1. Algebra — Matrices and Determinants

Invertible matrices

An invertible (or non-singular) matrix is a square matrix $A$ of order $n$ for which there exists a square matrix $B$ of the same order such that $AB = BA = I_n$, ensuring a unique bi-directional linear transformation. This topic covers invertibility conditions, adjoint properties, and determinant restrictions.

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

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Definition and statement

Theorem of Invertibility and Determinant Characterization

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

📐 Geometric Interpretation

Geometrically, an invertible matrix represents an invertible linear transformation of vector spaces (an automorphism) that neither collapses dimensions nor maps non-zero vectors to the origin, meaning the determinant represents a non-zero scaling factor of volume.

⚙️ Algebraic & Structural Meaning

Algebraically, the set of all invertible $n \times n$ matrices over a field forms a non-abelian group under matrix multiplication, known as the General Linear Group $GL(n, \mathbb{F})$. It guarantees the unique solvability of the matrix equation $AX = B$ via $X = A^{-1}B$.

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	`Dimension $n \times n$`	Strictly Mandatory / Necessary	Rectangular matrices ($m \times n$ where $m \neq n$) cannot have a two-sided inverse because left and right inverses would fail to match dimensions.
Non-zero Determinant Condition	`\det(A) \neq 0`	Strictly Mandatory	If $\det(A) = 0$ (singular matrix), the adjugate formula yields division by zero, and the columns/rows are linearly dependent, making the transformation irreversible.
Field Membership	`Entries in $\mathbb{R}$ or $\mathbb{C}$`	Essential for division by determinant	If entries belong to a ring without division (e.g., integers $\mathbb{Z}$), a matrix with determinant $\pm 1$ is invertible over $\mathbb{Z}$, but other non-zero determinants yield fractional entries not in the ring.

Useful results and shortcuts

Use these identities and results when their conditions apply:

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 the principal diagonal elements, change the signs of the off-diagonal elements, and divide by the determinant $ad-bc$ instantly without computing cofactor matrices.

Inverse of a Product of Matrices

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

Rule: Reverse the order of multiplication when inverting a product. Generalizes to $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$.

Application: Use this result in the stated type of problem.

Determinant and Transpose Inverse Properties

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

Rule: Inverting and transposing commute; the determinant of the inverse is the reciprocal of the determinant.

Adjugate Determinant Property

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

Rule: For an $n \times n$ matrix, the determinant of its adjugate is the $(n-1)$-th power of its determinant.

Inverse of Scalar Multiples

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

Rule: For a non-zero scalar $k$ and an $n \times n$ matrix $A$, scaling the matrix scales its inverse by the reciprocal, provided the dimension scaling factor is accounted for: $(kA)^{-1} = k^{-1}A^{-1}$.

Application: Tested in trick questions involving scalar factors in UP TGT.

Invertibility 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: A diagonal or triangular matrix is invertible if and only if all its diagonal entries are non-zero. The inverse of a diagonal matrix is simply obtained by reciprocating each diagonal entry.

Application: Allows instant identification of correct options without row operations.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

The Matrix Addition Inverse Fallacy

Common Pitfall: Candidates assume $(A + B)^{-1} = A^{-1} + B^{-1}$ or that invertibility distributes over addition.

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

Why it fails: $A+B = 0$, which is singular and has no inverse, whereas $A^{-1}+B^{-1} = 0$, but the sum matrix itself is not invertible.

🛡️ Exam Defense: Never distribute inverse over addition. Matrix addition and inversion do not commute.

⚠️ COMMON ERROR #02

Confusing Order in Product Inversion

Common Pitfall: Writing $(AB)^{-1} = A^{-1}B^{-1}$ instead of reversing the order.

Counterexample: $AB \neq BA$ in general for matrices.

Why it fails: $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$. The reversed order is algebraically mandatory.

🛡️ Exam Defense: Always reverse the sequence of matrices when distributing an inverse over a product.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Singular Matrix	Square matrix, $\det(A) = 0$	No inverse exists	Collapses space to a lower dimension	Keywords: 'non-invertible', 'singular', 'det equals zero'
Non-Singular Matrix	Square matrix, $\det(A) \neq 0$	$A^{-1} = \frac{\text{adj}(A)}{\det(A)}$	Bijective linear transformation	Keywords: 'invertible', 'unique solution', 'non-zero determinant'
Orthogonal Matrix	Square matrix, $A^T A = I$	$A^{-1} = A^T$	Rigid rotation or reflection (preserves length/angle)	Keywords: 'orthogonal', '$A^T = A^{-1}$'
Symmetric Matrix	Square matrix, $A^T = A$	$A^{-1}$ is also symmetric	Scaling along orthogonal principal axes	Keywords: 'symmetric', 'eigenvectors orthogonal'

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 the sum of the elements of $A^{-1}$.

(A) $\frac{1}{5}$(B) $\frac{2}{5}$(C) $-\frac{1}{5}$(D) $1$

📚 Method A: Formal Textbook Derivation

Step 1: Compute $\det(A) = (2)(4) - (3)(1) = 8 - 3 = 5$. Step 2: Find adj$(A) = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$. Step 3: $A^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix}$. Step 4: Sum of elements $= \frac{4 - 3 - 1 + 2}{5} = \frac{2}{5}$.

⚡ Method B: 25-Second Exam Shortcut

Use trace property and determinant: Sum of elements of $A^{-1} = \frac{\text{trace}(\text{adj}(A))}{\det(A)} = \frac{4+2}{5} = \frac{6}{5}$? Wait, trace of adj is $4+2=6$, but sum of all elements is $\frac{4-3-1+2}{5} = \frac{2}{5}$. Direct application of 2x2 inverse formula is safer.

💡 Tip: Be careful whether the question asks for the trace of the inverse or the sum of all its elements.

CANONICAL PROBLEM 02 Dual-Method Breakdown

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

(A) $5^2$(B) $5^4$(C) $5^6$(D) $5^8$

📚 Method A: Formal Textbook Derivation

We know $\det(\text{adj}(A)) = (\det(A))^{n-1}$. For order $n=3$, $\det(\text{adj}(A)) = 5^{3-1} = 5^2$. Applying the property again for adjugate of adjugate: $\det(\text{adj}(\text{adj}(A))) = (\det(A))^{(n-1)^2} = 5^{(2)^2} = 5^4$.

⚡ Method B: 25-Second Exam Shortcut

Direct formula: $\det(\text{adj}(\text{adj}(A))) = (\det(A))^{(n-1)^2}$. For $n=3$, power is $(3-1)^2 = 4$. Answer is $5^4$.

💡 Tip: Memorize the general rule $\det(\text{adj}^k(A)) = (\det(A))^{(n-1)^k}$ for rapid marking.

Practice questions

Choose an answer to check it and see the solution:

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

A $\det(A)$

B $\frac{1}{\det(A)}$

C $n \det(A)$

D $0$

Derivation & Analysis:

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

For any two invertible matrices $A$ and $B$ of the same order, $(AB)^{-1}$ is equal to:

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

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

C $AB^{-1}$

D $BA^{-1}$

Derivation & Analysis:

By reversing order and checking product: $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = I$, confirming $(AB)^{-1} = B^{-1}A^{-1}.

If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $A^{-1}$ exists because:

A $\det(A) = -2 \neq 0$

B $\det(A) = 0$

C $A$ is symmetric

D $A$ is a diagonal matrix

Derivation & Analysis:

The determinant of $A$ is $(1)(4) - (2)(3) = 4 - 6 = -2$. Since $\det(A) \neq 0$, the inverse exists.

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

A Invertible

B Always singular (non-invertible)

C An identity matrix

D Symmetric

Derivation & Analysis:

For any skew-symmetric matrix of odd order, the determinant is zero ($\det(A) = 0$). Hence, it is singular and not invertible.

If $A$ is a non-singular matrix of order $n$, then $\text{adj}(A)$ is invertible if and only if:

A $\det(A) = 0$

B $\det(A) \neq 0$

C $A = I$

D $A$ is skew-symmetric

Derivation & Analysis:

We know $\det(\text{adj}(A)) = (\det(A))^{n-1}$. Since $A$ is non-singular ($\det(A) \neq 0$), $\det(\text{adj}(A)) \neq 0$, so adj$(A)$ is invertible.

What is the inverse of the matrix $A = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$?

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 matrix which corresponds to option (A).

If $A^2 - A + I = 0$, then $A^{-1}$ is equal to:

A $A - I$

B $I - A$

C $A$

D $-A$

Derivation & Analysis:

Given $A^2 - A + I = 0$, multiply by $A^{-1}$ to get $A - I + A^{-1} = 0 \implies A^{-1} = I - A$? Wait: $A - I + A^{-1} = 0 \implies A^{-1} = I - A$. Let's re-verify: $A(I - A) = A - A^2 = I$. Hence $A^{-1} = I - A$ (Option B).

If $A$ is an orthogonal matrix, then $A^{-1}$ is equal to:

A $A^T$

B $-A$

C $I$

D $A$

Derivation & Analysis:

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

If $\det(A) = 3$ and order of $A$ is $3 \times 3$, then $\det(3A)$ is:

A $9$

B $27$

C $81$

D $243$

Derivation & Analysis:

For an $n \times n$ matrix, $\det(kA) = k^n \det(A)$. Here $k = 3, n = 3$, so $\det(3A) = 3^3 \times 3 = 27 \times 3 = 81$.

Q10

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

A $A+B$ is invertible

B $AB$ is invertible

C $A-B$ is invertible

D $\det(A+B) = \det(A) + \det(B)$

Derivation & Analysis:

The product of two invertible matrices is always invertible, with $(AB)^{-1} = B^{-1}A^{-1}$. Matrix addition does not preserve invertibility or determinant linearity.

Common questions

No. By definition, a matrix inverse requires $AB = BA = I$. This product commutativity and dimension compatibility are only possible when matrices are square of the same order.

Students frequently make order errors when inverting matrix products, writing $(AB)^{-1} = A^{-1}B^{-1}$ instead of $B^{-1}A^{-1}$, or miscalculating the scalar multiple determinant scaling factor $k^n \det(A)$.

Memorize standard shortcut formulas such as $\det(\text{adj}(A)) = (\det(A))^{n-1}$, $(kA)^{-1} = \frac{1}{k}A^{-1}$, and the 2x2 direct inversion shortcut to save crucial seconds.

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