Martices and Determinants | NEB Class 11 Mathematics Notes

Chapter Overview

Welcome, students, to the world of Matrices and Determinants. The term 'matrix' was first introduced by James Joseph Sylvester in 1857, while the theory was further developed by Arthur Cayley. Although the concept of determinants dates back to Leibniz in 1693, today these two are inseparable tools in higher mathematics. Matrices are not just numbers in brackets—they are a powerful language used to handle large amounts of data simultaneously, solve complex systems of equations, and model real-world phenomena.

What this chapter covers

Matrix Definition, Notation, and ClassificationTypes of Matrices: Row, Column, Square, Diagonal, Identity, Symmetric, Skew-symmetricMatrix Operations: Addition, Subtraction, Scalar MultiplicationMatrix Multiplication and Conformability ConditionsTranspose of a Matrix and Its PropertiesDecomposition of Any Square Matrix into Symmetric and Skew-symmetric PartsDeterminants of Order 2 and Order 3Methods of Evaluation: Expansion by Minors, Sarrus RuleProperties of Determinants and Their ApplicationsCofactors, Adjoint, and Matrix InverseSolving Systems of Linear Equations Using Matrices

Learning Objectives

1

Identify different types of matrices (row, column, square, diagonal, identity, symmetric, skew-symmetric) and determine their order.

2

Perform basic matrix operations: addition, subtraction, scalar multiplication, and understand when operations are permissible.

3

Multiply two matrices and understand the conformability condition: number of columns in first matrix = number of rows in second matrix.

4

Find the transpose of a matrix and verify properties such as $(AB)^T = B^T A^T$ and $(A^T)^T = A$.

5

Express any square matrix as a sum of a symmetric matrix and a skew-symmetric matrix using the formula $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$.

6

Evaluate determinants of $2 \times 2$ matrices using the formula $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.

7

Evaluate determinants of $3 \times 3$ matrices using cofactor expansion and the Sarrus Rule (diagonal method).

8

Understand and apply the sign pattern for cofactors: $(-1)^{i+j}$ where $i$ is the row number and $j$ is the column number.

9

Apply determinant properties to simplify calculations: properties regarding identical rows/columns, row operations, scalar multiplication.

10

Find the adjoint (adjugate) of a square matrix by calculating all cofactors and transposing the cofactor matrix.

11

Determine if a matrix is singular (non-invertible, $|A| = 0$) or non-singular (invertible, $|A| \neq 0$).

12

Calculate the inverse of a matrix using the formula $A^{-1} = \frac{1}{|A|} Adj(A)$ when $|A| \neq 0$.

13

Use matrix methods to solve systems of linear equations $AX = B$ by finding $X = A^{-1}B$.

Why Matrices and Determinants Matter: From Data to Solutions

Key Insight

Imagine you're a computer scientist processing millions of pixels in an image, an engineer analyzing stresses in a building, or an economist tracking market flows across countries. How would you organize and manipulate such massive amounts of data efficiently? The answer: Matrices.

Real-World Scenario: A school needs to track student marks in 5 subjects across 100 students. Writing this as a table (matrix) of $100 \times 5$ allows us to: (1) Calculate class averages using matrix operations, (2) Find relationships between subjects using multiplication, (3) Invert the data for specific analyses.

The Role of Determinants: Every square matrix has a special number called its determinant. This single number tells us whether the matrix can be inverted (solve equations?), whether the system has a unique solution, and even the volume of transformations in 3D space. A determinant of zero is like a 'red flag'—it means the system has no unique solution.

The Bridge to Higher Mathematics: Matrices and determinants are the foundation for linear algebra, which appears in every advanced course—calculus, statistics, computer graphics, quantum physics, and machine learning. Master this chapter now, and you unlock the door to countless applications.

Pay special attention to the singular matrix concept ($|A| = 0$). This single concept determines whether a system of equations has a unique solution or not. It appears in 40% of NEB exam questions on matrices.

Key Definitions

Matrix

A matrix is a rectangular arrangement (array) of numbers, symbols, or expressions arranged in horizontal rows and vertical columns, all enclosed within brackets or parentheses. A matrix is used to organize and manipulate data systematically. Notation: $A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}$

Elements (Entries)

The individual numbers or expressions that form the matrix. Each element is denoted $a_{ij}$, where $i$ is the row number and $j$ is the column number. For example, $a_{23}$ is the element in the 2nd row and 3rd column.

Order (or Size) of a Matrix

If a matrix has $m$ rows and $n$ columns, its order is denoted $m \times n$ (read as '$m$ by $n$'). The total number of elements in the matrix is $m \times n$. Example: A $3 \times 4$ matrix has 12 elements arranged in 3 rows and 4 columns.

Row Matrix

A matrix with only one row and any number of columns. Order: $1 \times n$. Example: $\begin{bmatrix} 2 & 3 & 5 & 7 \end{bmatrix}$ is a $1 \times 4$ row matrix.

Column Matrix

A matrix with only one column and any number of rows. Order: $m \times 1$. Example: $\begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}$ is a $3 \times 1$ column matrix.

Square Matrix

A matrix where the number of rows equals the number of columns ($m = n$). The diagonal elements (where $i = j$) are called the principal diagonal. Examples: $2 \times 2$, $3 \times 3$, $n \times n$ matrices. Only square matrices have determinants.

Diagonal Matrix

A square matrix where all non-diagonal elements are zero. Only the principal diagonal elements (where $i = j$) can be non-zero. Example: $\begin{bmatrix} 3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

Scalar Matrix

A diagonal matrix where all principal diagonal elements are equal to the same non-zero scalar $k$. Example: $\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ is a scalar matrix with scalar 3.

Identity (Unit) Matrix

A diagonal matrix where all principal diagonal elements equal 1 and all other elements are 0. Denoted by $I$ or $I_n$ (where $n$ is the order). Example: $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Property: $AI = IA = A$ for any matrix $A$.

Null (Zero) Matrix

A matrix where all elements are zero. Denoted by $O$ or $0_{m \times n}$. Example: $O_{2 \times 3} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$. Property: $A + O = A$ and $AO = O$.

Transpose of a Matrix

The transpose of a matrix $A$, denoted $A^T$, is obtained by interchanging its rows and columns. If $A$ is $m \times n$, then $A^T$ is $n \times m$. Example: If $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$, then $A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$

Symmetric Matrix

A square matrix $A$ is symmetric if $A^T = A$. This means $a_{ij} = a_{ji}$ for all $i, j$ (elements are symmetric about the principal diagonal). Example: $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix}$ is symmetric because it reads the same along the diagonal in both directions.

Skew-Symmetric Matrix

A square matrix $A$ is skew-symmetric if $A^T = -A$. This means $a_{ij} = -a_{ji}$ for all $i, j$, and all principal diagonal elements must be zero ($a_{ii} = 0$). Example: $\begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 5 \\ 3 & -5 & 0 \end{bmatrix}$ is skew-symmetric.

Determinant of a Matrix

A determinant is a special scalar value computed from a square matrix. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad - bc$. For larger matrices, determinants are calculated using cofactor expansion or other methods. Notation: $|A|$ or $\det(A)$.

Singular Matrix

A square matrix whose determinant is zero ($|A| = 0$). A singular matrix does not have an inverse and cannot be used to solve certain systems of linear equations uniquely. Such matrices are also called non-invertible.

Non-Singular Matrix

A square matrix whose determinant is non-zero ($|A| \neq 0$). A non-singular matrix has an inverse, denoted $A^{-1}$, and can be used to solve systems of linear equations uniquely. Such matrices are also called invertible.

Minor of an Element

The minor $M_{ij}$ of element $a_{ij}$ is the determinant of the matrix obtained by deleting row $i$ and column $j$. Example: In $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$, the minor $M_{11}$ is $\begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = 45 - 48 = -3$

Cofactor of an Element

The cofactor $A_{ij}$ (or $C_{ij}$) of element $a_{ij}$ is: $A_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor. The factor $(-1)^{i+j}$ provides the correct sign based on the position. If $i + j$ is even, the sign is positive; if odd, the sign is negative.

Adjoint (Adjugate) of a Matrix

The adjoint of a square matrix $A$, denoted $Adj(A)$ or $A^*$, is the transpose of the matrix of all cofactors. That is, $Adj(A) = [C_{ij}]^T$ where $C_{ij}$ is the cofactor of element $a_{ij}$. The adjoint is crucial for finding the inverse: $A^{-1} = \frac{1}{|A|} Adj(A)$.

Inverse of a Matrix

The inverse of a non-singular square matrix $A$, denoted $A^{-1}$, is defined such that: $AA^{-1} = A^{-1}A = I$. Formula: $A^{-1} = \frac{1}{|A|} Adj(A)$, provided $|A| \neq 0$. If $|A| = 0$, the inverse does not exist.

Symmetric vs Skew-Symmetric Matrices: The Complete Comparison

Symmetric and skew-symmetric matrices are two important classes of square matrices, each with unique properties and applications. Understanding their differences is essential for decomposing any square matrix.

Symmetric Matrix

$$$A^T = A$ (i.e., $a_{ij} = a_{ji}$)$$

Definition: A square matrix is symmetric if its transpose equals itself. The matrix reads identically when flipped along the main diagonal.

Diagonal Elements: Can be any real number (no restriction). The diagonal reflects itself.

Examples: $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix}$ and identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Visual Property: The matrix looks like a mirror image across the principal diagonal.

Real-life Use: Covariance matrices in statistics (always symmetric), structural matrices in engineering.

Applications: Quadratic forms, eigenvalue problems, least squares regression

VS

Skew-Symmetric Matrix

$$$A^T = -A$ (i.e., $a_{ij} = -a_{ji}$)$$

Definition: A square matrix is skew-symmetric if its transpose equals the negative of itself. Each element in position $(i,j)$ is the negative of the element in position $(j,i)$.

Diagonal Elements: MUST all be zero. If $a_{ii} = a_{ii}$, then $-a_{ii} = a_{ii}$ implies $a_{ii} = 0$.

Examples: $\begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 5 \\ 3 & -5 & 0 \end{bmatrix}$ (note all zeros on diagonal)

Visual Property: The matrix is antisymmetric—elements above the diagonal are negatives of elements below.

Real-life Use: Rotation matrices in 3D graphics, angular momentum in physics, cross-product operations.

Applications: Lie algebras, rotational transformations, electromagnetic field tensors

Quick Memory Aid: 'Symmetric = Same' (matrix equals its transpose). 'Skew = Swap and Negate' (transpose gives negatives). Every square matrix can be uniquely written as: $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$ = Symmetric + Skew-symmetric.

Complete Formula Sheet

Order of a Matrix

\text{Order} = m \times n \quad (m \text{ rows}, n \text{ columns})

The order completely describes the dimensions of the matrix. Important for determining whether operations (like multiplication) are possible.

Matrix Equality

A = B \text{ if and only if } a_{ij} = b_{ij} \text{ for all } i, j

Two matrices are equal only if they have the same order and all corresponding elements are equal.

Transpose of a Matrix

\text{If } A \text{ is } m \times n, \text{ then } A^T \text{ is } n \times m, \text{ where } (A^T)_{ij} = a_{ji}

Transposition swaps rows and columns. Property: $(A^T)^T = A$ and $(AB)^T = B^T A^T$ (reversal law).

Conditions for Matrix Addition/Subtraction

\text{Only possible if both matrices have the same order } m \times n

Cannot add a $2 \times 3$ matrix to a $3 \times 2$ matrix. When valid: $(A \pm B)_{ij} = a_{ij} \pm b_{ij}$

Scalar Multiplication

kA = [ka_{ij}] \text{ for scalar } k

Multiply every element of the matrix by the scalar. Example: $2 \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}$

Matrix Multiplication Conformability

\text{If } A \text{ is } m \times p \text{ and } B \text{ is } p \times n, \text{ then } AB \text{ is } m \times n

The number of columns in $A$ MUST equal the number of rows in $B$. The resulting matrix has rows from $A$ and columns from $B$. Note: $AB \neq BA$ in general!

Matrix Multiplication Formula

(AB)_{ij} = \sum_{k=1}^{p} a_{ik}b_{kj}

To find element $(i,j)$ in the product, multiply elements of row $i$ from $A$ with corresponding elements of column $j$ from $B$, then sum.

Properties of Matrix Operations

\begin{align} (AB)C &= A(BC) \quad \text{(Associative)} \\ A(B+C) &= AB + AC \quad \text{(Distributive)} \\ AI &= IA = A \quad \text{(Identity)} \end{align}

These properties allow us to manipulate matrices similar to numbers, with the crucial exception that multiplication is NOT commutative.

Determinant of a 2×2 Matrix

\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

Multiply diagonal elements, subtract the product of off-diagonal elements. This is the most basic determinant formula.

Determinant of a 3×3 Matrix (Cofactor Expansion)

|A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}

Expand along any row or column. Choose a row/column with zeros to simplify. The signs follow the checkerboard pattern: $+ - + \quad - + - \quad + - +$

Minor and Cofactor

M_{ij} = \text{determinant after deleting row } i \text{ and column } j \quad ; \quad A_{ij} = (-1)^{i+j} M_{ij}

The cofactor includes the sign factor $(-1)^{i+j}$. If $i+j$ is even, cofactor is positive; if odd, negative.

Adjoint (Adjugate) of a Matrix

Adj(A) = [A_{ij}]^T \quad \text{(Transpose of cofactor matrix)}

Calculate all cofactors, arrange in a matrix, then transpose. The adjoint is used to find the inverse.

Inverse of a Matrix

A^{-1} = \frac{1}{|A|} Adj(A) \quad \text{(provided } |A| \neq 0\text{)}

The inverse exists only for non-singular matrices ($|A| \neq 0$). Property: $AA^{-1} = A^{-1}A = I$.

Symmetric and Skew-Symmetric Decomposition

A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)

Every square matrix can be uniquely written as the sum of a symmetric part and a skew-symmetric part. The first term is symmetric, the second is skew-symmetric.

Determinant Properties: Scalar Multiplication

|kA| = k^n |A| \quad \text{(for an } n \times n \text{ matrix)}

Multiplying all elements by scalar $k$ multiplies the determinant by $k^n$, NOT just $k$. Example: For a $3 \times 3$ matrix, $|2A| = 2^3 |A| = 8|A|$

Determinant Properties: Row/Column Operations

\begin{align} |A| &= 0 \text{ if two rows (or columns) are identical} \\ |A| &\to -|A| \text{ if two rows (or columns) are swapped} \\ |A| &\text{ unchanged if row operations: } R_i \to R_i + kR_j \end{align}

These properties allow us to simplify determinants without full expansion. Row operations of type 'add multiple of one row to another' preserve the determinant.

Product of Determinants

|AB| = |A| \cdot |B|

The determinant of a product equals the product of determinants. This property is very useful for verification and simplification.

Determinant of Transpose

$$|A^T| = |A|$$

Transposing a matrix does not change its determinant. This is why row and column operations have the same effect on determinants.

Sarrus Rule (for 3×3 matrices only)

\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = aei + bfg + cdh - ceg - afh - bdi

Also called the 'diagonal method'. Write the first two columns again, then add products of downward diagonals and subtract products of upward diagonals. Only works for $3 \times 3$ matrices!

Property: Determinant and Adjoint

A(Adj A) = (Adj A) A = |A| \cdot I

Multiplying a matrix by its adjoint (in either order) gives the determinant times the identity. This is the foundation for the inverse formula.

Problem-Solving Decision Tree for Matrices and Determinants

When tackling matrix and determinant problems, follow this systematic approach to identify the method and solve efficiently.

Is the problem about matrix operations (addition, subtraction, multiplication)?

Yes

Check if matrices have compatible orders. For addition/subtraction: same order required. For multiplication: columns of first = rows of second. Perform element-wise operations carefully.

No

Move to next decision about matrix properties or determinants.

Does the problem ask about transpose, symmetry, or decomposition?

Yes

For transpose: swap rows and columns, then verify properties like $(AB)^T = B^T A^T$. For symmetry: check if $A^T = A$ (symmetric) or $A^T = -A$ (skew). For decomposition: use $\frac{1}{2}(A+A^T)$ and $\frac{1}{2}(A-A^T)$.

No

The problem likely involves determinants, cofactors, or matrix inverse.

Does the problem ask you to FIND or VERIFY a determinant?

Yes

For $2 \times 2$: use $ad - bc$ directly. For $3 \times 3$: use cofactor expansion (choose row/column with most zeros) or Sarrus Rule. To VERIFY: use determinant properties to simplify without full expansion.

No

Continue to inverse or system-solving questions.

Is the determinant value given or have you calculated it? Is it zero?

Yes

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No

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Does the problem ask to find the inverse $A^{-1}$?

Yes

Follow steps: (1) Calculate $|A|$, (2) Find all cofactors $A_{ij}$, (3) Write cofactor matrix, (4) Transpose to get adjoint, (5) Divide by $|A|$: $A^{-1} = \frac{1}{|A|} Adj(A)$.

No

The problem may ask to solve a system or verify matrix properties.

Does the problem ask to solve a system like $AX = B$?

Yes

Verify $|A| \neq 0$. If yes, find $A^{-1}$ and compute $X = A^{-1}B$. If $|A| = 0$, the system has no unique solution.

No

You've identified most common problem types. Check if the question asks to prove a determinant property or verify a relationship.

Is the problem asking to PROVE a determinant property or relationship?

Yes

Use determinant properties: row/column operations, factoring, identical rows/columns, or expansion. Show both sides equal without fully expanding.

No

Review the problem statement carefully for any missed requirements.

Top 10 Memory Tricks and Time-Saving Strategies

Trick: These shortcuts and mnemonics will save valuable exam time and prevent calculation errors. Master them to excel in matrix and determinant problems.

Steps to Remember:

Trick 1 - 'RC Cola' Rule: Always remember: Rows then Columns. Use this for: (1) Matrix order: $3 \times 4$ means 3 rows, 4 columns. (2) Element notation: $a_{23}$ means row 2, column 3. (3) Matrix multiplication: row of first times column of second. Never reverse this order!
Trick 2 - 'Multiplication Checklist': Before multiplying matrices $A_{m \times p}$ and $B_{p \times n}$, verify: 'Does $p$ match both sides?' If columns of $A$ ≠ rows of $B$, STOP—multiplication is undefined. This single check prevents hours of wasted work.
Trick 3 - 'Cofactor Sign Pattern' (Checkerboard): Instead of calculating $(-1)^{i+j}$ each time, memorize the sign pattern: Top-left starts with $+$, then alternates: $\begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix}$. Use this pattern as your reference.
Trick 4 - 'Expand Where Zeros Are': When calculating a $3 \times 3$ determinant, ALWAYS expand along the row or column with the most zeros. Why? Each zero element eliminates one cofactor term, making your calculation much shorter. This is not just a tip—it's a game-changer.
Trick 5 - '2×2 Determinant Never Forgets': For $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, remember 'mainline minus sideline': $ad - bc$. The main diagonal minus the side (off-) diagonal. Practice this 100 times until it's automatic!
Trick 6 - 'Scalar Multiplication in Determinants': CRITICAL DIFFERENCE: In matrices, $k$ multiplies every element. In determinants, $k$ multiplies ONE row or ONE column only (not all). So: $|kA| = k^n |A|$ (for $n \times n$ matrix), NOT $k|A|$. Mixing these up loses 3-4 marks.
Trick 7 - 'Symmetric Check (Mirror Image)': For symmetric matrices, ask: 'Does it look the same when I flip it along the main diagonal?' If yes, it's symmetric. For skew-symmetric: 'Does flipping give me the negative?' If yes, it's skew. No formula needed—use visual intuition!
Trick 8 - 'Decomposition Two-Line Shortcut': To decompose $A$ into symmetric and skew parts: (1) Calculate $A + A^T$ (2) Calculate $A - A^T$ (3) Divide both by 2. Done! The first is symmetric, the second is skew. No need to verify—it's always true.
Trick 9 - 'Determinant = 0 is Your Red Flag': Whenever you calculate $|A| = 0$, immediately remember: (1) Matrix is SINGULAR (no inverse), (2) System $AX = B$ has no unique solution, (3) Two or more rows/columns are linearly dependent. This single value tells you a lot!
Trick 10 - 'Transpose Reversal Law (Never Forget!)': $(AB)^T = B^T A^T$ (Note the REVERSAL). This is different from $(A+B)^T = A^T + B^T$ (no reversal in addition). The reversal in multiplication catches many students. Write it 10 times until it's muscle memory!

Mnemonic: Remember: 'MaDe DeCoMPo'—Matrix operations, Determinant calculations, Decomposition. Focus on these three areas, and you'll master the chapter. Also: 'Singular = Zero determinant = No inverse = System has no unique solution.'

Common Exam Mistakes and How to Avoid Them

These errors appear in 70% of student submissions. Study each mistake carefully and implement the fixes to protect your marks.

Mistake 1 - Reversing Row and Column in Order: A student writes: 'A $3 \times 4$ matrix has 4 rows and 3 columns.' WRONG! It has 3 rows and 4 columns. Always remember: Order = (Rows) $\times$ (Columns). Confusing this cascades into errors in all subsequent operations. Impact: 2-3 marks lost per problem.
Mistake 2 - Attempting Matrix Multiplication Without Checking Conformability: A student multiplies a $2 \times 3$ matrix by another $2 \times 3$ matrix. This is IMPOSSIBLE! You cannot multiply matrices with non-conformable orders. Always verify: columns of first = rows of second. Fix: Write out dimensions explicitly before starting calculation.
Mistake 3 - Calculating $(AB)^T$ as $(A)^T(B)^T$ (Forgetting Reversal): The CORRECT property is $(AB)^T = B^T A^T$ (note the reversal). Students often write $(AB)^T = A^T B^T$. This is a conceptual error that ruins verification problems. Impact: Entire problem marked wrong (5 marks lost).
Mistake 4 - Confusing Determinant Scalar Multiplication: Given a $3 \times 3$ matrix with $|A| = 2$, a student calculates $|3A| = 3 \times 2 = 6$. WRONG! Correct: $|3A| = 3^3 |A| = 27 \times 2 = 54$. The power is the ORDER of the matrix, not 1. This is THE most common mistake on determinant properties.
Mistake 5 - Forgetting the Sign in Cofactor Expansion: When expanding a determinant, a student calculates all minors but forgets to apply the $(-1)^{i+j}$ sign. Result: Completely wrong answer despite correct calculation method. Fix: Use the checkerboard sign pattern from the start: $+ - + \quad - + - \quad + - +$
Mistake 6 - Trying to Find Inverse of a Singular Matrix: Given a matrix with $|A| = 0$, a student proceeds to calculate $A^{-1} = \frac{1}{|A|} Adj(A)$. This involves division by zero—UNDEFINED! Always check $|A| \neq 0$ BEFORE attempting to find the inverse. If $|A| = 0$, write 'Inverse does not exist' and stop.
Mistake 7 - Expansion Error in 3×3 Determinants: A student expands along a row but uses minors from the wrong row/column. Or they expand two different ways and get different answers (sign errors). Fix: Expand only along ONE row or column. Verify by expanding along a different row/column if unsure.
Mistake 8 - Adding Matrices of Different Orders: Attempting $A_{2 \times 3} + B_{3 \times 2}$ is IMPOSSIBLE. Matrices must have the same order for addition. A student does this, writes nonsense, and loses marks. Fix: Check orders match BEFORE performing any addition or subtraction.
Mistake 9 - Wrong Formula for Adjoint (Confusing with Transpose): Adjoint is NOT the transpose! Adjoint = Transpose of Cofactor Matrix. A student calculates all cofactors, forgets to transpose, and gets the wrong adjoint, leading to wrong inverse. Impact: Complete loss of 5-7 marks.
Mistake 10 - Sarrus Rule Used for Matrices Other Than 3×3: A student tries to use Sarrus Rule (diagonal method) for a $2 \times 2$ or $4 \times 4$ determinant. Sarrus Rule is ONLY for $3 \times 3$ matrices! Using it elsewhere gives nonsense. Fix: For $2 \times 2$, use $ad - bc$. For larger matrices, use cofactor expansion.

Solved Examples (Easy → Medium → Hard)

1

Construct a $2 \times 2$ matrix $A = [a_{ij}]$ such that $a_{ij} = 2i - j$.

Easy

Solution

We need to construct a $2 \times 2$ matrix where each element is defined by the formula $a_{ij} = 2i - j$.

Step 1: Understand the structure of a 2×2 matrix

A $2 \times 2$ matrix has the form:
$$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$

We need to find the value of each of these four elements.

Step 2: Calculate $a_{11}$ (row 1, column 1)

Using $a_{ij} = 2i - j$ with $i = 1, j = 1$:
$$a_{11} = 2(1) - 1 = 2 - 1 = 1$$

Step 3: Calculate $a_{12}$ (row 1, column 2)

Using $a_{ij} = 2i - j$ with $i = 1, j = 2$:
$$a_{12} = 2(1) - 2 = 2 - 2 = 0$$

Step 4: Calculate $a_{21}$ (row 2, column 1)

Using $a_{ij} = 2i - j$ with $i = 2, j = 1$:
$$a_{21} = 2(2) - 1 = 4 - 1 = 3$$

Step 5: Calculate $a_{22}$ (row 2, column 2)

Using $a_{ij} = 2i - j$ with $i = 2, j = 2$:
$$a_{22} = 2(2) - 2 = 4 - 2 = 2$$

Step 6: Write the complete matrix

$$A = \begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix}$$

Verification: Check each element:
- $a_{11} = 1 = 2(1) - 1$ ✓
- $a_{12} = 0 = 2(1) - 2$ ✓
- $a_{21} = 3 = 2(2) - 1$ ✓
- $a_{22} = 2 = 2(2) - 2$ ✓

∴ $A = \begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix}$

2

Evaluate the determinant: $\begin{vmatrix} 3 & 5 \\ 2 & 4 \end{vmatrix}$

Easy

Solution

For a $2 \times 2$ matrix, we use the formula $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.

Step 1: Identify the elements

In our determinant:
- $a = 3$ (top-left)
- $b = 5$ (top-right)
- $c = 2$ (bottom-left)
- $d = 4$ (bottom-right)

Step 2: Apply the determinant formula

Remember: 'Main diagonal minus side diagonal'

$$\begin{vmatrix} 3 & 5 \\ 2 & 4 \end{vmatrix} = (3)(4) - (5)(2)$$

Step 3: Calculate the products

$$= 12 - 10$$

Step 4: Final result

$$= 2$$

Interpretation: The determinant is $2 \neq 0$, so the matrix is NON-SINGULAR and has an inverse.

∴ $\begin{vmatrix} 3 & 5 \\ 2 & 4 \end{vmatrix} = 2$

3

Show that the matrix $A = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix}$ is skew-symmetric.

Easy

Solution

A matrix is skew-symmetric if $A^T = -A$. We'll verify this by calculating both sides.

Step 1: Write the original matrix

$$A = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix}$$

Step 2: Calculate $A^T$ (transpose)

To find the transpose, swap rows and columns:
- Row 1 of $A$ becomes Column 1 of $A^T$: $[0, 5] \to \begin{bmatrix} 0 \\ 5 \end{bmatrix}$
- Row 2 of $A$ becomes Column 2 of $A^T$: $[-5, 0] \to \begin{bmatrix} -5 \\ 0 \end{bmatrix}$

$$A^T = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix}$$

Step 3: Calculate $-A$ (negative of original)

Multiply every element by $-1$:

$$-A = -\begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix}$$

Step 4: Compare $A^T$ and $-A$

$$A^T = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} = -A$$

Since $A^T = -A$, the matrix $A$ is SKEW-SYMMETRIC. ✓

Additional observation: Notice that all diagonal elements are 0 (which is always true for skew-symmetric matrices), and off-diagonal elements are negatives of each other: $a_{12} = 5 = -a_{21}$ and $a_{21} = -5 = -a_{12}$.

∴ Proved: $A$ is skew-symmetric because $A^T = -A$

4

Find the minor $M_{12}$ and cofactor $A_{12}$ of the element $a_{12} = 5$ in the determinant $\begin{vmatrix} 1 & 5 & 3 \\ 2 & 4 & 1 \\ 3 & 2 & 2 \end{vmatrix}$

Medium

Solution

To find the minor and cofactor of element $a_{12}$, we delete row 1 and column 2, then evaluate the remaining determinant.

Step 1: Understand what minor means

The minor $M_{12}$ is obtained by deleting row 1 and column 2 from the original matrix. The element we're analyzing is $a_{12} = 5$ (row 1, column 2).

Step 2: Write the original matrix and highlight what to delete

$$\begin{vmatrix} 1 & \textbf{5} & 3 \\ 2 & 4 & 1 \\ 3 & 2 & 2 \end{vmatrix}$$

Delete row 1: $\begin{bmatrix} 1 & 5 & 3 \end{bmatrix}$ (crossed out)
Delete column 2: The middle column is removed

Remaining matrix:
$$\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$$

Step 3: Calculate the minor $M_{12}$

$$M_{12} = \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = (2)(2) - (1)(3) = 4 - 3 = 1$$

Step 4: Calculate the cofactor $A_{12}$

The cofactor includes the sign factor $(-1)^{i+j}$:

$$A_{12} = (-1)^{1+2} \cdot M_{12} = (-1)^3 \cdot 1 = -1 \cdot 1 = -1$$

Note: $1 + 2 = 3$ (odd), so $(-1)^3 = -1$

Step 5: Verify using the sign pattern

The sign pattern for a $3 \times 3$ determinant is:
$$\begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix}$$

Position $(1,2)$ has a $-$ sign, confirming our cofactor is $-M_{12} = -1$. ✓

∴ $M_{12} = 1$ and $A_{12} = -1$

5

Express $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ as the sum of a symmetric matrix and a skew-symmetric matrix.

Medium

Solution

We use the formula: $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$

The first part is symmetric (call it $P$) and the second is skew-symmetric (call it $Q$).

Step 1: Calculate $A^T$ (transpose of A)

Original matrix: $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Transpose (swap rows and columns):
$$A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$$

Step 2: Calculate $A + A^T$

$$A + A^T = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 1+1 & 2+3 \\ 3+2 & 4+4 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix}$$

Step 3: Calculate symmetric part $P = \frac{1}{2}(A + A^T)$

$$P = \frac{1}{2} \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} = \begin{bmatrix} 1 & 2.5 \\ 2.5 & 4 \end{bmatrix}$$

Step 4: Calculate $A - A^T$

$$A - A^T = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 1-1 & 2-3 \\ 3-2 & 4-4 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$

Step 5: Calculate skew-symmetric part $Q = \frac{1}{2}(A - A^T)$

$$Q = \frac{1}{2} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -0.5 \\ 0.5 & 0 \end{bmatrix}$$

Step 6: Verify the decomposition

$$P + Q = \begin{bmatrix} 1 & 2.5 \\ 2.5 & 4 \end{bmatrix} + \begin{bmatrix} 0 & -0.5 \\ 0.5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = A$$ ✓

Step 7: Verify that P is symmetric

$$P^T = \begin{bmatrix} 1 & 2.5 \\ 2.5 & 4 \end{bmatrix}^T = \begin{bmatrix} 1 & 2.5 \\ 2.5 & 4 \end{bmatrix} = P$$ ✓ (Symmetric!)

Step 8: Verify that Q is skew-symmetric

$$Q^T = \begin{bmatrix} 0 & -0.5 \\ 0.5 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & 0.5 \\ -0.5 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -0.5 \\ 0.5 & 0 \end{bmatrix} = -Q$$ ✓ (Skew-symmetric!)

∴ $$A = \begin{bmatrix} 1 & 2.5 \\ 2.5 & 4 \end{bmatrix} + \begin{bmatrix} 0 & -0.5 \\ 0.5 & 0 \end{bmatrix}$$ where the first matrix is symmetric and the second is skew-symmetric.

6

Evaluate the determinant $\begin{vmatrix} 2 & 3 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}$ using cofactor expansion.

Medium

Solution

We'll expand along row 2 (which has a zero, making calculations easier).

Step 1: Write the determinant and choose the row/column for expansion

$$\begin{vmatrix} 2 & 3 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}$$

Row 2 is $[0, 1, 2]$. Since the first element is 0, expanding along this row will reduce calculations.

Step 2: Write the expansion formula

For row 2, the expansion is:
$$|A| = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$$
$$= 0 \cdot A_{21} + 1 \cdot A_{22} + 2 \cdot A_{23}$$
$$= A_{22} + 2A_{23}$$

Step 3: Calculate $A_{22}$ (cofactor of element 1 in row 2, column 2)

Delete row 2 and column 2:
$$A_{22} = (-1)^{2+2} \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = (+1) [(2)(0) - (1)(1)] = 0 - 1 = -1$$

Step 4: Calculate $A_{23}$ (cofactor of element 2 in row 2, column 3)

Delete row 2 and column 3:
$$A_{23} = (-1)^{2+3} \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (-1) [(2)(2) - (3)(1)] = (-1)[4 - 3] = (-1)(1) = -1$$

Step 5: Calculate the determinant

$$|A| = A_{22} + 2A_{23} = -1 + 2(-1) = -1 - 2 = -3$$

Step 6: Verify using a different expansion (optional)

Expand along column 1 for verification:
$$|A| = 2 \cdot A_{11} + 0 \cdot A_{21} + 1 \cdot A_{31}$$

$A_{11} = (+1)\begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} = (1)(0) - (2)(2) = -4$

$A_{31} = (+1)\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = (3)(2) - (1)(1) = 6 - 1 = 5$

$|A| = 2(-4) + 0 + 1(5) = -8 + 5 = -3$ ✓ (Same answer!)

∴ $\begin{vmatrix} 2 & 3 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = -3$

7

Find the inverse of the matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ using the formula $A^{-1} = \frac{1}{|A|} Adj(A)$.

Hard

Solution

To find the inverse, we follow four steps: (1) Calculate determinant, (2) Find cofactors, (3) Calculate adjoint, (4) Divide by determinant.

Step 1: Calculate $|A|$ (determinant)

$$|A| = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = (1)(4) - (2)(3) = 4 - 6 = -2$$

Since $|A| = -2 \neq 0$, the matrix is NON-SINGULAR and has an inverse. ✓

Step 2: Find all cofactors

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the cofactors are straightforward:

$A_{11} = (-1)^{1+1} \begin{vmatrix} 4 \end{vmatrix} = (+1)(4) = 4$

$A_{12} = (-1)^{1+2} \begin{vmatrix} 3 \end{vmatrix} = (-1)(3) = -3$

$A_{21} = (-1)^{2+1} \begin{vmatrix} 2 \end{vmatrix} = (-1)(2) = -2$

$A_{22} = (-1)^{2+2} \begin{vmatrix} 1 \end{vmatrix} = (+1)(1) = 1$

Step 3: Write the cofactor matrix and transpose to get adjoint

Cofactor matrix: $[A_{ij}] = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$

Adjoint (transpose of cofactor matrix):
$$Adj(A) = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}^T = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$

Step 4: Calculate $A^{-1}$

$$A^{-1} = \frac{1}{|A|} Adj(A) = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}$$

Or in fractional form:
$$A^{-1} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}$$

Step 5: Verify $AA^{-1} = I$

$$AA^{-1} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}$$

First row, first column: $(1)(-2) + (2)(3/2) = -2 + 3 = 1$ ✓
First row, second column: $(1)(1) + (2)(-1/2) = 1 - 1 = 0$ ✓
Second row, first column: $(3)(-2) + (4)(3/2) = -6 + 6 = 0$ ✓
Second row, second column: $(3)(1) + (4)(-1/2) = 3 - 2 = 1$ ✓

$$AA^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$ ✓

∴ $$A^{-1} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} \text{ or } \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}$$

8

Prove (without full expansion) that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$

Hard

Solution

We'll use determinant properties to simplify without expanding, then factor cleverly.

Step 1: Write the original determinant

$$D = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix}$$

Step 2: Apply row operations to create zeros

Perform operations $R_2 \to R_2 - aR_1$ and $R_3 \to R_3 - a^2R_1$:

The new Row 2: $[a-a, b-a, c-a] = [0, b-a, c-a]$
The new Row 3: $[a^2-a^2, b^2-a^2, c^2-a^2] = [0, b^2-a^2, c^2-a^2]$

$$D = \begin{vmatrix} 1 & 1 & 1 \\ 0 & b-a & c-a \\ 0 & b^2-a^2 & c^2-a^2 \end{vmatrix}$$

Step 3: Expand along column 1 (which has two zeros)

$$D = 1 \cdot \begin{vmatrix} b-a & c-a \\ b^2-a^2 & c^2-a^2 \end{vmatrix} - 0 + 0$$

$$D = \begin{vmatrix} b-a & c-a \\ b^2-a^2 & c^2-a^2 \end{vmatrix}$$

Step 4: Recognize and factor the 2×2 determinant

Note that:
- $b^2 - a^2 = (b-a)(b+a)$
- $c^2 - a^2 = (c-a)(c+a)$

So:
$$D = \begin{vmatrix} b-a & c-a \\ (b-a)(b+a) & (c-a)(c+a) \end{vmatrix}$$

Step 5: Factor out common terms from rows

Factor $(b-a)$ from row 2 and $(c-a)$ from column 2:r>$$D = (b-a)(c-a) \begin{vmatrix} 1 & 1 \\ b+a & c+a \end{vmatrix}$$

Step 6: Evaluate the 2×2 determinant

$$\begin{vmatrix} 1 & 1 \\ b+a & c+a \end{vmatrix} = (1)(c+a) - (1)(b+a) = c + a - b - a = c - b$$

But we need $(b - c) = -(c - b)$, so:
$$\begin{vmatrix} 1 & 1 \\ b+a & c+a \end{vmatrix} = -(b - c)$$

Step 7: Combine all factors

$$D = (b-a)(c-a) \cdot (-(b-c))$$
$$D = -(b-a)(c-a)(b-c)$$
$$D = (b-a)(c-a)(c-b) \quad \text{[since } -(b-c) = (c-b)\text{]}$$

Rearrange the factors:r>$$D = (a-b) \cdot (-1) \cdot (c-a) \cdot (c-b) \cdot (-1)$$
$$D = (a-b)(c-a)\frac{(c-b)}{1}$$

Wait, let me recalculate more carefully. We have:
$$D = -(b-a)(c-a)(b-c)$$

Note: $-(b-a) = (a-b)$ and $-(b-c) = (c-b)$

$$D = (a-b)(c-a) \cdot (-(b-c))$$
$$D = -(a-b)(c-a)(b-c)$$
$$D = (a-b)(c-a) \cdot (c-b)$$

Rearrange as:r>$$D = (a-b)(b-c)(c-a)$$

Therefore: $$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$$ ✓ (Proved)

∴ Proved using row operations and factoring without full expansion

MCQ Practice (10 Questions)

Mixed Level — MCQ

1

If a matrix has 6 elements, which of the following CANNOT be its order?

Correct!

Incorrect. Correct: $4 \times 2$

Solution

For a matrix with a given number of elements, the order (rows $\times$ columns) must have a product equal to the total number of elements.

Given: 6 elements total

Check each option:
- $1 \times 6 = 6$ ✓
- $2 \times 3 = 6$ ✓
- $3 \times 2 = 6$ ✓
- $4 \times 2 = 8$ ✗ (Not 6)

Therefore, $4 \times 2$ cannot be the order.

∴ $4 \times 2$

2

A square matrix $A$ is skew-symmetric if:

Correct!

Incorrect. Correct: $A^T = -A$

Solution

By definition, a matrix $A$ is skew-symmetric if and only if $A^T = -A$. This is equivalent to saying $a_{ij} = -a_{ji}$ for all elements.

∴ $A^T = -A$

3

For a $3 \times 3$ matrix $A$ with $|A| = 2$, what is $|3A|$?

Correct!

Incorrect. Correct: $54$

Solution

For a $3 \times 3$ matrix, the property is:

$$|kA| = k^n |A|$$

where $n = 3$ is the order of the matrix.

$$|3A| = 3^3 |A| = 27 \times 2 = 54$$

∴ $54$

4

If $|A| = -2$ and $|B| = 3$, what is $|AB|$?

Correct!

Incorrect. Correct: $-6$

Solution

The fundamental property is:

$$|AB| = |A| \times |B|$$

Substituting:
$$|AB| = (-2) \times 3 = -6$$

∴ $-6$

5

The minor of element $a_{11}$ in $\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix}$ is:

Correct!

Incorrect. Correct: $4$

Solution

The minor $M_{11}$ is found by deleting row 1 and column 1 from the matrix.

Original matrix: $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

After deleting row 1 and column 1, only element $a_{22} = 4$ remains.

Therefore: $M_{11} = 4$

∴ $4$

6

A matrix is singular if its determinant is:

Correct!

Incorrect. Correct: $0$

Solution

By definition: A square matrix $A$ is singular if and only if $|A| = 0$. Such matrices do not have inverses and cannot solve certain systems of linear equations uniquely.

∴ $0$

7

The transpose of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$ is:

Correct!

Incorrect. Correct: $\begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$

Solution

To find $A^T$, swap rows and columns:

Original (Row 1): $[1, 2, 3]$ becomes Column 1: $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Original (Row 2): $[4, 5, 6]$ becomes Column 2: $\begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$

Therefore: $A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$

∴ $\begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$

8

For matrix multiplication $AB$ to be possible, what condition must be satisfied?

Correct!

Incorrect. Correct: Columns of $A$ = Rows of $B$

Solution

The conformability condition states: Two matrices $A_{m \times p}$ and $B_{p \times n}$ can be multiplied if and only if the number of columns in $A$ equals the number of rows in $B$. The resulting product is $m \times n$.

∴ Columns of $A$ = Rows of $B$

9

If $A$ is a $2 \times 3$ matrix and $B$ is a $3 \times 4$ matrix, what is the order of $AB$?

Correct!

Incorrect. Correct: $2 \times 4$

Solution

For matrices $A_{m \times p}$ and $B_{p \times n}$:

$$\text{Order of } AB = m \times n$$

Here: $A$ is $2 \times 3$ and $B$ is $3 \times 4$

The product $AB$ is $2 \times 4$ (take rows from $A$ and columns from $B$).

Note: The middle dimension (3) must match for multiplication to be possible.

∴ $2 \times 4$

10

What is the value of $I^4$ where $I$ is the identity matrix?

Correct!

Incorrect. Correct: $I$

Solution

Key property of identity matrix: $I \times I = I$ (for any order)

Therefore:
$$I^2 = I \times I = I$$
$$I^4 = I^2 \times I^2 = I \times I = I$$

In general, $I^n = I$ for any positive integer $n$.

∴ $I$

Short Answer Questions (10 Questions)

1

Construct a $2 \times 3$ matrix where $a_{ij} = |i - 2j|$.

[2 marks]

Solution

A $2 \times 3$ matrix has 2 rows and 3 columns. Using $a_{ij} = |i - 2j|$:

$a_{11} = |1 - 2(1)| = |1 - 2| = |-1| = 1$
$a_{12} = |1 - 2(2)| = |1 - 4| = |-3| = 3$
$a_{13} = |1 - 2(3)| = |1 - 6| = |-5| = 5$

$a_{21} = |2 - 2(1)| = |2 - 2| = |0| = 0$
$a_{22} = |2 - 2(2)| = |2 - 4| = |-2| = 2$
$a_{23} = |2 - 2(3)| = |2 - 6| = |-4| = 4$

Therefore: $A = \begin{bmatrix} 1 & 3 & 5 \\ 0 & 2 & 4 \end{bmatrix}$

∴ $A = \begin{bmatrix} 1 & 3 & 5 \\ 0 & 2 & 4 \end{bmatrix}$

2

Find the value of $k$ if $\begin{bmatrix} k+3 & 4 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 4 \\ 2 & 1 \end{bmatrix}$.

[2 marks]

Solution

For two matrices to be equal, all corresponding elements must be equal.

Comparing the $(1,1)$ element:
$k + 3 = 7$
$k = 4$

Verify with other elements:
- $(1,2)$: $4 = 4$ ✓
- $(2,1)$: $2 = 2$ ✓
- $(2,2)$: $1 = 1$ ✓

∴ $k = 4$

3

If $A = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 \end{bmatrix}$, find $AB$.

[2 marks]

Solution

$A$ is $2 \times 1$ and $B$ is $1 \times 2$. Check conformability: Columns of $A$ (1) = Rows of $B$ (1) ✓

Result $AB$ will be $2 \times 2$:

$$AB = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \begin{bmatrix} 1 & 4 \end{bmatrix} = \begin{bmatrix} 2(1) & 2(4) \\ 3(1) & 3(4) \end{bmatrix} = \begin{bmatrix} 2 & 8 \\ 3 & 12 \end{bmatrix}$$

∴ $AB = \begin{bmatrix} 2 & 8 \\ 3 & 12 \end{bmatrix}$

4

Show that $A = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix}$ is skew-symmetric.

[2 marks]

Solution

Find $A^T$:
$$A^T = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix}$$

Find $-A$:
$$-A = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix}$$

Since $A^T = -A$, the matrix is skew-symmetric. ✓

∴ Proved: $A^T = -A$, so $A$ is skew-symmetric.

5

Evaluate: $\begin{vmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{vmatrix}$

[2 marks]

Solution

Using $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$:

$$= (\sin\theta)(\sin\theta) - (-\cos\theta)(\cos\theta)$$
$$= \sin^2\theta + \cos^2\theta$$
$$= 1$$

∴ $\begin{vmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{vmatrix} = 1$

6

Find the adjoint of $A = \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}$.

[2 marks]

Solution

For a $2 \times 2$ matrix, the cofactor matrix is obtained by swapping diagonal elements and negating off-diagonal elements:

Cofactor matrix: $\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}$

Adjoint = Transpose of cofactor matrix:
$$Adj(A) = \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix}$$

∴ $Adj(A) = \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix}$

7

Is the matrix $A = \begin{bmatrix} 4 & 6 \\ 2 & 3 \end{bmatrix}$ singular?

[2 marks]

Solution

Calculate the determinant:
$$|A| = (4)(3) - (6)(2) = 12 - 12 = 0$$

Since $|A| = 0$, the matrix is SINGULAR (non-invertible).

∴ Yes, $A$ is singular because $|A| = 0$.

8

If $A$ is order $3 \times 2$ and $B$ is $2 \times 4$, what is the order of $(AB)^T$?

[2 marks]

Solution

$AB$ is a matrix of order $3 \times 4$ (rows from $A$, columns from $B$).

When transposed, rows and columns swap:
$(AB)^T$ is order $4 \times 3$.

∴ $(AB)^T$ is order $4 \times 3$

9

State the property: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = - \begin{vmatrix} c & d \\ a & b \end{vmatrix}$

[2 marks]

Solution

This demonstrates the Row Interchange Property: If two rows of a determinant are interchanged, the determinant changes sign (multiplied by $-1$).

Here, rows 1 and 2 are swapped, so the sign changes.

∴ Property 3: Interchanging two rows changes the sign of the determinant.

10

Find $A^{-1}$ if $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

[2 marks]

Solution

Since $A = I$ (identity matrix):

$|I| = 1$ and $Adj(I) = I$

Therefore:
$$I^{-1} = \frac{1}{1} \cdot I = I$$

The inverse of the identity matrix is itself!

∴ $A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$

Long Answer Questions (5 Questions)

1

Solve for $x$: $\begin{vmatrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix} = 0$

[6 marks]

Solution

We need to find values of $x$ that make the determinant zero.

Step 1: Apply row operation to simplify

Perform $R_1 \to R_1 + R_2 + R_3$:

Row 1 becomes: $[(x-1)+1+1, 1+(x-1)+1, 1+1+(x-1)]$
$= [x+1, x+1, x+1]$
$= (x+1)[1, 1, 1]$

The determinant becomes:
$$\begin{vmatrix} x+1 & x+1 & x+1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix}$$

Step 2: Factor out $(x+1)$ from row 1

$$= (x+1) \begin{vmatrix} 1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix}$$

Step 3: Apply column operations to simplify the remaining determinant

Perform $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:

$$= (x+1) \begin{vmatrix} 1 & 0 & 0 \\ 1 & x-2 & 0 \\ 1 & 0 & x-2 \end{vmatrix}$$

Step 4: Expand along row 1

$$= (x+1) \cdot 1 \cdot \begin{vmatrix} x-2 & 0 \\ 0 & x-2 \end{vmatrix}$$

$$= (x+1)(x-2)(x-2)$$

$$= (x+1)(x-2)^2$$

Step 5: Set determinant equal to zero and solve

$(x+1)(x-2)^2 = 0$

This gives us:
- $x + 1 = 0 \Rightarrow x = -1$
- $(x-2)^2 = 0 \Rightarrow x = 2$ (with multiplicity 2)

Step 6: Verification (optional)

For $x = -1$: All rows become $[0, 0, 0]$, so determinant = 0 ✓
For $x = 2$: The matrix becomes $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ (all rows identical), so determinant = 0 ✓

∴ $x = -1$ or $x = 2$

2

Express $A = \begin{bmatrix} 1 & 4 & 5 \\ 2 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix}$ as the sum of a symmetric and skew-symmetric matrix.

[6 marks]

Solution

Use the decomposition formula: $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$

Step 1: Find $A^T$

$$A^T = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 2 & 1 \\ 5 & 3 & 0 \end{bmatrix}$$

Step 2: Calculate $A + A^T$

$$A + A^T = \begin{bmatrix} 1 & 4 & 5 \\ 2 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 3 \\ 4 & 2 & 1 \\ 5 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 6 & 8 \\ 6 & 4 & 4 \\ 8 & 4 & 0 \end{bmatrix}$$

Step 3: Find symmetric part $P = \frac{1}{2}(A + A^T)$

$$P = \begin{bmatrix} 1 & 3 & 4 \\ 3 & 2 & 2 \\ 4 & 2 & 0 \end{bmatrix}$$

Step 4: Calculate $A - A^T$

$$A - A^T = \begin{bmatrix} 1 & 4 & 5 \\ 2 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 4 & 2 & 1 \\ 5 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 2 \\ -2 & 0 & 2 \\ -2 & -2 & 0 \end{bmatrix}$$

Step 5: Find skew-symmetric part $Q = \frac{1}{2}(A - A^T)$

$$Q = \begin{bmatrix} 0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix}$$

Step 6: Verify

$P + Q = \begin{bmatrix} 1 & 3 & 4 \\ 3 & 2 & 2 \\ 4 & 2 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 4 & 5 \\ 2 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix} = A$ ✓

∴ $$A = \begin{bmatrix} 1 & 3 & 4 \\ 3 & 2 & 2 \\ 4 & 2 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix}$$ (symmetric + skew-symmetric)

3

Prove using determinant properties (without full expansion) that: $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$

[7 marks]

Solution

Step 1: Apply row operation $R_2 \to R_2 - aR_1$, $R_3 \to R_3 - a^2R_1$

$$\begin{vmatrix} 1 & 1 & 1 \\ a-a & b-a & c-a \\ a^2-a^2 & b^2-a^2 & c^2-a^2 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 0 & b-a & c-a \\ 0 & b^2-a^2 & c^2-a^2 \end{vmatrix}$$

Row operations don't change the determinant.

Step 2: Expand along column 1 (has two zeros)

$$= 1 \cdot \begin{vmatrix} b-a & c-a \\ b^2-a^2 & c^2-a^2 \end{vmatrix}$$

Step 3: Factor the $2 \times 2$ determinant

Note: $b^2 - a^2 = (b-a)(b+a)$ and $c^2 - a^2 = (c-a)(c+a)$

$$= \begin{vmatrix} b-a & c-a \\ (b-a)(b+a) & (c-a)(c+a) \end{vmatrix}$$

Step 4: Factor out $(b-a)$ from row 2 and $(c-a)$ from the determinant strategically

$$= (b-a)(c-a) \begin{vmatrix} 1 & 1 \\ b+a & c+a \end{vmatrix}$$

Step 5: Evaluate the $2 \times 2$ determinant

$$\begin{vmatrix} 1 & 1 \\ b+a & c+a \end{vmatrix} = (c+a) - (b+a) = c - b = -(b-c)$$

Step 6: Combine all factors

$$= (b-a)(c-a) \cdot (-(b-c))$$
$$= -(b-a)(c-a)(b-c)$$
$$= (a-b)(c-a)(c-b)$$
$$= (a-b)(b-c)(c-a)$$

Therefore: $$\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$$ ✓ (Proved)

∴ Proved using row operations and factoring without full expansion

4

Verify $(AB)^T = B^T A^T$ for $A = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$

[6 marks]

Solution

We need to show both sides are equal.

Left Side: $(AB)^T$

Step 1: Calculate $AB$

$$AB = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$$

$(AB)_{11} = (1)(2) + (2)(1) = 2 + 2 = 4$
$(AB)_{12} = (1)(1) + (2)(2) = 1 + 4 = 5$
$(AB)_{21} = (3)(2) + (0)(1) = 6 + 0 = 6$
$(AB)_{22} = (3)(1) + (0)(2) = 3 + 0 = 3$

$$AB = \begin{bmatrix} 4 & 5 \\ 6 & 3 \end{bmatrix}$$

Step 2: Find $(AB)^T$

$$$(AB)^T = \begin{bmatrix} 4 & 6 \\ 5 & 3 \end{bmatrix}$$

Right Side: $B^T A^T$

Step 3: Find $A^T$ and $B^T$

$$A^T = \begin{bmatrix} 1 & 3 \\ 2 & 0 \end{bmatrix}, \quad B^T = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$$

Step 4: Calculate $B^T A^T$

$$B^T A^T = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 0 \end{bmatrix}$$

$(B^T A^T)_{11} = (2)(1) + (1)(2) = 2 + 2 = 4$
$(B^T A^T)_{12} = (2)(3) + (1)(0) = 6 + 0 = 6$
$(B^T A^T)_{21} = (1)(1) + (2)(2) = 1 + 4 = 5$
$(B^T A^T)_{22} = (1)(3) + (2)(0) = 3 + 0 = 3$

$$B^T A^T = \begin{bmatrix} 4 & 6 \\ 5 & 3 \end{bmatrix}$$

Step 5: Compare

$$(AB)^T = \begin{bmatrix} 4 & 6 \\ 5 & 3 \end{bmatrix} = B^T A^T$$ ✓

∴ Verified: $(AB)^T = B^T A^T = \begin{bmatrix} 4 & 6 \\ 5 & 3 \end{bmatrix}$

5

Find the inverse of $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & 2 \\ 1 & 2 & 3 \end{bmatrix}$ using the formula $A^{-1} = \frac{1}{|A|} Adj(A)$.

[8 marks]

Solution

Step 1: Calculate the determinant $|A|$

Expand along row 1:
$$|A| = 1 \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} - 1 \begin{vmatrix} 2 & 2 \\ 1 & 3 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix}$$

$$= 1(3-4) - 1(6-2) + 1(4-1)$$
$$= 1(-1) - 1(4) + 1(3)$$
$$= -1 - 4 + 3 = -2$$

Since $|A| = -2 \neq 0$, the inverse exists. ✓

Step 2: Calculate all cofactors

$C_{11} = (-1)^{1+1} \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = +(3-4) = -1$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 2 \\ 1 & 3 \end{vmatrix} = -(6-2) = -4$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = +(4-1) = 3$

$C_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = -(3-2) = -1$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = +(3-1) = 2$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = -(2-1) = -1$

$C_{31} = (-1)^{3+1} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = +(2-1) = 1$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = -(2-2) = 0$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = +(1-2) = -1$

Step 3: Write cofactor matrix

$$[C_{ij}] = \begin{bmatrix} -1 & -4 & 3 \\ -1 & 2 & -1 \\ 1 & 0 & -1 \end{bmatrix}$$

Step 4: Find adjoint (transpose of cofactor matrix)

$$Adj(A) = \begin{bmatrix} -1 & -1 & 1 \\ -4 & 2 & 0 \\ 3 & -1 & -1 \end{bmatrix}$$

Step 5: Calculate $A^{-1}$

$$A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & -1 & 1 \\ -4 & 2 & 0 \\ 3 & -1 & -1 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 & -1/2 \\ 2 & -1 & 0 \\ -3/2 & 1/2 & 1/2 \end{bmatrix}$$

Step 6: Verification (checking one element of $AA^{-1}$)

The $(1,1)$ element of $AA^{-1}$ should be 1:
$(1)(1/2) + (1)(2) + (1)(-3/2) = 1/2 + 2 - 3/2 = 1$ ✓

∴ $$A^{-1} = \begin{bmatrix} 1/2 & 1/2 & -1/2 \\ 2 & -1 & 0 \\ -3/2 & 1/2 & 1/2 \end{bmatrix}$$

One-Page Revision Cheat Sheet

1

Order of Matrix: Always write as Rows $\times$ Columns. A $m \times n$ matrix has $m \times n$ elements.

2

Matrix Notation: $a_{ij}$ = element in row $i$, column $j$. Notation 'RC' = Rows then Columns.

3

Addition/Subtraction: Only possible for matrices of the SAME order. Element-wise operation: $(A \pm B)_{ij} = a_{ij} \pm b_{ij}$.

4

Matrix Multiplication: $A_{m \times p} \times B_{p \times n} = (AB)_{m \times n}$. Conformability: columns of $A$ = rows of $B$. Result has rows from $A$ and columns from $B$.

5

Transpose: Swap rows and columns. Properties: $(A^T)^T = A$, $(A+B)^T = A^T + B^T$, $(AB)^T = B^T A^T$ (REVERSAL!), $(kA)^T = kA^T$.

6

Symmetric Matrix: $A^T = A$ (i.e., $a_{ij} = a_{ji}$). Diagonal elements are free. Off-diagonals mirror each other.

7

Skew-Symmetric Matrix: $A^T = -A$ (i.e., $a_{ij} = -a_{ji}$). ALL diagonal elements are 0. Off-diagonals are negatives.

8

Decomposition: Any square matrix: $A = \frac{1}{2}(A+A^T) + \frac{1}{2}(A-A^T)$ = Symmetric + Skew.

9

Determinant 2×2: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ (Main diagonal minus side diagonal).

10

Determinant 3×3: Use cofactor expansion along row/column with most zeros, or use Sarrus Rule (diagonals). Sign pattern: $+ - + \quad - + - \quad + - +$

11

Minor: $M_{ij}$ = determinant of remaining matrix after deleting row $i$ and column $j$.

12

Cofactor: $C_{ij} = (-1)^{i+j} M_{ij}$ (includes sign based on position).

13

Adjoint: Transpose of cofactor matrix: $Adj(A) = [C_{ij}]^T$.

14

Inverse Formula: $A^{-1} = \frac{1}{|A|} Adj(A)$ (provided $|A| \neq 0$).

15

Singular vs Non-Singular: Singular: $|A| = 0$ (no inverse). Non-singular: $|A| \neq 0$ (has inverse $A^{-1}$).

16

Determinant Properties: $|kA| = k^n|A|$ (not $k|A|$!). $|AB| = |A||B|$. Two identical rows/cols → determinant = 0. Interchanging two rows → sign change. Row operations $R_i \to R_i + kR_j$ → determinant unchanged.

17

Identity Matrix: All diagonal = 1, rest = 0. $AI = IA = A$. $I^n = I$. $I^{-1} = I$.

18

Zero (Null) Matrix: All elements = 0. $A + O = A$. $AO = O$.

19

CRITICAL EXAM TIPS: (1) Check matrix orders BEFORE multiplying. (2) Use cofactor expansion along zeros to save time. (3) Don't forget sign in cofactors. (4) Always check $|A| \neq 0$ before finding inverse. (5) For determinant scalar multiplication, the power is the matrix ORDER, not 1.