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1. Algebra — Matrices and Determinants

Non-singular matrices

A square matrix $A$ of order $n$ is termed non-singular if and only if its determinant is non-zero, i.e., $\det(A) \neq 0$. Geometrically, this signifies that the linear transformation represented by $A$ is bijective and preserves full dimensional volume without collapsing space. This topic covers non-singular matrices.

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Definition and statement

Invertibility and Determinant Criterion for Non-Singular Matrices
Let $A$ be a square matrix of order $n$ over a field $F$. The following statements are equivalent: (1) $A$ is non-singular, i.e., $|A| \neq 0$; (2) $A$ is invertible, meaning there exists a unique matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I_n$; (3) The rank of $A$ is equal to $n$ ($\text{rank}(A) = n$); (4) The homogeneous system $AX = 0$ possesses only the trivial solution $X = 0$.

📐 Geometric Interpretation

Geometrically, a non-singular matrix represents an invertible linear transformation of $\mathbb{R}^n$ into itself. It maps linearly independent vectors to linearly independent vectors, scales volumes by a non-zero factor $|\det(A)|$, and never collapses an $n$-dimensional region into a lower-dimensional subspace.

⚙️ Algebraic & Structural Meaning

Algebraically, non-singular matrices form a group under matrix multiplication known as the General Linear Group $GL(n, F)$. They ensure uniqueness of solutions to linear systems via Cramer's Rule and matrix inversion.

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 Order $n \times n$ Strictly Mandatory / Necessary Rectangular matrices cannot have determinants nor can they possess standard two-sided inverses, rendering the concept of non-singularity inapplicable.
Non-zero Determinant $|A| \neq 0$ Strictly Mandatory If $|A| = 0$, the matrix is singular, its inverse $A^{-1}$ does not exist, and its rows/columns are linearly dependent.
Field Closure / Scalar Field Entries in $\mathbb{R}$ or $\mathbb{C}$ Essential for inversion formulas Division by zero occurs during cofactor scaling if elements or determinants do not belong to an appropriate field.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Inverse of a 2x2 Matrix

$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \implies A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$

Rule: Swap principal diagonal elements, change signs of off-diagonal elements, and divide by the determinant $ad - bc \neq 0$.

Application: Frequently asked in UP PGT for quick calculation of matrix inverses or satisfying polynomial equations.
02

Determinant of Product and Powers

$$|AB| = |A||B|, \quad |A^k| = (|A|)^k, \quad |A^{-1}| = \frac{1}{|A|}$$

Rule: Never compute matrices explicitly for powers or products when evaluating determinants or non-singularity; evaluate scalar determinants independently.

03

Adjoint Matrix Properties

$$|\text{adj}(A)| = |A|^{n-1}, \quad \text{adj}(\text{adj}(A)) = |A|^{n-2}A$$

Rule: Use the order reduction formula for the determinant of an adjugate matrix to solve parameter values instantly.

Application: Directly tested in UP TGT/PGT matching matrix dimensions.
04

Inverse of Diagonal and Triangular Matrices

$$A = \text{diag}(d_1, d_2, \dots, d_n) \implies A^{-1} = \text{diag}(d_1^{-1}, d_2^{-1}, \dots, d_n^{-1})$$

Rule: For diagonal or triangular matrices, the determinant is the product of diagonal entries. Non-singularity requires all diagonal entries to be non-zero.

05

Trace and Eigenvalue Determinant Link

$$|A| = \prod_{i=1}^{n} \lambda_i \neq 0 \iff \lambda_i \neq 0 \quad \forall i$$

Rule: A matrix is non-singular if and only if none of its eigenvalues are zero.

Application: Crucial for advanced PGT questions involving characteristic equations.
06

Rapid Elimination via Zero Rows/Columns

$$\text{If any row/column of } A \text{ is zero, } |A| = 0$$

Rule: Immediately classify matrices with proportional rows or zero rows as singular without full expansion.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusion Between Singular and Zero Matrix

Common Pitfall: Candidates assume a singular matrix must be a zero matrix.

Counterexample: $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$

Why it fails: Here $|A| = (1)(4) - (2)(2) = 0$, so $A$ is singular, yet $A \neq 0$.

🛡️ Exam Defense: Remember that singularity only requires $|A| = 0$, not that all entries are zero.
⚠️ COMMON ERROR #02

Linearity Property of Determinants

Common Pitfall: Assuming $|A + B| = |A| + |B|$.

Counterexample: $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, B = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$

Why it fails: $|A| = 1, |B| = 1$, but $|A + B| = |0| = 0 \neq |A| + |B| = 2$.

🛡️ Exam Defense: Only multiplicative property $|AB| = |A||B|$ holds universally for determinants.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Non-singular MatrixSquare matrix, $|A| \neq 0$$A^{-1}$ exists, $|A^{-1}| = |A|^{-1}$Full-rank bijection, volume scalingWords: 'invertible', 'non-zero determinant'
Singular MatrixSquare matrix, $|A| = 0$$A^{-1}$ does not exist, $\text{rank} < n$Collapses space, lower dimensionWords: 'singular', 'linearly dependent rows'
Orthogonal Matrix$A^T A = I$, square$A^{-1} = A^T$, $|A| = \pm 1$Isometry (rotation/reflection)Words: 'orthogonal', 'preserves inner product'
Symmetric Matrix$A = A^T$Real eigenvalues, orthogonal eigenvectorsPrincipal axes alignmentWords: 'symmetric', '$A = A^T$'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

For what value of $k$ is the matrix $A = \begin{pmatrix} k & 2 \\ 3 & 6 \end{pmatrix}$ singular?

(A) $k = 1$(B) $k = 2$(C) $k = 0$(D) $k = 3$
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Set determinant to zero for singularity. Step 2: $|A| = (k)(6) - (2)(3) = 6k - 6$. Step 3: $6k - 6 = 0 \implies k = 1$. Wait, let's re-verify: $(k)(6) - (2)(3) = 6k - 6 = 0 \implies k = 1$. Let's check options: if $k=1$, matrix is $\begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix}$, determinant is $6 - 6 = 0$. Hence option (A) is correct.
⚡ Method B: 25-Second Exam Shortcut
Observe that column 2 is twice column 1 if $k=1$ (since $\begin{pmatrix} 2 \\ 6 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 3 \end{pmatrix}$), so $k=1$.
💡 Tip: Check for proportional rows/columns for instant answers in $2 \times 2$ matrices.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $A$ is a non-singular square matrix of order $3$ with $|A| = 5$, then find $|3A|$.

(A) $15$(B) $45$(C) $135$(D) $27$
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Use the property $|kA| = k^n |A|$ where $n$ is the order of the matrix. Here $k = 3$, $n = 3$, and $|A| = 5$. Thus, $|3A| = 3^3 \times |A| = 27 \times 5 = 135$.
⚡ Method B: 25-Second Exam Shortcut
Instantly compute $3^3 \times 5 = 27 \times 5 = 135$ using scalar multiple determinant rule.
💡 Tip: Do not multiply every entry by 3 before taking the determinant; apply the scalar power formula directly.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $A$ and $B$ are non-singular square matrices of the same order, then $(AB)^{-1}$ is equal to:
A $A^{-1}B^{-1}$
B $B^{-1}A^{-1}$
C $AB$
D $B A$
Derivation & Analysis:

By standard matrix inverse reversal law, $(AB)^{-1} = B^{-1}A^{-1}$ because $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$.

Q2
MCQ 2: If $A$ is a square matrix such that $A^2 - A + I = 0$, then $A^{-1}$ equals:
A $A$
B $I - A$
C $A - I$
D $-A$
Derivation & Analysis:

Given $A^2 - A + I = 0$. Multiply by $A^{-1}$: $A - I + A^{-1} = 0 \implies A^{-1} = I - A$. Wait, let's check: $A - I + A^{-1} = 0 \implies A^{-1} = I - A$. Let's verify with options: if option is $I - A$, then $A^{-1} = I - A$.

Q3
MCQ 3: The determinant of an orthogonal matrix is always:
A $0$
B $1$
C $-1$
D $1$ or $-1$
Derivation & Analysis:

For an orthogonal matrix $A$, $A^T A = I$. Taking determinants: $|A^T A| = |I| \implies |A^T||A| = 1 \implies |A|^2 = 1 \implies |A| = \pm 1$.

Q4
MCQ 4: If $A$ is a non-singular matrix of order $n$, then $|\text{adj}(A)|$ is equal to:
A $|A|$
B $|A|^{n-1}$
C $|A|^n$
D $|A|^{1-n}$
Derivation & Analysis:

The standard adjugate determinant property states that $|\text{adj}(A)| = |A|^{n-1}$ for an $n \times n$ non-singular matrix.

Q5
MCQ 5: Which of the following matrices is singular?
A $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$
B $\begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}$
C $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
D $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
Derivation & Analysis:

For $\begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}$, determinant is $(2)(6) - (3)(4) = 12 - 12 = 0$. Thus it is singular.

Q6
MCQ 6: If $A$ is a non-singular matrix, then $\text{rank}(A)$ is:
A $0$
B $n-1$
C $n$
D Depends on entries
Derivation & Analysis:

A non-singular matrix of order $n$ has full rank, i.e., $\text{rank}(A) = n$.

Q7
MCQ 7: If $A$ is a skew-symmetric matrix of odd order $n$, then $A$ is always:
A Non-singular
B Singular
C Identity
D Diagonal
Derivation & Analysis:

The determinant of a skew-symmetric matrix of odd order is always zero ($|A| = |-A^T| = (-1)^n |A| = -|A| \implies 2|A| = 0 \implies |A| = 0$). Hence it is always singular.

Q8
MCQ 8: If $|A| = 5$ and order of $A$ is $3$, then $|\text{adj}(\text{adj}(A))|$ is:
A $5$
B $5^2$
C $5^4$
D $5^8$
Derivation & Analysis:

Using formula $|\text{adj}(\text{adj}(A))| = |A|^{(n-1)^2}$. For $n=3$, $(3-1)^2 = 4$. Thus $|A|^4 = 5^4$.

Q9
MCQ 9: If $A$ is a non-singular matrix, then $(A^T)^{-1}$ is equal to:
A $A$
B $A^T$
C $(A^{-1})^T$
D $-A^{-1}$
Derivation & Analysis:

Transposition and inversion commute: $(A^T)^{-1} = (A^{-1})^T$.

Q10
MCQ 10: Let $A$ be a $3 \times 3$ matrix such that $|A| = 4$. Then the value of $|A \cdot \text{adj}(A)|$ is:
A $4$
B $16$
C $64$
D $256$
Derivation & Analysis:

We know $A \cdot \text{adj}(A) = |A|I_n$. Taking determinant of both sides: $|A \cdot \text{adj}(A)| = ||A|I_n| = |A|^n |I_n| = 4^3 \times 1 = 64$.

Common questions

Yes! Non-singularity depends strictly on the determinant value, not on whether individual entries are zero. For example, the permutation matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ has determinant $-1 \neq 0$ and is non-singular despite containing zeros.

Candidates frequently confuse scalar multiplication formulas such as $|kA| = k^n|A|$ with matrix addition or misapply adjugate determinant properties like $|\text{adj}(A)| = |A|^{n-1}$ by forgetting to adjust for the matrix order $n$.

Candidates must memorize standard formulas for adjugates, inverse properties, and block matrix determinants, and actively look for row/column dependencies to spot singular matrices instantly without brute-force expansion.

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