Chapter 4: Determinants
Complete Board Exam Focused Notes with Applications and Matrix Methods
Introduction to Determinants
What is a Determinant?
To every square matrix $A = [a_{ij}]$ of order $n$, we can associate a number (real or complex) called the determinant of the matrix A.
It is denoted by $|A|$ or $\det(A)$ or $\Delta$
To every square matrix $A = [a_{ij}]$ of order $n$, we can associate a number (real or complex) called the determinant of the matrix A.
It is denoted by $|A|$ or $\det(A)$ or $\Delta$
Connection with System of Equations
Consider the system of linear equations:
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
This can be represented as: $\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$
The number $a_1b_2 - a_2b_1$ determines whether the system has a unique solution.
Key Point: If $a_1b_2 - a_2b_1 \neq 0$, then the system has a unique solution.
This number is the determinant of matrix $A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}$
This number is the determinant of matrix $A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}$
Important Notes:
• For matrix A, $|A|$ is read as "determinant of A" and NOT "modulus of A"
• Only square matrices have determinants
• Determinants are numbers (scalars), not matrices
• For matrix A, $|A|$ is read as "determinant of A" and NOT "modulus of A"
• Only square matrices have determinants
• Determinants are numbers (scalars), not matrices
Determinant of Order 1 and 2
Order 1
If $A = [a]$ is a matrix of order 1, then:
$|A| = a$
$|A| = a$
Order 2 (Most Important!) 🔥🔥🔥
If $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$, then:
$|A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{21}a_{12}$
Remember: Product of main diagonal minus product of other diagonal
$|A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{21}a_{12}$
Remember: Product of main diagonal minus product of other diagonal
Example 1: Evaluate $\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix}$
Solution:
$\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = 2(2) - 4(-1) = 4 + 4 = 8$
Solution:
$\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = 2(2) - 4(-1) = 4 + 4 = 8$
Example 2: Evaluate $\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix}$
Solution:
$= x(x) - (x+1)(x-1)$
$= x^2 - (x^2 - 1)$
$= x^2 - x^2 + 1 = 1$
Solution:
$= x(x) - (x+1)(x-1)$
$= x^2 - (x^2 - 1)$
$= x^2 - x^2 + 1 = 1$
Quick Memory Trick: For $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, think "ad minus bc"
Main diagonal product MINUS other diagonal product
Main diagonal product MINUS other diagonal product
Determinant of Order 3 × 3 ⭐⭐⭐⭐⭐
A determinant of order 3 can be expanded along any row or column.
Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$
Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$
Expansion Along First Row (R₁)
$|A| = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$
= $a_{11}(a_{22}a_{33} - a_{32}a_{23}) - a_{12}(a_{21}a_{33} - a_{31}a_{23}) + a_{13}(a_{21}a_{32} - a_{31}a_{22})$
= $a_{11}(a_{22}a_{33} - a_{32}a_{23}) - a_{12}(a_{21}a_{33} - a_{31}a_{23}) + a_{13}(a_{21}a_{32} - a_{31}a_{22})$
Sign Pattern
The sign pattern for a 3×3 determinant is:
$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$
Multiply element by $(-1)^{i+j}$ where $i$ = row number, $j$ = column number
Important Points:
• Expanding along any row or column gives the same value
• For easier calculation, expand along the row/column with maximum zeros
• Instead of $(-1)^{i+j}$, use the sign pattern: +, -, +, -, +, ...
• Expanding along any row or column gives the same value
• For easier calculation, expand along the row/column with maximum zeros
• Instead of $(-1)^{i+j}$, use the sign pattern: +, -, +, -, +, ...
Example: Evaluate $\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$
Solution: Expanding along C₃ (it has two zeros):
$\Delta = 4\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} - 0 + 0$
$= 4[(-1)(1) - (3)(4)]$
$= 4(-1 - 12) = 4(-13) = -52$
Solution: Expanding along C₃ (it has two zeros):
$\Delta = 4\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} - 0 + 0$
$= 4[(-1)(1) - (3)(4)]$
$= 4(-1 - 12) = 4(-13) = -52$
Area of a Triangle Using Determinants 🎯
Formula: The area of triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:
$\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$
$= \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$
$= \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Important Notes:
• Area is always positive, so take absolute value
• If area is given, use both ± values of determinant
• If three points are collinear, area = 0
• Area is always positive, so take absolute value
• If area is given, use both ± values of determinant
• If three points are collinear, area = 0
Example: Find the area of triangle with vertices $(3, 8)$, $(-4, 2)$, and $(5, 1)$
Solution:
$\Delta = \frac{1}{2}\begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix}$
Expanding along R₁:
$= \frac{1}{2}[3(2-1) - 8(-4-5) + 1(-4-10)]$
$= \frac{1}{2}[3(1) - 8(-9) + 1(-14)]$
$= \frac{1}{2}[3 + 72 - 14]$
$= \frac{1}{2}(61) = \frac{61}{2}$ square units
Solution:
$\Delta = \frac{1}{2}\begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix}$
Expanding along R₁:
$= \frac{1}{2}[3(2-1) - 8(-4-5) + 1(-4-10)]$
$= \frac{1}{2}[3(1) - 8(-9) + 1(-14)]$
$= \frac{1}{2}[3 + 72 - 14]$
$= \frac{1}{2}(61) = \frac{61}{2}$ square units
Collinearity Test: Three points $A$, $B$, $C$ are collinear if and only if:
$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$
$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$
Minors and Cofactors 🔥🔥🔥
Minor of an Element
Definition: The minor of an element $a_{ij}$ is the determinant obtained by deleting the $i^{th}$ row and $j^{th}$ column.
It is denoted by $M_{ij}$
It is denoted by $M_{ij}$
Cofactor of an Element
Definition: The cofactor of an element $a_{ij}$ is:
$A_{ij} = (-1)^{i+j}M_{ij}$
where $M_{ij}$ is the minor of $a_{ij}$
$A_{ij} = (-1)^{i+j}M_{ij}$
where $M_{ij}$ is the minor of $a_{ij}$
Example: Find minor and cofactor of element 6 in $\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$
Solution: Element 6 is at position (2,3), i.e., $a_{23} = 6$
Minor: Delete R₂ and C₃:
$M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = 8 - 14 = -6$
Cofactor:
$A_{23} = (-1)^{2+3}M_{23} = (-1)^5(-6) = -(-6) = 6$
Solution: Element 6 is at position (2,3), i.e., $a_{23} = 6$
Minor: Delete R₂ and C₃:
$M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = 8 - 14 = -6$
Cofactor:
$A_{23} = (-1)^{2+3}M_{23} = (-1)^5(-6) = -(-6) = 6$
Important Formula
The value of determinant can be calculated as:
$|A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$ (along R₁)
= Sum of (elements × their cofactors) along any row or column
$|A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$ (along R₁)
= Sum of (elements × their cofactors) along any row or column
Key Property: If elements of one row (or column) are multiplied with cofactors of another row (or column), their sum is zero.
Example: $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$
Example: $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$
Adjoint of a Matrix ⭐⭐⭐⭐⭐
Definition: The adjoint of a square matrix $A = [a_{ij}]_{n \times n}$ is the transpose of the matrix of cofactors.
If $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$
Then $\text{adj}(A) = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$
(Note: Transpose of cofactor matrix!)
If $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$
Then $\text{adj}(A) = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$
(Note: Transpose of cofactor matrix!)
For 2×2 Matrix (Quick Method)
For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$
$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$
Steps:
1. Interchange $a$ and $d$
2. Change signs of $b$ and $c$
$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$
Steps:
1. Interchange $a$ and $d$
2. Change signs of $b$ and $c$
Example: Find adj(A) for $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$
Solution: Using the quick method:
$\text{adj}(A) = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$
Solution: Using the quick method:
$\text{adj}(A) = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$
Important Theorem
For any square matrix A of order n:
$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| \cdot I$
where $I$ is the identity matrix of order n
$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| \cdot I$
where $I$ is the identity matrix of order n
Key Result: For a square matrix of order n:
$|\text{adj}(A)| = |A|^{n-1}$
$|\text{adj}(A)| = |A|^{n-1}$
Inverse of a Matrix 🎯🎯🎯
Singular and Non-Singular Matrices
Singular Matrix: A square matrix A is singular if $|A| = 0$
Non-Singular Matrix: A square matrix A is non-singular if $|A| \neq 0$
Non-Singular Matrix: A square matrix A is non-singular if $|A| \neq 0$
Inverse of a Matrix
A square matrix A is invertible if and only if A is non-singular.
Formula:
$A^{-1} = \frac{1}{|A|} \text{adj}(A)$, provided $|A| \neq 0$
Formula:
$A^{-1} = \frac{1}{|A|} \text{adj}(A)$, provided $|A| \neq 0$
Important Points:
• If $|A| = 0$, then $A^{-1}$ does not exist
• If $A^{-1}$ exists, it is unique
• $A \cdot A^{-1} = A^{-1} \cdot A = I$
• $(A^{-1})^{-1} = A$
• $(AB)^{-1} = B^{-1}A^{-1}$ (order reversed!)
• If $|A| = 0$, then $A^{-1}$ does not exist
• If $A^{-1}$ exists, it is unique
• $A \cdot A^{-1} = A^{-1} \cdot A = I$
• $(A^{-1})^{-1} = A$
• $(AB)^{-1} = B^{-1}A^{-1}$ (order reversed!)
Example: Find $A^{-1}$ if $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$
Solution:
Step 1: Find $|A| = 2(4) - 3(1) = 8 - 3 = 5 \neq 0$
So $A^{-1}$ exists.
Step 2: Find adj(A) = $\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$
Step 3: $A^{-1} = \frac{1}{5}\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{bmatrix}$
Solution:
Step 1: Find $|A| = 2(4) - 3(1) = 8 - 3 = 5 \neq 0$
So $A^{-1}$ exists.
Step 2: Find adj(A) = $\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$
Step 3: $A^{-1} = \frac{1}{5}\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{bmatrix}$
Properties of Inverse
1. $|A^{-1}| = \frac{1}{|A|}$ or $|A^{-1}| = |A|^{-1}$
2. If A and B are non-singular matrices of same order, then:
$(AB)^{-1} = B^{-1}A^{-1}$
3. $(A')^{-1} = (A^{-1})'$
2. If A and B are non-singular matrices of same order, then:
$(AB)^{-1} = B^{-1}A^{-1}$
3. $(A')^{-1} = (A^{-1})'$
System of Linear Equations (Matrix Method) 🔥🔥🔥
Consider the system of equations:
$a_1x + b_1y + c_1z = d_1$
$a_2x + b_2y + c_2z = d_2$
$a_3x + b_3y + c_3z = d_3$
This can be written as $AX = B$, where:
$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$
$a_1x + b_1y + c_1z = d_1$
$a_2x + b_2y + c_2z = d_2$
$a_3x + b_3y + c_3z = d_3$
This can be written as $AX = B$, where:
$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$
Solution Methods
Case 1: If $|A| \neq 0$ (Non-singular)
The system has a unique solution:
$X = A^{-1}B$
Steps to solve:
1. Calculate $|A|$
2. Find adj(A)
3. Calculate $A^{-1} = \frac{1}{|A|}\text{adj}(A)$
4. Find $X = A^{-1}B$
The system has a unique solution:
$X = A^{-1}B$
Steps to solve:
1. Calculate $|A|$
2. Find adj(A)
3. Calculate $A^{-1} = \frac{1}{|A|}\text{adj}(A)$
4. Find $X = A^{-1}B$
Case 2: If $|A| = 0$ (Singular)
• If $(\text{adj }A) \cdot B \neq O$: No solution (Inconsistent)
• If $(\text{adj }A) \cdot B = O$: Infinitely many solutions or no solution
• If $(\text{adj }A) \cdot B \neq O$: No solution (Inconsistent)
• If $(\text{adj }A) \cdot B = O$: Infinitely many solutions or no solution
Consistency of System:
• Consistent: Solution exists (one or more)
• Inconsistent: No solution exists
• Consistent: Solution exists (one or more)
• Inconsistent: No solution exists
Example: Solve using matrix method:
$2x + 5y = 1$
$3x + 2y = 7$
Solution: $AX = B$ where $A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}$
$|A| = 4 - 15 = -11 \neq 0$
$\text{adj}(A) = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$
$A^{-1} = \frac{-1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$
$X = A^{-1}B = \frac{-1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 7 \end{bmatrix} = \frac{-1}{11}\begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$
Therefore: $x = 3$, $y = -1$
$2x + 5y = 1$
$3x + 2y = 7$
Solution: $AX = B$ where $A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}$
$|A| = 4 - 15 = -11 \neq 0$
$\text{adj}(A) = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$
$A^{-1} = \frac{-1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$
$X = A^{-1}B = \frac{-1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 7 \end{bmatrix} = \frac{-1}{11}\begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$
Therefore: $x = 3$, $y = -1$
Solved Examples (Board Pattern)
Example 1: Find minors and cofactors of all elements of $\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}$
Solution:
$M_{11} = 3$, $M_{12} = 4$, $M_{21} = -2$, $M_{22} = 1$
$A_{11} = (-1)^{1+1}(3) = 3$
$A_{12} = (-1)^{1+2}(4) = -4$
$A_{21} = (-1)^{2+1}(-2) = 2$
$A_{22} = (-1)^{2+2}(1) = 1$
Example 2: If $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$, verify that $A(\text{adj }A) = |A|I$
Solution:
$|A| = 1(16-9) - 3(4-3) + 3(3-4) = 7 - 3 - 3 = 1 \neq 0$
Finding cofactors:
$A_{11} = 7$, $A_{12} = -1$, $A_{13} = -1$
$A_{21} = -3$, $A_{22} = 1$, $A_{23} = 0$
$A_{31} = -3$, $A_{32} = 0$, $A_{33} = 1$
$\text{adj}(A) = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
$A \cdot \text{adj}(A) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I = |A| \cdot I$
Hence verified.
Example 3: Solve the system: $3x - 2y + 3z = 8$, $2x + y - z = 1$, $4x - 3y + 2z = 4$
Solution: $AX = B$ where:
$A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$
$|A| = 3(2-3) + 2(4+4) + 3(-6-4) = -3 + 16 - 30 = -17 \neq 0$
Finding adj(A) and then $A^{-1} = \frac{1}{-17}\text{adj}(A)$
$X = A^{-1}B = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$
Answer: $x = 1$, $y = 2$, $z = 3$
Formula Sheet (Quick Revision) 📋
Determinants
1. Order 2: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$2. Order 3: Expand along any row or column
3. Area of triangle: $\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$
Minors and Cofactors
4. Cofactor: $A_{ij} = (-1)^{i+j}M_{ij}$5. Determinant: $|A| = \sum a_{ij}A_{ij}$ (along any row/column)
6. $a_{i1}A_{j1} + a_{i2}A_{j2} + a_{i3}A_{j3} = 0$ (if $i \neq j$)
Adjoint and Inverse
7. adj(A) = Transpose of cofactor matrix8. $A \cdot \text{adj}(A) = |A| \cdot I$
9. $A^{-1} = \frac{1}{|A|}\text{adj}(A)$ (if $|A| \neq 0$)
10. $(AB)^{-1} = B^{-1}A^{-1}$
11. $|A^{-1}| = \frac{1}{|A|}$
12. $|\text{adj}(A)| = |A|^{n-1}$ (for n×n matrix)
System of Equations
13. $AX = B \Rightarrow X = A^{-1}B$ (if $|A| \neq 0$)14. Unique solution exists if $|A| \neq 0$
Common Mistakes & Exam Tips
Common Mistakes 🚨
Mistake 1: Forgetting to take absolute value when finding area of triangle.
Mistake 2: Wrong sign pattern in 3×3 determinant expansion. Remember: +, -, +, -, +, ...
Mistake 3: Confusing adj(A) with transpose. Adjoint is transpose of COFACTOR matrix, not the matrix itself!
Mistake 4: Calculating $A^{-1}$ when $|A| = 0$. Always check if $|A| \neq 0$ first!
Mistake 5: Writing $(AB)^{-1} = A^{-1}B^{-1}$. Correct is $(AB)^{-1} = B^{-1}A^{-1}$ (order reversed!)
Exam Tips 🏆
Tip 1: For 3×3 determinants, always expand along the row/column with most zeros to save time.
Tip 2: For 2×2 inverse, use the quick formula: interchange diagonal, change signs of off-diagonal, divide by determinant.
Tip 3: In system of equations, first check if $|A| \neq 0$ before attempting to find inverse.
Tip 4: Always verify your answer by substituting back into original equations when solving systems.
Tip 5: Show all steps clearly - write "Finding $|A|$", "Finding adj(A)", "Finding $A^{-1}$" for full marks.
Tip 6: For area problems, write formula first, then substitute values for step-wise marking.
Summary
• Determinant is a number associated with every square matrix
• Only square matrices have determinants
• For 2×2 matrix: $|A| = ad - bc$
• For 3×3 matrix: Expand along any row or column
• Area of triangle = $\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$
• Minor $M_{ij}$ = determinant after deleting $i^{th}$ row and $j^{th}$ column
• Cofactor $A_{ij} = (-1)^{i+j}M_{ij}$
• Adjoint = Transpose of cofactor matrix
• $A \cdot \text{adj}(A) = |A| \cdot I$
• Singular matrix: $|A| = 0$; Non-singular: $|A| \neq 0$
• $A^{-1} = \frac{1}{|A|}\text{adj}(A)$ (exists only if $|A| \neq 0$)
• System $AX = B$ has unique solution $X = A^{-1}B$ if $|A| \neq 0$
• Consistent system: solution exists; Inconsistent: no solution
• $(AB)^{-1} = B^{-1}A^{-1}$
• $|A^{-1}| = \frac{1}{|A|}$
• $|\text{adj}(A)| = |A|^{n-1}$ for n×n matrix