1. Algebra — Matrices and Determinants

Multiplication of matrices

Matrix multiplication is a binary operation that produces a new matrix from two conformable matrices via row-by-column inner products.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Matrix Conformability and Multiplication

Let $A = [a_{ij}]$ be an $m \times n$ matrix and $B = [b_{jk}]$ be an $n \times p$ matrix. The product $C = AB$ is an $m \times p$ matrix $C = [c_{ik}]$, where the entries are defined by the scalar inner product: $$c_{ik} = \sum_{j=1}^n a_{ij} b_{jk} = a_{i1}b_{1k} + a_{i2}b_{2k} + \dots + a_{in}b_{nk}$$ for $i = 1, 2, \dots, m$ and $k = 1, 2, \dots, p$.

📐 Geometric Interpretation

Geometrically, matrix multiplication corresponds to the composition of linear transformations. If $T_A: \mathbb{R}^n \to \mathbb{R}^m$ and $T_B: \mathbb{R}^p \to \mathbb{R}^n$ are linear transformations, then the product matrix $AB$ represents the composite transformation $T_A \circ T_B$.

⚙️ Algebraic & Structural Meaning

Algebraically, the set of square matrices of order $n$ over a field forms a non-commutative ring with identity (under standard addition and multiplication). Multiplication is associative and distributes over addition, but fails the commutative property ($AB \neq BA$ in general).

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Column-Row Conformability	`Dimension matching ($n_A = m_B$)`	Strictly Mandatory / Necessary	If the number of columns in $A$ does not equal the number of rows in $B$, the product $AB$ is undefined.
Square Matrix Property for Powers/Trace	`$m = n$ ($n \times n$ order)`	Essential for powers and commutativity tests	Non-square matrices cannot be multiplied by themselves ($A^2$ is undefined), and commutativity $AB = BA$ cannot even be tested unless dimensions match symmetrically.
Field / Scalar Domain Closure	`Entries in $\mathbb{R}$ or $\mathbb{C}$`	Mandatory for arithmetic operations	Elements outside the defined field or ring structure lead to undefined scalar additions during the inner product summation.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Powers of Special Matrices (Idempotent & Involutory)

$$A^2 = A \implies A^k = A \quad (k \ge 1), \quad B^2 = I \implies B^{2k} = I, B^{2k+1} = B$$

Rule: For idempotent matrices ($A^2 = A$), all higher powers equal $A$. For involutory matrices ($B^2 = I$), even powers are identity $I$ and odd powers equal $B$.

Application: Directly asked in UP PGT to compute high powers like $A^{50}$ or $(A+I)^{100}$.

Nilpotent Matrix Index Rule

$$A^k = 0 \implies (I + A)^n = I + nA + \frac{n(n-1)}{2!}A^2 + \dots + \binom{n}{k-1}A^{k-1}$$

Rule: When $A$ is nilpotent of index $k$ (i.e., $A^k = 0$), the binomial expansion for $(I + A)^n$ terminates strictly after $k-1$ terms.

Application: Use this to evaluate polynomial expressions of matrices without full multiplication.

Diagonal Matrix Multiplication Property

$$D = \text{diag}(d_1, d_2, \dots, d_n) \implies D^k = \text{diag}(d_1^k, d_2^k, \dots, d_n^k)$$

Rule: Multiplying diagonal matrices or raising them to power $k$ requires only raising the individual diagonal entries to power $k$.

Trace Invariance under Cyclic Permutation

$$\text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB)$$

Rule: The trace of a product of matrices is invariant under cyclic permutations, even though the matrices themselves do not commute.

Application: Asked in UP PGT to find traces of complex matrix products without expanding the matrix multiplication.

Transpose and Inverse of Products

$$(AB)^T = B^T A^T, \quad (AB)^{-1} = B^{-1} A^{-1}$$

Rule: Reversal law applies to both transpose and inverse of a matrix product. Always reverse the order of factors.

Application: Standard recurring property tested in TGT/PGT matching lists and assertion-reason questions.

Determinant Multiplicativity

$$\det(AB) = \det(A) \det(B)$$

Rule: To check if $AB$ is invertible or to find its determinant, never multiply the matrices; compute individual determinants and multiply scalars.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming Commutativity ($AB = BA$)

Common Pitfall: Candidates apply algebraic expansions like $(A+B)^2 = A^2 + 2AB + B^2$ as if matrices commute.

Counterexample: $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \; B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \implies AB \neq BA$

Why it fails: Matrix multiplication is non-commutative. $(A+B)^2 = A^2 + AB + BA + B^2$, which only equals $A^2 + 2AB + B^2$ if $AB = BA$.

🛡️ Exam Defense: Never write $2AB$ unless explicitly given that $AB = BA$ or $A$ and $B$ commute.

⚠️ COMMON ERROR #02

Zero Divisors in Matrix Ring

Common Pitfall: Assuming that if $AB = 0$, then either $A = 0$ or $B = 0$.

Counterexample: A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \; B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \implies AB = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

Why it fails: The ring of matrices contains non-zero zero divisors. Both $A$ and $B$ are non-zero matrices, yet their product is the zero matrix.

🛡️ Exam Defense: Never deduce $A = 0$ or $B = 0$ from $AB = 0$ unless $\det(A) \neq 0$ or $\det(B) \neq 0$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Idempotent Matrix	$A^2 = A$, Square	$A^k = A$ for all $k \ge 1$	Projection operator onto subspace	Keywords: 'satisfies $A^2 = A$'
Involutory Matrix	$A^2 = I$, Square	$A^{2k} = I, \; A^{2k+1} = A$	Reflection operator across origin/hyperplane	Keywords: 'satisfies $A^2 = I$'
Nilpotent Matrix	$A^k = 0$ for some $k$, Square	$A^m = 0$ for all $m \ge k$	Degenerate transformation collapsing space	Keywords: ' nilpotent of index $k$'
Orthogonal Matrix	$A^T A = A A^T = I$	$A^{-1} = A^T, \; \det(A) = \pm 1$	Rigid rotation or reflection (preserves inner product)	Keywords: 'orthogonal matrix'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, then for any positive integer $n$, the matrix $A^n$ is given by:

(A) \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}(B) \begin{pmatrix} n & n \\ 0 & n \end{pmatrix}(C) \begin{pmatrix} 1 & n^2 \\ 0 & 1 \end{pmatrix}(D) \begin{pmatrix} n & 1 \\ 0 & n \end{pmatrix}

📚 Method A: Formal Textbook Derivation

Step 1: Compute $A^2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$. Step 2: Compute $A^3 = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$. Step 3: By mathematical induction, assume $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$. Step 4: Show $A^{k+1} = \begin{pmatrix} 1 & k+1 \\ 0 & 1 \end{pmatrix}$.

⚡ Method B: 25-Second Exam Shortcut

Write $A = I + N$ where $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ is nilpotent ($N^2 = 0$). By binomial theorem: $A^n = (I + N)^n = I + nN = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & n \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$.

💡 Tip: Recognize matrices of the form $I + N$ where $N$ is nilpotent to bypass lengthy induction.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $A$ and $B$ be two $3 \times 3$ matrices such that $AB = 0$ and $\det(A) \neq 0$. Which of the following must be true?

(A) $B = 0$(B) $B$ is invertible(C) $\det(B) = 0$(D) $BA = 0$

📚 Method A: Formal Textbook Derivation

Step 1: Given $AB = 0$. Step 2: Since $\det(A) \neq 0$, the inverse $A^{-1}$ exists. Step 3: Pre-multiply both sides by $A^{-1}$: $A^{-1}(AB) = A^{-1}0 \implies (A^{-1}A)B = 0 \implies IB = 0 \implies B = 0$.

⚡ Method B: 25-Second Exam Shortcut

Use cancellation law: if product is zero and one factor is non-singular (invertible), the other factor must be the zero matrix.

💡 Tip: Check invertibility before applying cancellation laws in matrix equations.

Practice questions

Choose an answer to check it and see the solution:

If $A$ is a $3 \times 4$ matrix and $B$ is a matrix such that $AB$ is a $3 \times 5$ matrix, then the order of $B$ is:

A $3 \times 5$

B $4 \times 5$

C $5 \times 4$

D $3 \times 4$

Derivation & Analysis:

For $AB$ to be defined and of order $3 \times 5$, the inner dimensions must match and outer dimensions give the result. Since $A$ is $3 \times 4$, let $B$ be $m \times n$. Then $4 = m$ and $n = 5$. Thus $B$ is $4 \times 5$.

If $A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$, then $A^n$ for any positive integer $n$ is:

A $\begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$

B $\begin{pmatrix} \cos^n \theta & \sin^n \theta \\ -\sin^n \theta & \cos^n \theta \end{pmatrix}$

C $\begin{pmatrix} n\cos \theta & n\sin \theta \\ -n\sin \theta & n\cos \theta \end{pmatrix}$

D $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Derivation & Analysis:

Using De Moivre's theorem or mathematical induction, $A^2 = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix}$. Extending to $n$ yields $A^n = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$.

If $A$ and $B$ are square matrices of order $n$ such that $AB = A$ and $BA = B$, then $A^2 + B^2$ equals:

A $AB$

B $A + B$

C $I$

D $2(A + B)$

Derivation & Analysis:

Compute $A^2 = A(A) = A(AB) = (AA)B = A^2 B$ or more simply use idempotent property: $A^2 = A(AB) = (AB)B = AB = A$. Similarly, $B^2 = (BA)A = B(AA) = BA = B$. Thus $A^2 + B^2 = A + B$.

If $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, then $A^{2024}$ is equal to:

A $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

B $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$

C $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

D $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

Derivation & Analysis:

Compute powers of $A$: $A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I$. Then $A^4 = (-I)^2 = I$. Since $2024 = 4 \times 506$, $A^{2024} = (A^4)^{506} = I^{506} = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

Which of the following statements is always true for any two square matrices $A$ and $B$ of the same order?

A $(AB)^2 = A^2 B^2$

B $AB = BA$

C $(A+B)(A-B) = A^2 - B^2$

D $(AB)^T = B^T A^T$

Derivation & Analysis:

The transpose of a product reverses the order: $(AB)^T = B^T A^T$ is a universal matrix identity. Options A, B, and C require commutativity ($AB = BA$), which does not hold in general.

If $A = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$ and $A^2 - 4A + kI = 0$, then the value of $k$ is:

A $3$

B $-3$

C $4$

D $0$

Derivation & Analysis:

Compute $A^2 = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix}$. Substitute into equation: $\begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix} - 4\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} + k\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. From top-left entry: $5 - 8 + k = 0 \implies k = 3$.

If $A$ is a skew-symmetric matrix of odd order $n$, then $\det(A)$ is:

A $1$

B $-1$

C $0$

D Depends on $A$

Derivation & Analysis:

For any square matrix $A$, $\det(A^T) = \det(A)$. For a skew-symmetric matrix, $A^T = -A$, so $\det(A^T) = \det(-A) = (-1)^n \det(A)$. If $n$ is odd, $\det(A) = -\det(A) \implies 2\det(A) = 0 \implies \det(A) = 0$.

If $A$ is a symmetric matrix, then $B^T A B$ is:

A Skew-symmetric

B Symmetric

C Diagonal

D Zero matrix

Derivation & Analysis:

Let $X = B^T A B$. Taking transpose: $X^T = (B^T A B)^T = B^T A^T (B^T)^T = B^T A B = X$ (since $A^T = A$). Thus $X$ is symmetric.

If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $AX = \begin{pmatrix} 3 & 5 \\ 5 & 11 \end{pmatrix}$, then the matrix $X$ is:

A $\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$

B $\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$

C $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$

D $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$

Derivation & Analysis:

Let $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Multiply $AX$: $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a+2c & b+2d \\ 3a+4c & 3b+4d \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 5 & 11 \end{pmatrix}$. Solving linear equations yields $a=1, c=1, b=1, d=2$, so $X = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$.

Q10

If $A$ and $B$ are non-zero square matrices of the same order such that $AB = 0$, then:

A Both $A$ and $B$ must be singular

B Both $A$ and $B$ must be non-singular

C $A$ must be singular and $B$ non-singular

D $A$ must be non-singular and $B$ singular

Derivation & Analysis:

If either $A$ or $B$ were non-singular (invertible), we could cancel it to force the other matrix to be zero. Since both are non-zero yet $AB = 0$, neither can be invertible, meaning both must be singular ($\det(A) = 0$ and $\det(B) = 0$).

Common questions

Matrix multiplication represents the composition of linear transformations. Performing transformation $B$ first and then $A$ ($AB$) generally produces a completely different geometric effect than applying $A$ first and then $B$ ($BA$). Algebraically, the row-by-column inner products involve entirely different components.

Candidates frequently commit algebraic expansion errors by applying formulas like $(A+B)^2 = A^2 + 2AB + B^2$ or $(A+B)(A-B) = A^2 - B^2$ without verifying whether $AB = BA$. Without commutativity, these standard expansions are mathematically invalid.

Always inspect the matrix for special properties like idempotency ($A^2 = A$), involutory nature ($A^2 = I$), nilpotency ($A^k = 0$), or check if it can be expressed as $I + N$. Utilizing these canonical forms reduces 3-minute calculations to 15-second shortcuts.

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