1. Algebra — Matrices and Determinants

Inverse of a square matrix

The inverse of a square matrix $A$ of order $n$, denoted by $A^{-1}$, is the unique matrix satisfying $A A^{-1} = A^{-1} A = I_n$, existing if and only if $A$ is non-singular ($\det(A) \neq 0$). This topic covers matrix inversion properties, adjugate relations, and elementary transformations.

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

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Definition and statement

Theorem of Matrix Invertibility and Adjugate Formula

Let $A$ be a square matrix of order $n$. $A$ is invertible (non-singular) if and only if $\det(A) \neq 0$. Furthermore, the inverse is given explicitly by $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$$ where $\text{adj}(A)$ is the transpose of the cofactor matrix of $A$.

📐 Geometric Interpretation

Geometrically, a square matrix represents a linear transformation of vector space $\mathbb{R}^n$. The determinant measures the scaling factor of volumes. A non-zero determinant ensures the transformation is bijective (one-to-one and onto), meaning no dimensions collapse. The inverse matrix $A^{-1}$ represents the undoing or reversal of this geometric transformation.

⚙️ Algebraic & Structural Meaning

Algebraically, invertible matrices form a group under matrix multiplication known as the General Linear Group $\text{GL}_n(\mathbb{F})$. Inverting matrices allows for the systematic solution of consistent systems of linear equations $AX = B$ via $X = A^{-1}B$ when $\det(A) \neq 0$.

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 Condition	`Order $n \times n$`	Strictly Mandatory / Necessary	Rectangular matrices ($m \times n$ where $m \neq n$) cannot have a two-sided inverse; they may possess one-sided inverses (left or right) only under full rank conditions.
Non-Singular Matrix Condition	$\det(A) \neq 0$	Strictly Mandatory	If $\det(A) = 0$, the matrix is singular, its rows/columns are linearly dependent, and division by zero makes the adjugate formula invalid as $A^{-1}$ does not exist.
Field of Scalars	`Field $\mathbb{R}$ or $\mathbb{C}$`	Essential for division by determinant	If entries belong to a ring without division (such as integers $\mathbb{Z}$), the inverse matrix entries will not exist within the same algebraic structure unless the determinant is $\pm 1$.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Inverse of a 2x2 Matrix

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

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

Application: Directly tested in UP TGT/PGT for rapid calculation of matrix inverses and parameter determination.

Inverse of Diagonal Matrices

$$ \text{If } D = \text{diag}(d_1, d_2, \dots, d_n), \text{ then } D^{-1} = \text{diag}\left(\frac{1}{d_1}, \frac{1}{d_2}, \dots, \frac{1}{d_n}\right) $$

Rule: Invert each diagonal entry directly, provided none of them equals zero.

Reversal Law for Inverses

$$ (AB)^{-1} = B^{-1}A^{-1} $$

Rule: The inverse of a product is the product of the inverses in reverse order. Extends to $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$.

Determinant and Transpose Inverse Properties

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

Rule: Taking the inverse commutes with transposition and inverts the determinant value.

Application: Used in properties-based assertion-reason and rapid calculation questions.

Inverse of Scalar Multiples and Powers

$$ (kA)^{-1} = \frac{1}{k}A^{-1}, \quad (A^k)^{-1} = (A^{-1})^k = A^{-k} $$

Rule: Scalar scales reciprocally; matrix power inverts alongside the matrix.

Application: Appears in polynomial matrix equations (Cayley-Hamilton theorem applications).

Adjugate Determinant & Power Rules

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

Rule: Direct exponent rules for adjugates of order $n$ matrices saving minutes of manual expansion.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Distributive Fallacy of Inverses

Common Pitfall: Assuming $(A + B)^{-1} = A^{-1} + B^{-1}$, which is generally false in matrix algebra.

Counterexample: Let $A = I, B = -I$, then $A+B = 0$ (singular, non-invertible), but $A^{-1} + B^{-1}$ exists.

Why it fails: Matrix addition and multiplication do not satisfy standard real number distributive laws for inverses.

🛡️ Exam Defense: Never split the inverse across a sum. Only apply $(AB)^{-1} = B^{-1}A^{-1}$ across products.

⚠️ COMMON ERROR #02

Singular Matrix Division Trap

Common Pitfall: Attempting to find $A^{-1}$ when $\det(A) = 0$ or canceling matrices without checking if they are invertible.

Counterexample: If $AB = AC$ and $A \neq 0$, we cannot conclude $B = C$ unless $A$ is invertible.

Why it fails: Zero divisors exist in matrix rings; singular matrices destroy information, making left/right cancellation invalid.

🛡️ Exam Defense: Always compute $\det(A)$ first before attempting any inversion or cancellation step.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Singular vs Non-Singular	Square matrix $A$	$\det(A) = 0$ vs $\det(A) \neq 0$	Collapse of dimension vs Bijection	Keywords: 'invertible', 'unique solution'
Adjugate Inverse Formula	$\det(A) \neq 0$	$A^{-1} = \frac{\text{adj}(A)}{\det(A)}$	Scaling by cofactor reciprocals	Explicit matrix components given
Reversal Law	Invertible $A, B$	$(AB)^{-1} = B^{-1}A^{-1}$	Sequential geometric transformations	Product of matrices inverted
Cayley-Hamilton Method	Square matrix $A$	$A^{-1}$ from characteristic equation	Polynomial root annihilation	Find inverse using matrix polynomial

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$, find $A^{-1}$.

(A) $\frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$(B) $\frac{1}{5} \begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}$(C) $\begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$(D) $\frac{1}{5} \begin{pmatrix} -4 & 3 \\ 1 & -2 \end{pmatrix}$

📚 Method A: Formal Textbook Derivation

Step 1: Compute $\det(A) = (2)(4) - (3)(1) = 8 - 3 = 5 \neq 0$. Step 2: Find adjugate by swapping diagonal elements and negating off-diagonal elements: $\text{adj}(A) = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$. Step 3: Apply $A^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$.

⚡ Method B: 25-Second Exam Shortcut

Use 2x2 formula instantly: determinant is 5, swap 2 and 4, negate 3 and 1.

💡 Tip: Check determinant sign carefully to avoid sign errors in options.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $A$ is a square matrix of order $3$ such that $\det(A) = 5$, find $\det(\text{adj}(A))$.

(A) 5(B) 25(C) 125(D) 1/5

📚 Method A: Formal Textbook Derivation

Use the standard property $\det(\text{adj}(A)) = (\det(A))^{n-1}$. Here $n = 3$, so $\det(\text{adj}(A)) = (5)^{3-1} = 5^2 = 25$.

⚡ Method B: 25-Second Exam Shortcut

Direct formula application: $(\det(A))^{n-1} = 5^2 = 25$. Takes 5 seconds.

💡 Tip: Memorize the order $n$ dependency in adjugate determinant formulas.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $A$ is an invertible square matrix of order $n$, then $\det(A^{-1})$ is equal to:

A $\det(A)$

B $(\det(A))^{-1}$

C $n \det(A)$

D $0$

Derivation & Analysis:

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

MCQ 2: The inverse of the matrix $A = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ is:

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

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

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

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

Derivation & Analysis:

The determinant is $\cos^2 \theta + \sin^2 \theta = 1$. Using the 2x2 inverse formula, we swap diagonal elements and negate off-diagonal elements, yielding the transpose which is $A^T$ (orthogonal matrix property $A^{-1} = A^T$).

MCQ 3: If $A$ is a non-singular matrix of order $n$, then $\text{adj}(\text{adj}(A))$ is equal to:

A $(\det(A))^{n-1} A$

B $(\det(A))^{n-2} A$

C $(\det(A))^{n} A^{-1}$

D $\text{adj}(A)$

Derivation & Analysis:

By standard matrix identity, $\text{adj}(\text{adj}(A)) = (\det(A))^{n-2} A$ for any non-singular square matrix of order $n \ge 2$.

MCQ 4: If $A^2 - 3A + I = 0$, then the inverse matrix $A^{-1}$ is:

A $3I - A$

B $A - 3I$

C $3I + A$

D $I - 3A$

Derivation & Analysis:

Given $A^2 - 3A + I = 0$, multiply by $A^{-1}$: $A - 3I + A^{-1} = 0 \implies A^{-1} = 3I - A$.

MCQ 5: Which of the following matrices is always symmetric if $A$ is invertible and symmetric?

A $\text{adj}(A)$

B $A^{-1}$

C Both (A) and (B)

D None of these

Derivation & Analysis:

If $A$ is symmetric ($A^T = A$), then $A^{-1}$ is also symmetric because $(A^{-1})^T = (A^T)^{-1} = A^{-1}$, and $\text{adj}(A)$ is symmetric as well.

MCQ 6: If $A$ and $B$ are invertible matrices of the same order, then $(AB)^{-1}$ is:

A $A^{-1}B^{-1}$

B $B^{-1}A^{-1}$

C $AB$

D $BA$

Derivation & Analysis:

By the reversal law of matrix inversion, $(AB)^{-1} = B^{-1}A^{-1}$ since $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$.

MCQ 7: If $\det(A) = 3$ and order of $A$ is $3$, then $\det(5A^{-1})$ is:

A $\frac{125}{3}$

B $\frac{25}{3}$

C $125$

D $\frac{5}{3}$

Derivation & Analysis:

For an order $n$ matrix, $\det(kA) = k^n \det(A)$. Thus $\det(5A^{-1}) = 5^3 \det(A^{-1}) = 125 \times \frac{1}{3} = \frac{125}{3}$.

MCQ 8: A square matrix $A$ is orthogonal if and only if:

A $A^{-1} = A^T$

B $A^{-1} = A$

C $A^2 = I$

D $\det(A) = 0$

Derivation & Analysis:

By definition, an orthogonal matrix satisfies $A^T A = I$, which implies $A^{-1} = A^T$.

MCQ 9: If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, what is the trace of $A^{-1}$?

A $-2$

B $2$

C $-\frac{5}{2}$

D $\frac{5}{2}$

Derivation & Analysis:

$\det(A) = 4 - 6 = -2$. $A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$. Trace = $-2 - 1/2 = -5/2$.

Q10

MCQ 10: If $A$ is a skew-symmetric matrix of odd order $n$, then $A$ is:

A Invertible

B Singular / Non-invertible

C Identity matrix

D Orthogonal

Derivation & Analysis:

The determinant of a skew-symmetric matrix of odd order is always $0$. Since $\det(A) = 0$, $A$ is singular and thus non-invertible.

Common questions

No. By definition, matrix inversion requires the matrix to be square ($n \times n$) so that left and right inverses coincide and the determinant is defined.

If time permits, use the adjugate formula $A^{-1} = \frac{\text{adj}(A)}{\det(A)}$. However, in competitive exams, questions on 3x3 inverses are usually structured around the Cayley-Hamilton theorem or simple structural properties where full cofactors are not required.

Matrix multiplication is non-commutative ($AB \neq BA$). To undo the combined transformation $AB$, you must reverse the operations in opposite order: first undo $B$ (using $B^{-1}$), then undo $A$ (using $A^{-1}$).

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