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1. Algebra — Matrices and Determinants

Properties of determinants

This section covers the fundamental and advanced properties of determinants.

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

Fundamental Properties of Determinants
Let $A = [a_{ij}]$ be a square matrix of order $n$. The determinant $\det(A)$ or $|A|$ satisfies: (1) $|A^T| = |A|$, (2) If any two rows (or columns) are interchanged, the sign of the determinant changes: $|A'| = -|A|$, (3) If two rows (or columns) are identical or proportional, $|A| = 0$, (4) If a multiple of one row is added to another, the value remains invariant: $|A| = |A_{\text{new}}|$.

📐 Geometric Interpretation

Geometrically, the absolute value of the determinant of a $2 \times 2$ matrix represents the scaling factor of area of a parallelogram spanned by its column vectors, while for a $3 \times 3$ matrix, it represents the volume scaling factor of a parallelepiped. A zero determinant indicates collapse of dimension (linear dependence).

⚙️ Algebraic & Structural Meaning

Determinants act as a homomorphism from the multiplicative monoid of $n \times n$ matrices over a field $\mathbb{F}$ to the multiplicative group of $\mathbb{F}$. They characterize invertibility ($|A| \neq 0 \iff A^{-1}$ exists) and eigenvalue distributions via the characteristic polynomial $\det(A - \lambda I) = 0$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Square Matrix Requirement Matrix order $n \times n$ Strictly Mandatory / Necessary Determinants are undefined for rectangular matrices ($m \times n$ where $m \neq n$).
Field of Entries Entries in $\mathbb{R}$ or $\mathbb{C}$ Strictly Mandatory Operations like row scaling and addition require algebraic closure and field axioms to preserve determinant properties.
Scalar Multiplication Rule $k \in \mathbb{R}$ Essential for scalar factoring Multiplying an entire matrix $A$ of order $n$ by scalar $k$ yields $|kA| = k^n |A|$, not $k|A|$.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Scalar Factor Extraction per Row/Column

$$ \begin{vmatrix} ka & kb & kc \\ d & e & f \\ g & h & i \end{vmatrix} = k \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} $$

Rule: A scalar can be factored out from any single row or column independently, unlike matrix scalar multiplication where it factors out with power $n$.

Application: Use this to evaluate large polynomial determinants by factoring out common terms from rows/columns.
02

Properties of Skew-Symmetric Determinants

$$ \det(A) = 0 \quad \text{if } A^T = -A \text{ and order } n \text{ is odd.} $$

Rule: Every skew-symmetric matrix of odd order has a determinant identically equal to zero.

Application: Direct theoretical and numerical questions in DSSSB and UP TGT asking for determinants of $3 \times 3$ skew-symmetric matrices.
03

Triangular and Diagonal Determinants

$$ \det(A) = \prod_{i=1}^{n} a_{ii} = a_{11} a_{22} \cdots a_{nn} $$

Rule: The determinant of an upper triangular, lower triangular, or diagonal matrix is simply the product of its main diagonal entries.

Application: Used after applying elementary row operations to convert complex matrices into triangular form.
04

Determinant of Product and Powers

$$ \det(AB) = \det(A)\det(B), \quad \det(A^n) = (\det(A))^n $$

Rule: Multiplicative property allows splitting determinants of matrix products and directly raising scalar determinant values to powers.

05

Cyclic / Circulant Determinant Form

$$ \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} = (a+b+c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) $$

Rule: Apply $C_1 \to C_1 + C_2 + C_3$ to factor out $(a+b+c)$ instantly for cyclic structures.

Application: Very popular in UP PGT Mathematics for cyclic algebraic expressions.
06

Rapid Substitution for Variables

$$ \text{If } \Delta(x) = \text{poly}(x), \text{ set } x = 0, 1, -1 \text{ to match options.} $$

Rule: When a determinant contains variables, substitute simple integer values (like $0, 1$) to evaluate numerical value and match with parametric options.

Application: Saves 2 minutes in lengthy polynomial determinant problems in UP TGT/PGT.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Scalar Multiplication Power Confusion

Common Pitfall: Candidates evaluate $|kA|$ as $k|A|$ regardless of matrix order $n$.

Counterexample: For $A$ of order $3 \times 3$, $|2A| = 2^3 |A| = 8|A|$

Why it fails: Each of the $3$ rows gets multiplied by $2$, so the scalar is factored out $3$ times.

🛡️ Exam Defense: Always remember the scaling formula: $|kA| = k^n |A|$ for an $n \times n$ matrix.
⚠️ COMMON ERROR #02

Addition Property Misapplication

Common Pitfall: Assuming $\det(A + B) = \det(A) + \det(B)$.

Counterexample: Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$. $|A| = 1, |B| = 1$, but $A+B = 0$, so $|A+B| = 0 \neq |A| + |B| = 2$.

Why it fails: Determinant is a linear function of individual rows/columns separately, but not a linear transformation over matrix addition.

🛡️ Exam Defense: Never split determinants across matrix addition: $\det(A+B) \neq \det(A) + \det(B)$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Transpose PropertySquare Matrix $A$$|A^T| = |A|$Volume scaling remains invariant under transpositionSimplifying rows to columns
Scalar ScalingOrder $n$ square matrix, scalar $k$$|kA| = k^n |A|$Uniform scaling in $n$-dimensional spaceConstant factor outside matrix
Row InterchangeAny two rows/columns swapped$|A'| = -|A|$Orientation reversal of geometric figureOdd number of row/column swaps
Skew-Symmetric (Odd)$A^T = -A$, order $n$ is odd$|A| = 0$Degenerate volume in odd dimensionsOdd order skew-symmetric matrix

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $A$ is a square matrix of order $3$ such that $|A| = 5$, find the value of $|2 \text{adj}(3A^Tdak)|$ -- simplified to: find the value of $|2 \text{adj}(A)|$.

(A) 40(B) 50(C) 200(D) 250
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: Use property $|\text{adj}(A)| = |A|^{n-1} = 5^{3-1} = 25$. Step 2: Use scalar property for order $3$: $|2 \text{adj}(A)| = 2^3 |\text{adj}(A)| = 8 \times 25 = 200$.
⚡ Method B: 25-Second Exam Shortcut
Direct formula: $|k \text{ adj}(A)| = k^n |A|^{n-1}$. Here $k=2, n=3, |A|=5 \implies 2^3 \times 5^2 = 8 \times 25 = 200$.
💡 Tip: Memorize adjugate determinant relations: $|\text{adj}(A)| = |A|^{n-1}$ and $|\text{adj}(\text{adj}(A))| = |A|^{(n-1)^2}$.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Evaluate the determinant $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix}$.

(A) $(a-b)(b-c)(c-a)$(B) $(a+b)(b+c)(c+a)$(C) $abc(a-b)(b-c)$(D) Zero
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Apply column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$. Step 2: Factor out $(b-a)$ and $(c-a)$. Step 3: Expand the remaining $2 \times 2$ determinant to get $(a-b)(b-c)(c-a)$.
⚡ Method B: 25-Second Exam Shortcut
Vandermonde determinant standard result. Recognize immediately as $\prod_{1 \le i < j \le n}(x_j - x_i)$.
💡 Tip: Vandermonde determinants appear frequently in UP PGT; memorize the standard form and result.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $A$ is a non-singular square matrix of order $3$ with $|A| = 4$, then $|\text{adj}(A)|$ is equal to:
A 16
B 64
C 12
D 4
Derivation & Analysis:

By the property of adjugate determinants, $|\text{adj}(A)| = |A|^{n-1}$. For $n=3$ and $|A|=4$, $|\text{adj}(A)| = 4^{3-1} = 4^2 = 16$. Thus option (A) is correct.

Q2
MCQ 2: The value of the determinant $\begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}$ is:
A $x+y+z$
B $0$
C $xyz$
D $x^3+y^3+z^3$
Derivation & Analysis:

Apply $R_1 \to R_1 + R_2$: the first row becomes $(x+y+z, x+y+z, x+y+z)$. Factoring out $(x+y+z)$ leaves the first and third rows as multiples/identical after combining, or notice $R_1$ is the sum of $R_2$ and $R_3$ multiplied by constants. Specifically, $R_1 = R_2 + (z,x,y)$ wait, $R_1 = [x+y, y+z, z+x]$, $R_2 = [z, x, y]$, $R_3 = [1, 1, 1]$. If we apply $R_1 \to R_1 + R_2$, $R_1 = [x+y+z, x+y+z, x+y+z]$. Factoring $(x+y+z)$ out leaves row 1 as $[1,1,1]$, which is identical to $R_3$. Hence, the determinant is $0$.

Q3
MCQ 3: If $A$ is a skew-symmetric matrix of odd order $n$, then $\det(A)$ is always:
A 1
B -1
C 0
D $n$
Derivation & Analysis:

For any skew-symmetric matrix $A$, $A^T = -A$. Taking determinants on both sides for order $n$: $|A^T| = |-A| \implies |A| = (-1)^n |A|$. When $n$ is odd, $|A| = -|A| \implies 2|A| = 0 \implies |A| = 0$.

Q4
MCQ 4: If $A$ and $B$ are square matrices of order $3$ suchytically with $|A| = 3$ and $|B| = 5$, then $|2AB^{-1}|$ is equal to:
A $\frac{24}{5}$
B $24$
C $\frac{8}{5}$
D $40$
Derivation & Analysis:

$|2AB^{-1}| = 2^3 |A| |B^{-1}| = 8 \times |A| \times \frac{1}{|B|} = 8 \times 3 \times \frac{1}{5} = \frac{24}{5}$.

Q5
MCQ 5: What is the value of $\begin{vmatrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{vmatrix}$?
A $abc$
B $abc$ squared
C $0$
D $a^2+b^2+c^2$
Derivation & Analysis:

This is a $3 \times 3$ skew-symmetric matrix (odd order). By property, its determinant is identically $0$.

Q6
MCQ 6: If $\omega$ is a complex cube root of unity, then $\begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix}$ is equal to:
A 1
B 0
C $3\omega$
D $3\omega^2$
Derivation & Analysis:

Applying $C_1 \to C_1 + C_2 + C_3$ gives elements $(1+\omega+\omega^2)$ in the first column. Since $1+\omega+\omega^2 = 0$, the entire first column is zero, making the determinant $0$.

Q7
MCQ 7: If $A$ is an invertible matrix of order $n$, then $\det(A^{-1})$ is expressed as:
A $\det(A)$
B $-\det(A)$
C $(\det(A))^{-1}$
D $n \det(A)$
Derivation & Analysis:

Since $A A^{-1} = I$, taking determinants yields $\det(A)\det(A^{-1}) = \det(I) = 1$. Thus $\det(A^{-1}) = \frac{1}{\det(A)} = (\det(A))^{-1}$.

Q8
MCQ 8: Assertion (A): The determinant of a diagonal matrix is the product of its diagonal elements. Reason (R): Elementary row operations do not alter the determinant value when a multiple of one row is added to another.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

Both statements are individually true fundamental properties of determinants, but Reason (R) is not the direct deductive proof/explanation of Assertion (A).

Q9
MCQ 9: If the rows of a $3 \times 3$ matrix $A$ are given by vectors whose scalar triple product is $12$, then $|A|$ equals:
A 4
B 12
C 36
D 4
Derivation & Analysis:

The determinant of a matrix whose rows are formed by three vectors is precisely the scalar triple product of those vectors. Thus $|A| = 12$.

Q10
MCQ 10: For any square matrix $A$ of order $n$, which of the following is always true?
A $|A + B| = |A| + |B|$
B $|kA| = k|A|$
C $|A^T| = |A|$
D $|AB| \neq |A||B|$
Derivation & Analysis:

The determinant of a matrix equals the determinant of its transpose ($|A^T| = |A|$), which is a foundational property. The others contain standard misconception formulas.

Common questions

In matrix scalar multiplication, multiplying $A$ by scalar $k$ multiplies every entry of the matrix by $k$. In a determinant, factoring out a scalar is done row-by-row (or column-by-column). Since there are $n$ rows in an $n \times n$ matrix, factoring $k$ out from all $n$ rows yields $k \times k \times \cdots \times k$ ($n$ times), resulting in $k^n |A|$.

Candidates frequently confuse matrix scalar multiplication rules with determinant scalar properties, or misapply the adjugate determinant formula $|\text{adj}(A)| = |A|^{n-1}$ by using power $n$ instead of $n-1$.

Use symmetry and cyclic substitution. Applying $C_1 \to C_1 + C_2 + C_3$ or substituting $a=b=0$ (where permitted) or using root-finding methods with options saves valuable time.

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