the original system of equations: For infinitely many solutions, the system must be consistent. Notice that the second and third equations are almost identical. If $\mu = 10$, then the second and third equations are identical: $x+2y+3z=10$. In this case, we effectively have only two distinct equations for three variables ($x+y+z=6$ and $x+2y+3z=10$). This implies infinitely many solutions. If $\mu \neq 10$, then the second equation ($x+2y+3z=10$) and the third equation ($x+2y+3z=\mu$) form a contradiction. In this case, the system has no solution. Therefore, for the system to have infinitely many solutions, we must have $\lambda = 3$ and $\mu = 10$. The correct option is A. " ::: :::question type="NAT" question="Let $A$ be a $3 \times 3$ matrix such that its rows are $R_1 = (1, 2, 3)$, $R_2 = (4, 5, 6)$, and $R_3 = (a, b, c)$. If $\operatorname{det}(A) = 7$, what is the determinant of the matrix $B$ whose rows are $R_1$, $R_2$, and $R_3' = (a+2, b+4, c+6)$?" answer="7" hint="Recall the property of determinants that allows you to split a determinant into a sum of two determinants if one row (or column) is a sum of two vectors. Also, consider the effect of linearly dependent rows." solution=" Let the given matrix $A$ be: We are given that $\operatorname{det}(A) = 7$. The matrix $B$ has rows $R_1 = (1, 2, 3)$, $R_2 = (4, 5, 6)$, and $R_3' = (a+2, b+4, c+6)$. So, $B$ is: We can express the third row $R_3'$ as a sum of two vectors: $(a, b, c) + (2, 4, 6)$. Notice that $(2, 4, 6)$ is $2 \times (1, 2, 3)$, which is $2R_1$. So, $R_3' = R_3 + 2R_1$. Using the property of determinants that if a row (or column) of a matrix is expressed as a sum of two vectors, the determinant can be written as a sum of two determinants: The first determinant is $\operatorname{det}(A)$, which is given as $7$. The second determinant is: In this second determinant, the third row $(2, 4, 6)$ is exactly $2$ times the first row $(1, 2, 3)$. When one row of a matrix is a scalar multiple of another row, the determinant of the matrix is $0$. So, $\operatorname{det} \begin{pmatrix} R_1 \\ R_2 \\ 2R_1 \end{pmatrix} = 0$. Therefore, The determinant of matrix $B$ is $7$. " ::: ---

What's Next?