Matrices and Determinants

Overview

The study of matrices and determinants forms a cornerstone of linear algebra and is an indispensable tool in nearly every branch of engineering and computer science. Matrices provide a powerful and compact framework for representing and manipulating large sets of linear equations, which arise frequently when modeling complex systems. From representing transformations in computer graphics to encoding the structure of graphs in algorithm design, the applications of matrix theory are both vast and fundamental. A thorough understanding of matrix algebra is therefore not merely an academic exercise but a prerequisite for a deeper comprehension of advanced engineering concepts.

For the Graduate Aptitude Test in Engineering (GATE), proficiency in this chapter is of paramount importance. Questions related to matrix operations, the properties of determinants, and the computation of matrix inverses appear consistently and carry significant weight. Mastery of these topics is essential for solving problems directly related to linear algebra and also serves as a foundational skill for tackling more complex problems in areas such as differential equations, numerical methods, and optimization. In this chapter, we shall systematically develop the principles of matrix theory, focusing on the techniques and properties most relevant to the GATE examination.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Matrix Operations | Fundamental algebraic manipulations of matrices. |
| 2 | Determinants | Calculating the scalar value of matrices. |
| 3 | Matrix Inverse | Finding the multiplicative inverse of matrices. |

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Learning Objectives

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We now turn our attention to Matrix Operations...

Part 1: Matrix Operations

Introduction

In the study of linear algebra, the matrix serves as a fundamental object for representing and manipulating linear transformations and systems of linear equations. A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns, which provides a powerful framework for organizing and processing data. A thorough understanding of matrix operations is indispensable, as these operations form the computational bedrock for nearly all advanced topics within linear algebra, including the analysis of vector spaces, the solution of differential equations, and the implementation of numerous algorithms in computer science.

Our primary focus will be on the essential arithmetic of matrices—addition, scalar multiplication, and matrix multiplication—along with key unary operations such as the transpose and the trace. We will investigate the properties that govern these operations, such as associativity and distributivity, and pay special attention to properties that diverge from scalar arithmetic, most notably the non-commutativity of matrix multiplication. Mastery of these operational rules and their consequences is a prerequisite for tackling more complex concepts in the GATE examination.

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Key Concepts

1. Basic Matrix Arithmetic

The foundational operations of matrix algebra are defined element-wise or through more complex row-column interactions.

Matrix Addition and Subtraction

Two matrices $A = [a_{ij}]$ and $B = [b_{ij}]$ can be added or subtracted if and only if they have the same dimensions ($m \times n$). The resulting matrix $C = A \pm B$ is also of dimension $m \times n$, with elements given by $c_{ij} = a_{ij} \pm b_{ij}$.

Scalar Multiplication

The product of a scalar $k$ and a matrix $A = [a_{ij}]$ is a matrix $B = kA$ of the same dimension, where each element is multiplied by the scalar: $b_{ij} = k \cdot a_{ij}$.

Matrix Multiplication

The product of two matrices, $A_{m \times n}$ and $B_{n \times p}$, is a matrix $C_{m \times p}$. For the product $AB$ to be defined, the number of columns in the first matrix ($A$) must be equal to the number of rows in the second matrix ($B$). The element $c_{ij}$ in the $i$-th row and $j$-th column of the product matrix $C$ is computed as the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.

Properties of Matrix Multiplication:

• Associative: $(AB)C = A(BC)$


• Distributive: $A(B+C) = AB + AC$ and $(A+B)C = AC + BC$


• Not Commutative: In general, $AB \neq BA$.

Worked Example:

Problem: Given

and

compute the product $C = AB$.

Solution:

Step 1: Verify the dimensions for multiplication.
Matrix $A$ has dimensions $2 \times 3$. Matrix $B$ has dimensions $3 \times 2$. Since the inner dimensions (3 and 3) match, the product $AB$ is defined. The resulting matrix $C$ will have dimensions $2 \times 2$.

Step 2: Calculate the element $c_{11}$. This is the dot product of the first row of $A$ and the first column of $B$.

Step 3: Calculate the element $c_{12}$. This is the dot product of the first row of $A$ and the second column of $B$.

Step 4: Calculate the element $c_{21}$. This is the dot product of the second row of $A$ and the first column of $B$.

Step 5: Calculate the element $c_{22}$. This is the dot product of the second row of $A$ and the second column of $B$.

Step 6: Assemble the final matrix $C$.

Answer:

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2. The Trace of a Matrix

The trace is a simple yet powerful operator on square matrices that has several useful properties.

The trace possesses properties of linearity, meaning $\operatorname{tr}(A+B) = \operatorname{tr}(A) + \operatorname{tr}(B)$ and $k \cdot \operatorname{tr}(A) = k \cdot \operatorname{tr}(A)$. However, its most significant property in competitive examinations relates to the product of matrices.

Let us observe why this property holds. The $i$-th diagonal element of $AB$ is $(AB)_{ii} = \sum_{j=1}^{n} a_{ij} b_{ji}$.
Therefore,

Similarly, the $j$-th diagonal element of $BA$ is $(BA)_{jj} = \sum_{i=1}^{m} b_{ji} a_{ij}$.
Therefore,

Since scalar multiplication is commutative ($a_{ij}b_{ji} = b_{ji}a_{ij}$) and the order of finite summations can be interchanged, the two expressions are identical.

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3. Powers of a Matrix

For a square matrix $A$, we can define its positive integer powers recursively: $A^1 = A$ and $A^k = A \cdot A^{k-1}$ for $k > 1$. Computing high powers of a matrix by direct multiplication is often inefficient. The key is to compute the first few powers ($A^2, A^3, \dots$) and search for a recurring pattern.

Common patterns include:

• Idempotent Matrix: $A^2 = A$. It follows that $A^k = A$ for all $k \ge 1$.


• Nilpotent Matrix: $A^k = O$ (the zero matrix) for some integer $k$.


• Involutory Matrix: $A^2 = I$ (the identity matrix). It follows that $A^k$ cycles between $A$ (for odd $k$) and $I$ (for even $k$).


• Scalar Multiple of Identity: $A^k = cI$ for some scalar $c$ and integer $k$. This simplifies higher power calculations significantly. For instance, if $A^2 = 5I$, then $A^8 = (A^2)^4 = (5I)^4 = 5^4 I^4 = 625I$.

Worked Example:

Problem: Let

Compute $A^{101}$.

Solution:

Step 1: Compute the first few powers of $A$ to identify a pattern.

Step 2: Calculate $A^2$.

We observe that $A^2 = -I$.

Step 3: Use this pattern to simplify the calculation of $A^{101}$.

Step 4: Substitute $A^2 = -I$.

Step 5: Simplify the expression. Since $I^k = I$ for any positive integer $k$, and $(-1)^{50} = 1$.

Answer: $A^{101} = A =$

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4. Vector Operations and Orthogonality

A vector in $\mathbb{R}^n$ can be represented as an $n \times 1$ matrix (a column vector) or a $1 \times n$ matrix (a row vector). Matrix operations provide a natural framework for vector manipulations.

Two vectors are said to be orthogonal if their dot product is zero.

This principle is fundamental in constructing orthogonal bases for vector spaces and is directly tested in GATE. For example, in a 10-dimensional space, one can find at most 10 non-zero vectors that are all orthogonal to each other.

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5. Elementary Row Operations and Their Effects

Elementary row operations are transformations applied to the rows of a matrix. While their primary use is in solving linear systems and finding inverses, their effect on matrix properties like the determinant is a key concept.

| Operation | Description | Effect on Determinant |
| :--- | :--- | :--- |
| 1. Swapping | Interchange two rows, $R_i \leftrightarrow R_j$ | $\det(B) = -\det(A)$ |
| 2. Scaling | Multiply a row by a non-zero scalar, $R_i \to kR_i$ | $\det(B) = k \cdot \det(A)$ |
| 3. Addition | Add a multiple of one row to another, $R_i \to R_i + kR_j$ | $\det(B) = \det(A)$ |

A crucial consequence is that elementary row operations preserve the invertibility of a matrix. A matrix $A$ is invertible if and only if $\det(A) \neq 0$. Since these operations only multiply the determinant by non-zero scalars or leave it unchanged, if $\det(A) \neq 0$, the determinant of the resulting matrix will also be non-zero.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="NAT" question="Let matrices $P$ and $Q$ be defined as $P = \begin{pmatrix} 1 & 0 & 2 \\ 3 & -1 & 1 \end{pmatrix}$ and $Q = \begin{pmatrix} 4 & 1 \\ 0 & 2 \\ 1 & 5 \end{pmatrix}$. Calculate the $\operatorname{tr}$ of the matrix product $QP$." answer="12" hint="Recall the cyclic property of the trace. It may be easier to compute the $\operatorname{tr}$ of $PQ$ than $QP$." solution="
Step 1: State the relevant property of the trace.
The cyclic property of the trace states that for matrices $Q_{n \times m}$ and $P_{m \times n}$, we have $\operatorname{tr}(QP) = \operatorname{tr}(PQ)$.

Step 2: Determine the dimensions of the products.
$P$ is a $2 \times 3$ matrix and $Q$ is a $3 \times 2$ matrix.
The product $QP$ is a $3 \times 3$ matrix.
The product $PQ$ is a $2 \times 2$ matrix.
Calculating the trace via $PQ$ requires computing a smaller matrix.

Step 3: Calculate the matrix product $PQ$.

Step 4: Compute the trace of $PQ$.
The trace is the sum of the diagonal elements.

Result:
Since $\operatorname{tr}(QP) = \operatorname{tr}(PQ)$, the trace of $QP$ is 12.
Answer: \boxed{12}
"
:::

:::question type="MSQ" question="Let $A$ and $B$ be $n \times n$ symmetric matrices. Which of the following matrices are always symmetric?" options=["$A-B$", "$AB$", "$ABA$", "$A^2$"] answer="A-B,ABA,A^2" hint="A matrix $M$ is symmetric if $M^T = M$. Use the properties of transpose, including $(XY)^T = Y^T X^T$, and the fact that $A^T = A$ and $B^T = B$." solution="
A matrix $M$ is symmetric if $M^T = M$. We are given that $A^T = A$ and $B^T = B$. We test each option.

Option A: $A-B$
Let $M = A-B$.

Since $A$ and $B$ are symmetric, $A^T = A$ and $B^T = B$.

Thus, $A-B$ is always symmetric.

Option B: $AB$
Let $M = AB$.

Using the symmetric property, we get:

For $M$ to be symmetric, we need $M^T = M$, which means we need $BA = AB$. Since matrix multiplication is not generally commutative, this condition does not always hold. Thus, $AB$ is not always symmetric.

Option C: $ABA$
Let $M = ABA$.

Using the symmetric property, we get:

Thus, $ABA$ is always symmetric.

Option D: $A^2$
Let $M = A^2 = AA$.

Using the symmetric property, we get:

Thus, $A^2$ is always symmetric.

The correct options are $A-B$, $ABA$, and $A^2$.
"
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:::question type="MCQ" question="Consider the matrix $A = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$. Which of the following is equal to $A^{2025}$?" options=["$\begin{pmatrix} 0 & 0 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 1 & 0 \end{pmatrix}$","$\begin{pmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 1 & 0 & 0 \end{pmatrix}$","$\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \end{pmatrix}$","$\begin{pmatrix} 1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 1 & 0 & 1 \end{pmatrix}$"] answer="$\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \end{pmatrix}$" hint="Calculate the first few powers of $A$ ($A^2$, $A^3$) to find a repeating pattern. Use this pattern to simplify the high exponent." solution="
Step 1: Calculate $A^2$.

Step 2: Calculate $A^3$.

Step 3: Identify the pattern.
We observe that $A^3 = I$, the identity matrix. This means the powers of $A$ will cycle with a period of 3.

Step 4: Use the pattern to evaluate $A^{2025}$.
We can write the exponent 2025 in terms of the cycle length 3. We find the remainder of 2025 when divided by 3. The sum of digits of 2025 is $2+0+2+5=9$, which is divisible by 3. Thus, $2025 = 3k$ for some integer $k$. Specifically, $2025 = 3 \times 675$.

Step 5: Simplify the expression.

Since $A^3 = I$:

Result:

"
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:::question type="NAT" question="Given matrices $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ and $B = \begin{pmatrix} 7 & 8 \\ 9 & 1 \\ 2 & 3 \end{pmatrix}$, if $C = AB$, find the value of the element $c_{21}$." answer="85" hint="The element $c_{ij}$ of a product matrix $C = AB$ is found by taking the dot product of the $i$-th row of $A$ and the $j$-th column of $B$. You do not need to compute the entire matrix $C$." solution="
Step 1: Recall the formula for an element of a product matrix.
The element in the $i$-th row and $j$-th column of the product matrix $C = AB$ is given by $c_{ij} = \sum_{k} a_{ik} b_{kj}$. For $c_{21}$, we need the elements from the 2nd row of $A$ and the 1st column of $B$.

Step 2: Identify the 2nd row of $A$ and the 1st column of $B$.
Row 2 of $A$ is $\begin{pmatrix} 4 & 5 & 6 \end{pmatrix}$.
Column 1 of $B$ is $\begin{pmatrix} 7 \\ 9 \\ 2 \end{pmatrix}$.

Step 3: Compute the dot product.

Step 4: Perform the arithmetic.

Result:
The value of the element $c_{21}$ is 85.
Answer: $\boxed{85}$
"
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:::question type="MCQ" question="In the vector space $\mathbb{R}^5$, what is the maximum possible number of non-zero vectors in a set $\mathcal{L}$ such that for every pair of distinct vectors $u, v \in \mathcal{L}$, their dot product $u^T v$ is zero?" options=["4","5","6","25"] answer="5" hint="This question concerns the properties of mutually orthogonal vectors in an $n$-dimensional space." solution="
Step 1: Interpret the given condition.
The condition $u^T v = 0$ for any two distinct vectors $u, v$ in the set $\mathcal{L}$ means that $\mathcal{L}$ is a set of mutually orthogonal vectors.

Step 2: Recall the relationship between orthogonality and linear independence.
A fundamental theorem in linear algebra states that any set of non-zero, mutually orthogonal vectors is also linearly independent.

Step 3: Relate linear independence to the dimension of the space.
The vector space is $\mathbb{R}^5$, which has a dimension of 5. The dimension of a vector space is defined as the number of vectors in any basis for that space. It also represents the maximum number of linearly independent vectors that can exist in that space.

Step 4: Conclude the maximum size of the set $\mathcal{L}$.
Since the set $\mathcal{L}$ must be linearly independent, and the space $\mathbb{R}^5$ can contain at most 5 linearly independent vectors, the maximum possible cardinality (number of vectors) for $\mathcal{L}$ is 5.

Result:
The maximum number of such vectors is 5.
"
:::

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Summary

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What's Next?

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Part 2: Determinants

Introduction

In the study of linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix. It serves as a fundamental concept with wide-ranging applications, from solving systems of linear equations to understanding the geometric transformations represented by matrices. For a given square matrix, the determinant encodes essential information about its properties; most notably, it provides a criterion to determine if the matrix is invertible. A non-zero determinant signifies that the matrix is non-singular and that a unique inverse exists, which is a cornerstone for many computational and theoretical problems.

Our exploration of determinants will be grounded in the context of the GATE examination. We will begin by establishing the formal definition and methods of computation. Subsequently, we will delve into the critical properties of determinants, as these are frequently the basis for problem-solving in a competitive exam setting. Mastery of these properties allows for the elegant and efficient solution of complex matrix problems, often without the need for laborious direct computation. The determinant also forms a conceptual bridge to more advanced topics such as eigenvalues and eigenvectors, making its thorough understanding indispensable.

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Key Concepts

1. Calculation of Determinants

While the formula for a $2 \times 2$ matrix is straightforward, calculating determinants for higher-order matrices requires a more systematic approach. The most common method is the Laplace expansion, which relies on the concepts of minors and cofactors.

Minors and Cofactors

Consider an $n \times n$ matrix $A = [a_{ij}]$.

• The minor of the element $a_{ij}$, denoted $M_{ij}$, is the determinant of the $(n-1) \times (n-1)$ submatrix formed by deleting the $i$-th row and $j$-th column of $A$.
• The cofactor of the element $a_{ij}$, denoted $C_{ij}$, is related to the minor by the following expression:

The term $(-1)^{i+j}$ creates a "checkerboard" pattern of signs.

Laplace Expansion

The determinant of an $n \times n$ matrix $A$ can be computed by expanding along any row or any column.

• Expansion along row $i$:

• Expansion along column $j$:

Choosing a row or column with the most zeros significantly simplifies the computation.

Worked Example:

Problem: Calculate the determinant of the matrix

Solution:

We shall expand along the first row, as it contains a zero element, which simplifies the calculation.

Step 1: Apply the Laplace expansion formula for the first row ($i=1$).

Step 2: Calculate each term.

The first term involves $a_{11} = 2$. The cofactor is

The second term involves $a_{12} = 1$. The cofactor is

The third term involves $a_{13} = 0$.

Step 3: Sum the terms to find the determinant.

Step 4: Compute the final answer.

Answer: $\boxed{-1}$

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2. Properties of Determinants

For the GATE examination, understanding the properties of determinants is paramount. These properties often provide a direct path to a solution, bypassing tedious calculations. Let $A$ and $B$ be square matrices of order $n$.

Determinant of Transpose: The determinant of a matrix is equal to the determinant of its transpose.

Product Rule: The determinant of a product of matrices is the product of their individual determinants.

Power of a Matrix: A direct consequence of the product rule. For any non-negative integer $k$:

Scalar Multiplication: If a matrix $A$ is multiplied by a scalar $k$, the determinant is scaled by $k^n$.

Determinant of Inverse: The determinant of the inverse of a non-singular matrix is the reciprocal of its determinant.

This follows from $A A^{-1} = I$, so $\det(A) \det(A^{-1}) = \det(I) = 1$.

Effect of Row/Column Operations:

* Swapping two rows or two columns multiplies the determinant by $-1$.
* Multiplying a single row or column by a scalar $k$ multiplies the determinant by $k$.
* Adding a multiple of one row (or column) to another row (or column) does not change the value of the determinant. This property is extremely useful for simplifying matrices before calculation.

Zero Determinant Conditions: The determinant of a matrix is zero if:

* An entire row or column consists of zeros.
* Two rows or two columns are identical.
* One row or column is a scalar multiple of another.
* The matrix is singular (i.e., not invertible).

3. Determinant of Special Matrices

• Triangular and Diagonal Matrices: The determinant of an upper triangular, lower triangular, or diagonal matrix is simply the product of its diagonal elements.

If $A = [a_{ij}]$ is triangular, then $\det(A) = a_{11} a_{22} \dots a_{nn}$.

• Identity Matrix: The determinant of the identity matrix $I_n$ is $1$.

• Zero Matrix: The determinant of a zero matrix $O$ is $0$.

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Problem-Solving Strategies

For GATE, problems on determinants often test the application of properties rather than rote calculation.

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Common Mistakes

A clear understanding of what the properties do not state is as important as knowing what they do.

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Practice Questions

:::question type="MCQ" question="Let $A$ be a $3 \times 3$ non-singular matrix such that $A^3 = -I$, where $I$ is the identity matrix. If $B = (2A)^2$, what is the value of $\det(B)$?" options=["$4$","$8$","$16$","$64$"] answer="64" hint="Take the determinant of the given matrix equation to find $\det(A)$. Then use the properties of determinants to find $\det(B)$." solution="
Step 1: Start with the given matrix equation.

Step 2: Take the determinant of both sides.

Step 3: Apply the properties $\det(X^k) = (\det(X))^k$ and $\det(kX) = k^n \det(X)$. For a $3 \times 3$ matrix, $n=3$.

Since the determinant is a scalar, we have $\det(A) = -1$.

Step 4: Now, consider the expression for matrix $B$.

Step 5: Find the determinant of $B$.

Step 6: Apply the scalar multiplication and power properties for determinants.

Step 7: Substitute the value of $\det(A)$ found in Step 3.

Result:

The value of the determinant of $B$ is $64$.
Answer: $\boxed{64}$
"
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:::question type="NAT" question="Consider the matrix $P = \begin{pmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{pmatrix}$. Given that $a=1.5$, $b=2.5$, and $c=4.5$, the value of $\det(P)$ is:" answer="9" hint="This is a Vandermonde matrix. Its determinant has a well-known formula. Alternatively, use row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ to simplify." solution="
Step 1: The given matrix is a Vandermonde matrix.

The determinant of a $3 \times 3$ Vandermonde matrix of this form is given by the formula $\det(P) = (b-a)(c-a)(c-b)$.

Step 2: Substitute the given values $a=1.5$, $b=2.5$, and $c=4.5$ into the formula.

Step 3: Multiply these values to find the determinant.

Alternative Method (Row Operations):

Step 1: Apply row operations to introduce zeros in the first column. Let $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$. These operations do not change the determinant.

Step 2: Expand the determinant along the first column.

Step 3: Factor the terms in the second column.

Step 4: Factor out $(b-a)$ from the first row and $(c-a)$ from the second row.

Step 5: Calculate the remaining $2 \times 2$ determinant.

This confirms the formula.

Step 6: Substitute the values: $a=1.5, b=2.5, c=4.5$.

Result:

The value of $\det(P)$ is $6$.
Answer: $\boxed{6}$
"
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:::question type="MSQ" question="Let $A$ and $B$ be $n \times n$ invertible matrices, and let $k$ be a non-zero scalar. Which of the following statements are ALWAYS true?" options=["$\det(A+B) = \det(A) + \det(B)$","$\det((AB)^{-1}) = \det(A^{-1})\det(B^{-1})$","$\det(k A) = k \det(A)$","$\det(A B A^{-1}) = \det(B)$"] answer="B,D" hint="Carefully apply the fundamental properties of determinants. Test each statement for validity. For false statements, a simple counterexample is sufficient." solution="
Statement A: $\det(A+B) = \det(A) + \det(B)$.
This is a common misconception. Let $A = I_2 = <div class="math-display"><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mo fence="true">(</mo><mtable rowspacing="0.16em" columnalign="center center" columnspacing="1em"><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mn>1</mn></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="false"><mn>0</mn></mstyle></mtd></mtr><mtr><mtd><mstyle scriptlevel="0" displaystyle="false"><mn>0</mn></mstyle></mtd><mtd><mstyle scriptlevel="0" displaystyle="false"><mn>1</mn></mstyle></mtd></mtr></mtable><mo fence="true">)</mo></mrow><annotation encoding="application/x-tex">\begin{pmatrix} 1 & amp; 0 \\ 0 & amp; 1 \end{pmatrix}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:2.4em;vertical-align:-0.95em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;"><span class="delimsizing size3">(</span></span><span class="mord"><span class="mtable"><span class="col-align-c"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.45em;"><span style="top:-3.61em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">1</span></span></span><span style="top:-2.41em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">0</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.95em;"><span></span></span></span></span></span><span class="arraycolsep" style="width:0.5em;"></span><span class="arraycolsep" style="width:0.5em;"></span><span class="col-align-c"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.45em;"><span style="top:-3.61em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">0</span></span></span><span style="top:-2.41em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.95em;"><span></span></span></span></span></span></span></span><span class="mclose delimcenter" style="top:0em;"><span class="delimsizing size3">)</span></span></span></span></span></span></span></div> and$B = I_2 =

.


So


Since $4 \neq 2$, statement A is false.

Statement B: $\det((AB)^{-1}) = \det(A^{-1})\det(B^{-1})$.
Let's simplify both sides using determinant properties.
Left Hand Side (LHS):

Right Hand Side (RHS):

Since LHS = RHS, statement B is true.

Statement C: $\det(k A) = k \det(A)$.
The correct property is $\det(kA) = k^n \det(A)$, where $n$ is the order of the matrix. The statement is only true if $n=1$. For any $n > 1$ and $k \neq 1$, it is false. For example, if $A=I_2$ and $k=2$,

But

Since $4 \neq 2$, statement C is false.

Statement D: $\det(A B A^{-1}) = \det(B)$.
Using the product rule for determinants:

Since $\det(A^{-1}) = 1/\det(A)$:

As $A$ is invertible, $\det(A) \neq 0$, so we can cancel the terms.

Statement D is true. The matrix $ABA^{-1}$ is known as a similarity transformation of $B$.

Result: The correct statements are B and D.
Answer: $\boxed{\text{B, D}}$
"
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:::question type="MCQ" question="If $A$ is a $3 \times 3$ matrix with $\det(A) = 5$, what is the value of $\det(\operatorname{adj}(A))$?" options=["$5$","$1/5$","$25$","$125$"] answer="25" hint="Recall the relationship between the determinant of a matrix and the determinant of its adjoint: $\det(\operatorname{adj}(A)) = (\det(A))^{n-1}$." solution="
Step 1: State the formula for the determinant of the adjoint of a matrix.
For an $n \times n$ matrix $A$, the determinant of its adjoint is given by:

Step 2: Identify the given values from the problem statement.
The matrix $A$ is of order $n=3$.
The determinant of $A$ is $\det(A) = 5$.

Step 3: Substitute these values into the formula.

Result:
The value of $\det(\operatorname{adj}(A))$ is $25$.
Answer: $\boxed{25}$
"
:::

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Summary

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What's Next?

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Part 3: Matrix Inverse

Introduction

In the realm of linear algebra, the concept of an inverse matrix serves as a powerful analogue to the reciprocal of a scalar. Just as multiplying a number by its reciprocal yields the multiplicative identity, $1$, multiplying a matrix by its inverse yields the identity matrix, $I$. This operation is fundamental to solving systems of linear equations of the form $Ax = b$ and is pivotal in numerous applications across computer science, including computer graphics, cryptography, and network analysis.

Our study of the matrix inverse will focus on the conditions for its existence and the methods for its computation. We shall see that not all matrices possess an inverse. A matrix must be square and have a non-zero determinant to be invertible. We will systematically develop the procedure for finding the inverse using the determinant and the adjugate of a matrix, providing a robust tool for algebraic manipulation.

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Key Concepts

The computation and existence of a matrix inverse are intrinsically linked to the determinant and the adjugate of the matrix. Let us examine these components.

1. Condition for Existence

A fundamental theorem in linear algebra states that a square matrix $A$ is invertible if and only if its determinant is non-zero.

This condition serves as the first and most critical check before attempting to compute the inverse of any matrix. If the determinant is zero, our work is done; no inverse exists.

2. Formula for the Inverse

For a non-singular square matrix $A$, its inverse $A^{-1}$ is computed using its determinant and its adjugate (also known as the adjoint).

The adjugate of a matrix $A$, denoted $\operatorname{adj}(A)$, is found by taking the transpose of the matrix of cofactors, $C$. That is, $\operatorname{adj}(A) = C^T$. The element $C_{ij}$ of the cofactor matrix is given by $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

3. Inverse of a 2x2 Matrix

For a $2 \times 2$ matrix, the general formula simplifies to a very convenient shortcut, which is frequently useful in time-constrained examinations.

Worked Example:

Problem: Find the inverse of the matrix $A = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}$.

Solution:

Step 1: Calculate the determinant of $A$.

Since $\det(A) = 10 \neq 0$, the inverse exists.

Step 2: Apply the shortcut formula for a $2 \times 2$ matrix.

Step 3: Distribute the scalar term into the matrix.

Step 4: Simplify the matrix elements.

Answer: $\boxed{A^{-1} = \begin{pmatrix} 3/5 & -7/10 \\ -1/5 & 2/5 \end{pmatrix}}$

4. Inverse of a 3x3 Matrix

Calculating the inverse of a $3 \times 3$ matrix is more involved and requires a systematic application of the adjugate formula.

Worked Example:

Problem: Find the inverse of the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$.

Solution:

Step 1: Calculate the determinant of $A$. We expand along the first row.

Since $\det(A) = 1 \neq 0$, the inverse exists.

Step 2: Calculate the matrix of cofactors, $C$.

The cofactor matrix is:

Step 3: Find the adjugate of $A$ by transposing the cofactor matrix $C$.

Step 4: Apply the inverse formula $A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)$.

Answer: $\boxed{A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}}$

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="What is the inverse of the matrix $A = \begin{pmatrix} 3 & -1 \\ -4 & 2 \end{pmatrix}$?" options=["

","","","The inverse does not exist"] answer="" hint="Use the 2x2 shortcut: swap diagonal elements, negate off-diagonal elements, and divide by the determinant." solution="
Step 1: Calculate the determinant.

Since $\det(A) \neq 0$, the inverse exists.

Step 2: Apply the 2x2 inverse formula.

Step 3: Simplify the expression.

Answer: $\boxed{\begin{pmatrix} 1 & 1/2 \\ 2 & 3/2 \end{pmatrix}}$
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:::question type="NAT" question="Let $A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{pmatrix}$. If $B = A^{-1}$, what is the value of the element $B_{23}$?" answer="1" hint="The element $B_{23}$ of the inverse matrix is given by $\frac{1}{\det(A)} C_{32}$, where $C_{32}$ is the cofactor of the element at row 3, column 2 of matrix A." solution="
Step 1: Calculate the determinant of $A$.

Since $\det(A) \neq 0$, the inverse exists.

Step 2: The formula for an element of the inverse matrix is $(A^{-1})_{ij} = \frac{C_{ji}}{\det(A)}$. We need to find $B_{23}$, which is $(A^{-1})_{23}$.

Step 3: Calculate the cofactor $C_{32}$ of matrix $A$.

Step 4: Substitute the values into the formula for $B_{23}$.

Answer: $\boxed{1}$
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:::question type="MSQ" question="Let $A$ and $B$ be two invertible square matrices of the same order. Which of the following statements are ALWAYS true?" options=["$(AB)^{-1} = A^{-1}B^{-1}$","$\det(A^{-1}) = \frac{1}{\det(A)}$","$(A+B)^{-1} = A^{-1} + B^{-1}$","$(A^T)^{-1} = (A^{-1})^T$"] answer="$\det(A^{-1}) = \frac{1}{\det(A)}$,$(A^T)^{-1} = (A^{-1})^T$" hint="Consider the fundamental properties of matrix inverse, determinant, and transpose operations. Test the options with simple 2x2 matrices if unsure." solution="

• Option A: $(AB)^{-1} = A^{-1}B^{-1}$ is incorrect. The correct property is the 'reversal law' or 'socks-shoes property': $(AB)^{-1} = B^{-1}A^{-1}$.


• Option B: $\det(A^{-1}) = \frac{1}{\det(A)}$ is correct. We know that $AA^{-1} = I$. Taking the determinant of both sides gives $\det(AA^{-1}) = \det(I)$. Using the property $\det(XY) = \det(X)\det(Y)$, we get $\det(A)\det(A^{-1}) = 1$. Since $A$ is invertible, $\det(A) \neq 0$, so we can divide to get $\det(A^{-1}) = \frac{1}{\det(A)}$.


• Option C: $(A+B)^{-1} = A^{-1} + B^{-1}$ is incorrect. The inverse of a sum is not the sum of the inverses, in general. For example, let $A=I$ and $B=I$. Then $(A+B)^{-1} = (2I)^{-1} = \frac{1}{2}I$, whereas $A^{-1}+B^{-1} = I+I = 2I$.


• Option D: $(A^T)^{-1} = (A^{-1})^T$ is correct. This is a standard property of matrix inverses. We can prove it by taking the transpose of $AA^{-1} = I$, which gives $(AA^{-1})^T = I^T$. This simplifies to $(A^{-1})^T A^T = I$. By the definition of an inverse, $(A^{-1})^T$ is the inverse of $A^T$.

Answer: $\boxed{\text{Options B and D}}$
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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 non-singular $3 \times 3$ matrix with $|A|=2$. If $B = \operatorname{adj}(2A)$, what is the value of the determinant of $B$?" options=["64", "256", "1024", "4096"] answer="B" hint="Use the properties $|\operatorname{adj}(M)| = |M|^{n-1}$ and $|kM| = k^n|M|$." solution="
We are asked to find the determinant of $B = \operatorname{adj}(2A)$.
The matrix $A$ is of order $n=3$.
We will use two fundamental properties of determinants:

For a scalar $k$ and an $n \times n$ matrix $M$, $|kM| = k^n|M|$.

For an $n \times n$ matrix $M$, $|\operatorname{adj}(M)| = |M|^{n-1}$.

Let's apply the second property to our matrix $B$:

Now, we apply the first property to evaluate $|2A|$:

Given that $|A|=2$, we have:

Substituting this result back into our expression for $|B|$:

Therefore, the determinant of $B$ is 256.
Answer: $\boxed{256}$
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:::question type="NAT" question="A square matrix $A$ satisfies the equation $A^2 - 4A + 4I = 0$, where $I$ is the identity matrix. If the inverse of the matrix $A$ is given by $A^{-1} = \frac{1}{k}(A - cI)$, what is the numerical value of $k+c$?" answer="0" hint="Multiply the characteristic equation by $A^{-1}$ and rearrange the terms to match the form of the given inverse." solution="
The given matrix equation is:

Since the constant term ($4I$) is non-zero, the matrix $A$ is invertible. We can multiply the entire equation by $A^{-1}$:


Using the properties $A^{-1}A = I$ and $A^{-1}I = A^{-1}$:



Now, we rearrange the equation to solve for $A^{-1}$:


This expression is in the form $A^{-1} = \frac{1}{k}(A - cI)$. Let us rearrange our result to match this form:

Comparing $A^{-1} = \frac{1}{-4}(A - 4I)$ with $A^{-1} = \frac{1}{k}(A - cI)$, we find:
$k = -4$ and $c = 4$.
The question asks for the value of $k+c$.

Answer: $\boxed{0}$
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:::question type="MCQ" question="Let $P$ be a $3 \times 3$ skew-symmetric matrix. Which of the following statements must be true?" options=["$P$ is invertible", "$|P|=1$", "All eigenvalues of $P$ are zero", "The trace of $P$ is zero"] answer="D" hint="Recall the definition of a skew-symmetric matrix ($P^T = -P$) and its implications for the diagonal elements and the determinant." solution="
A matrix $P$ is skew-symmetric if it satisfies the property $P^T = -P$. Let the elements of $P$ be denoted by $p_{ij}$. This property means $p_{ji} = -p_{ij}$ for all $i, j$.

Diagonal Elements and Trace: For the diagonal elements, we set $i=j$. The property becomes $p_{ii} = -p_{ii}$, which implies $2p_{ii} = 0$, so $p_{ii} = 0$. Since all diagonal elements of any skew-symmetric matrix are zero, their sum, which is the trace, must also be zero.

Therefore, statement D is always true.

Determinant and Invertibility: Let's examine the determinant. We know that $|P^T| = |P|$.

For an $n \times n$ matrix, $|-P| = (-1)^n |P|$. Here, $n=3$, which is an odd number.

This implies $2|P| = 0$, which means $|P| = 0$.
Since the determinant is zero, the matrix $P$ is singular (not invertible). This makes statement A ("$P$ is invertible") and statement B ("$|P|=1$") false.

Eigenvalues: The eigenvalues of a real skew-symmetric matrix are either zero or purely imaginary. They are not all necessarily zero. For example, the matrix

is skew-symmetric. Its characteristic equation is $\lambda(\lambda^2+1)=0$, giving eigenvalues $\lambda = 0, i, -i$. Since not all eigenvalues are zero, statement C is false.

From our analysis, the only statement that must be true for any $3 \times 3$ skew-symmetric matrix is that its trace is zero.
Answer: $\boxed{\text{The trace of } P \text{ is zero}}$
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What's Next?