Determinants

This chapter rigorously introduces determinants, essential scalar values derived from square matrices, covering their definition and evaluation across various orders. Mastery of determinant calculation and properties is fundamental for solving systems of linear equations, understanding matrix invertibility, and is frequently tested in CMI examinations. These concepts form a cornerstone for advanced topics in linear algebra.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Determinants of order 2 | | 2 | Determinants of order 3 | | 3 | Evaluation techniques | | 4 | Properties of determinants |

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We begin with Determinants of order 2.

Part 1: Determinants of order 2

Determinants of Order 2

Overview

A determinant of order 2 is the simplest determinant, but it already carries important algebraic and geometric information. It tests invertibility, area scaling, row-operation effects, and parameter conditions. In exam problems, the challenge is not the formula itself, but knowing what the determinant tells you. ---

Learning Objectives

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Basic Formula

This must be memorized exactly. ---

Invertibility Criterion

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Geometric Meaning

The area of the triangle formed by the same two vectors is $\qquad \dfrac{1}{2}|ad-bc|$ ::: ---

Row and Column Effects

These are fundamental and often more useful than direct expansion. ---

Minimal Worked Examples

Example 1 Compute $\qquad \begin{vmatrix}2 & 3\\1 & 4\end{vmatrix}$ Using $ad-bc$: $\qquad 2\cdot 4 - 3\cdot 1 = 8-3=5$ --- Example 2 Find $k$ such that $\qquad \begin{vmatrix}k & 2\\3 & 6\end{vmatrix}=0$ We get $\qquad 6k-6=0$ So $\qquad k=1$ ---

Standard Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The value of $\begin{vmatrix}2 & 3\\1 & 4\end{vmatrix}$ is" options=["$5$","$11$","$-5$","$8$"] answer="A" hint="Use $ad-bc$." solution="Using the formula, $\qquad 2\cdot 4 - 3\cdot 1 = 8-3=5$ Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Find the value of $k$ for which $\begin{vmatrix}k & 2\\3 & 6\end{vmatrix}=0$." answer="1" hint="Set $6k-6=0$." solution="The determinant is $\qquad 6k-2\cdot 3 = 6k-6$ Setting this equal to zero: $\qquad 6k-6=0$ $\qquad k=1$ Hence the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following are true?" options=["Swapping two rows changes the sign of the determinant","If two rows are equal, the determinant is zero","The determinant of \begin{pmatrix}a & b\c & d\end{pmatrix} is $bc-ad$","A $2\times 2$ matrix is invertible exactly when its determinant is nonzero"] answer="A,B,D" hint="One formula is written with the wrong sign order." solution="1. True.

True.

False. The correct formula is $ad-bc$.

True.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Find all real values of $x$ for which the matrix $\begin{pmatrix}x & 1\\2 & x\end{pmatrix}$ is singular." answer="$x=\pm \sqrt{2}$" hint="Set the determinant equal to zero." solution="The matrix is singular when its determinant is zero. So $\qquad \begin{vmatrix}x & 1\\2 & x\end{vmatrix}=x^2-2$ Set this equal to zero: $\qquad x^2-2=0$ Hence $\qquad x=\pm \sqrt{2}$ Therefore the required values are $\boxed{x=\pm\sqrt{2}}$." ::: ---

Summary

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Part 2: Determinants of order 3

Determinants of Order 3

Overview

Determinants of order $3$ are the first nontrivial determinants where structure matters. Unlike order $2$, direct expansion can become messy if done carelessly. In exam problems, this topic is tested through direct evaluation, row/column observation, factor extraction, parameter values for zero determinant, and geometric interpretation such as coplanarity or area/volume-type reasoning. ---

Learning Objectives

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Core Definition

This is the standard expansion along the first row. ---

Standard Expansion Formula

The signs are: $\qquad +\quad -\quad +$ for first-row expansion. ---

Sarrus-Type Memory Aid

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When the Determinant is Zero Immediately

These are high-value observations in exam problems. ---

Expansion Strategy

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Minimal Worked Examples

Example 1 Evaluate $\qquad \begin{vmatrix} 1 & 2 & 3\\ 0 & 1 & 4\\ 5 & 6 & 0 \end{vmatrix} $ Expand along the first row: $\qquad 1\begin{vmatrix}1&4\\6&0\end{vmatrix} -2\begin{vmatrix}0&4\\5&0\end{vmatrix} +3\begin{vmatrix}0&1\\5&6\end{vmatrix} $ $\qquad =1(0-24)-2(0-20)+3(0-5) $ $\qquad =-24+40-15=1 $ So the determinant is $\boxed{1}$. --- Example 2 Evaluate $\qquad \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^2 & b^2 & c^2 \end{vmatrix} $ This is a standard Vandermonde-type determinant, and it equals $\qquad (b-a)(c-a)(c-b)$ up to sign convention. In the ordered form above it is $\qquad (b-a)(c-a)(c-b)$ This is important in factor-detection problems. ::: ---

Special Patterns

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Parameter Questions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The sign pattern in expansion along the first row of a $3\times 3$ determinant is" options=["$+,+,+$","$+,-,+$","$-,+,-$","$+,-,-$"] answer="B" hint="Recall cofactor signs." solution="The cofactor signs in the first row are $\qquad +\,-\,+$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find $\begin{vmatrix}1 & 0 & 0\\2 & 3 & 0\\4 & 5 & 6\end{vmatrix}$." answer="18" hint="Use triangular structure." solution="The matrix is lower triangular, so the determinant is the product of diagonal entries: $\qquad 1\cdot 3\cdot 6=18$ Hence the answer is $\boxed{18}$." ::: :::question type="MSQ" question="Which of the following conditions force a $3\times 3$ determinant to be zero?" options=["Two rows are equal","One row is all zero","Two columns are proportional","The diagonal entries are all nonzero"] answer="A,B,C" hint="Think about dependence of rows/columns." solution="1. True.

True.

True.

False. Nonzero diagonal entries do not by themselves force determinant zero.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Evaluate $\begin{vmatrix}1 & 2 & 3\\2 & 1 & 3\\3 & 2 & 1\end{vmatrix}$." answer="$8$" hint="Expand along the first row." solution="Expand along the first row: $\qquad \begin{vmatrix} 1&2&3\\ 2&1&3\\ 3&2&1 \end{vmatrix} = 1\begin{vmatrix}1&3\\2&1\end{vmatrix} -2\begin{vmatrix}2&3\\3&1\end{vmatrix} +3\begin{vmatrix}2&1\\3&2\end{vmatrix} $ Now compute the minors: $\qquad \begin{vmatrix}1&3\\2&1\end{vmatrix}=1-6=-5 $ $\qquad \begin{vmatrix}2&3\\3&1\end{vmatrix}=2-9=-7 $ $\qquad \begin{vmatrix}2&1\\3&2\end{vmatrix}=4-3=1 $ So $\qquad -5 -2(-7)+3(1) = -5+14+3=12$ Hence the determinant is $\boxed{12}$." ::: ---

Summary

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Part 3: Evaluation techniques

Evaluation Techniques

Overview

Many determinant questions are not meant to be expanded directly. Instead, they are designed to reward observation and transformation. Evaluation techniques include row and column operations, factor extraction, creating zeros, symmetry spotting, and reducing the determinant to a standard form. In exam problems, the best solution is often the one that avoids brute-force expansion. ---

Learning Objectives

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Main Principle

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High-Value Operations

These techniques are useful because they often preserve the determinant or change it in a controlled way. ---

Factor Extraction

This often reveals hidden polynomials or standard forms. ---

Creating Zeros

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Standard Patterns

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Minimal Worked Examples

Example 1 Evaluate $\qquad \begin{vmatrix} 1&1&1\\ 2&3&4\\ 5&7&9 \end{vmatrix} $ Use column differences: $\qquad C_2 \to C_2-C_1,\quad C_3 \to C_3-C_2$ A more systematic move is: $\qquad \begin{vmatrix} 1&1&1\\ 2&3&4\\ 5&7&9 \end{vmatrix} \overset{C_2-C_1,\ C_3-C_2}{\longrightarrow} \begin{vmatrix} 1&0&0\\ 2&1&1\\ 5&2&2 \end{vmatrix} $ Now expand along the first row: $\qquad 1\begin{vmatrix}1&1\\2&2\end{vmatrix}=0 $ So the determinant is $\boxed{0}$. --- Example 2 Evaluate $\qquad \begin{vmatrix} x&x&x\\ y&y&y\\ z&z&z \end{vmatrix} $ All columns are identical, so the determinant is immediately $\boxed{0}$. ---

Vandermonde-Type Recognition

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Parameter Evaluation

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="In determinant evaluation, the most efficient first step is often" options=["direct expansion every time","checking for simplification opportunities","multiplying all entries","changing signs randomly"] answer="B" hint="Think about why evaluation techniques exist." solution="The purpose of evaluation techniques is to simplify before expanding. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Evaluate $\begin{vmatrix}1 & 1 & 1\\2 & 2 & 2\\3 & 3 & 3\end{vmatrix}$." answer="0" hint="Look at the columns." solution="All three columns are identical, so the determinant is $\boxed{0}$." ::: :::question type="MSQ" question="Which of the following are useful evaluation techniques for determinants?" options=["Creating zeros","Extracting common factors","Recognising standard patterns","Ignoring row/column structure"] answer="A,B,C" hint="Think about what reduces computation." solution="1. True.

True.

True.

False.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Evaluate $\begin{vmatrix}1 & 1 & 1\a & b & c\a+b & b+c & c+a\end{vmatrix}$." answer="$(a-c)(b-c)$" hint="Use a column operation or first-row expansion after simplification." solution="Expand along the first row: $\qquad \begin{vmatrix} 1&1&1\\ a&b&c\\ a+b&b+c&c+a \end{vmatrix} = 1\begin{vmatrix}b&c\\ b+c&c+a\end{vmatrix} -1\begin{vmatrix}a&c\\ a+b&c+a\end{vmatrix} +1\begin{vmatrix}a&b\\ a+b&b+c\end{vmatrix} $ Now compute each minor: $\qquad b(c+a)-c(b+c)=ab-c^2$ $\qquad a(c+a)-c(a+b)=a^2-bc$ $\qquad a(b+c)-b(a+b)=ac-b^2$ So the determinant is $\qquad (ab-c^2) - (a^2-bc) + (ac-b^2)$ $\qquad = ab-c^2-a^2+bc+ac-b^2$ This can be regrouped as $\qquad -(a^2+b^2+c^2-ab-bc-ca) + 2ac$ A cleaner route is to simplify first: Take $\qquad C_3 \to C_3-C_2$ Then $\qquad C_2 \to C_2-C_1$ The determinant becomes $\qquad \begin{vmatrix} 1&0&0\\ a&b-a&c-b\\ a+b&c-a&a-b \end{vmatrix} $ Now expand along the first row: $\qquad \begin{vmatrix} b-a & c-b\\ c-a & a-b \end{vmatrix} $ This equals $\qquad (b-a)(a-b) - (c-b)(c-a) = -(b-a)^2 - (c-b)(c-a) $ Expanding and simplifying gives $\qquad ab+ac+bc-a^2-b^2-c^2 $ Hence the determinant is $\qquad \boxed{ab+ac+bc-a^2-b^2-c^2}$" ::: ---

Summary

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Part 4: Properties of determinants

Properties of Determinants

Overview

The power of determinants lies not only in computation, but in their properties. These properties explain how determinants change under row and column operations, when they become zero, and how they behave under scalar multiplication or matrix products. In exam problems, many determinant questions are solved more by properties than by expansion. ---

Learning Objectives

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Core Properties

These are the most frequently used determinant properties in exams. ---

Zero Conditions

These all express dependence. ---

Sign Change and Scaling

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Invariance Under Row Addition

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Triangular and Diagonal Cases

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Determinant of a Product

This is an important theoretical property and appears in proof-style questions. ---

Minimal Worked Examples

Example 1 If two rows of a determinant are equal, then the determinant is $0$. Reason: swapping those equal rows changes the sign, but the determinant remains the same matrix. So $\qquad \det = -\det$ hence $\qquad \det=0$ --- Example 2 Evaluate $\qquad \begin{vmatrix} 1&2&3\\ 2&4&6\\ 1&0&1 \end{vmatrix} $ The second row is $2$ times the first row, so the determinant is immediately $\boxed{0}$. ---

Linearity View

You do not always need to state linearity explicitly, but it is the idea behind many standard properties. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If two rows of a determinant are interchanged, the determinant" options=["does not change","changes sign","becomes zero","doubles"] answer="B" hint="Recall the basic row-swap property." solution="Interchanging two rows multiplies the determinant by $-1$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="If one row of a determinant is multiplied by $5$, the determinant is multiplied by" answer="5" hint="Use the scaling property." solution="Multiplying one row by $5$ multiplies the determinant by $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following are true?" options=["If two columns are equal, the determinant is zero","Adding a multiple of one row to another row changes the determinant","A triangular determinant equals the product of the diagonal entries","If a row is all zero, the determinant is zero"] answer="A,C,D" hint="Check which operations preserve the determinant." solution="1. True.

False. That operation leaves the determinant unchanged.

True.

True.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Prove that if two rows of a determinant are equal, then the determinant is zero." answer="Use the sign-change property under row interchange." hint="Swap the two equal rows." solution="Let $D$ be a determinant with two equal rows. Now interchange these two equal rows. Since the rows are equal, the numerical entries of the determinant do not change, so the determinant is still $D$. But by the row-interchange property, swapping two rows changes the sign of the determinant. Therefore the new determinant must be $-D$. Hence we have $\qquad D=-D$ So $\qquad 2D=0$ and therefore $\qquad D=0$ Thus, if two rows of a determinant are equal, the determinant must be $\boxed{0}$." ::: ---

Summary

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Given that $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = 5$, what is the value of $\begin{vmatrix} a+2d & b+2e & c+2f \\ d & e & f \\ g-d & h-e & i-f \end{vmatrix}$?" options=["-5","0","5","10"] answer="5" hint="Apply elementary row operations and their effect on the determinant value." solution="Let $D = \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$.
The given determinant is $D' = \begin{vmatrix} a+2d & b+2e & c+2f \\ d & e & f \\ g-d & h-e & i-f \end{vmatrix}$.
Apply the row operation $R_1 \to R_1 - 2R_2$: The value of the determinant remains unchanged.
$D' = \begin{vmatrix} a & b & c \\ d & e & f \\ g-d & h-e & i-f \end{vmatrix}$.
Apply the row operation $R_3 \to R_3 + R_2$: The value of the determinant remains unchanged.
$D' = \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = D$.
Since $D=5$, the value of the new determinant is 5."
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:::question type="NAT" question="Find the positive value of $x$ for which $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$." answer="6" hint="Evaluate both determinants and set them equal." solution="The left determinant is $x \cdot x - 2 \cdot 18 = x^2 - 36$.
The right determinant is $6 \cdot 6 - 2 \cdot 18 = 36 - 36 = 0$.
Setting them equal: $x^2 - 36 = 0 \implies x^2 = 36$.
Thus, $x = \pm 6$. The positive value of $x$ is 6."
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:::question type="MCQ" question="If $A$ is a $3 \times 3$ matrix such that $\det(A) = 4$, then $\det(2A)$ equals:" options=["4","8","16","32"] answer="32" hint="Recall the property $\det(kA) = k^n \det(A)$ for an $n \times n$ matrix $A$." solution="For an $n \times n$ matrix $A$ and a scalar $k$, we have $\det(kA) = k^n \det(A)$.
Here, $A$ is a $3 \times 3$ matrix, so $n=3$. The scalar is $k=2$.
Therefore, $\det(2A) = 2^3 \det(A) = 8 \times 4 = 32$."
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:::question type="NAT" question="If the points $(a, 0)$, $(0, b)$, and $(1, 1)$ are collinear, what is the value of $\frac{ab}{a+b}$? (Assume $a,b \neq 0$)" answer="1" hint="Points are collinear if the area of the triangle formed by them is zero. Use the determinant formula for area." solution="The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $\frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$.
For collinear points, the area is 0. So, $\begin{vmatrix} a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0$.
Expanding along the first row:
$a(b \cdot 1 - 1 \cdot 1) - 0(0 \cdot 1 - 1 \cdot 1) + 1(0 \cdot 1 - b \cdot 1) = 0$
$a(b-1) - 0 + 1(-b) = 0$
$ab - a - b = 0$
$ab = a + b$.
Since $a,b \neq 0$, we can divide by $a+b$ (which also implies $a+b \neq 0$ if $ab \neq 0$) and $ab$:
$\frac{ab}{a+b} = 1$."
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What's Next?