Determinants and Matrices — MHT-CET Maths Formulas and Traps

Determinants, Cofactors and the Adjoint Identities

Determinants and Cofactors: Expansion Along a Row

Expansion and cofactors

|A| = \sum_{j} a_{ij}A_{ij}, \qquad A_{ij} = (-1)^{i+j}M_{ij}, \qquad \sum_j a_{ij}A_{kj} = 0 \ (k \ne i)

The Adjoint and A·adj(A) = |A|·I

The adjoint identity

A\,\operatorname{adj}A = |A|\,I \qquad \operatorname{adj}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \qquad (\operatorname{adj}A)^{-1} = \frac{A}{|A|}

|adj A| = |A|ⁿ⁻¹ and |kA| = kⁿ|A|

Determinant identities

|\operatorname{adj}A| = |A|^{n-1} \qquad |kA| = k^n|A| \qquad |AB| = |A||B| \qquad |A^{-1}| = \frac{1}{|A|}

A·adj(A) = AAᵀ: Two Equations From the Diagonal and Off-Diagonal

The AAᵀ condition

A\,\operatorname{adj}A = AA^T \iff AA^T = |A|\,I \iff (\text{off-diagonal of } AA^T) = 0 \ \text{and}\ (\text{diagonal of } AA^T) = |A|

Determinant Equations: When Does |A| Vanish?

Vanishing determinant

|A| = 0 \iff A \text{ singular} \iff A^{-1} \text{ does not exist} \qquad 1 + \omega + \omega^2 = 0,\ \omega^3 = 1

Common traps

Reading (adj A)₂₃ as the cofactor A₂₃

The adjoint is the TRANSPOSE of the cofactor matrix, so its

(2, 3)

element is

A_{32}

. With

a_{ij} = 2i + j

that gives

4

, and the untransposed reading gives

-4

— both on the option list.

Substituting the relations before expanding

|A| = xyz - 8x - 4y - 3z + 28

contains BOTH given quantities and a constant. Using only

8x + 4y + 3z = 20

gives

20I

; the full expansion gives

68I

. Expand first, substitute last.

Using |adj A| = |A|

For

3 \times 3

,

|\operatorname{adj}A| = |A|^2

. Setting

|P| = 4

instead of

16

gives

\alpha = 5

, which is the option planted for that mistake.

Equating AAᵀ to |A| only on the diagonal

The off-diagonal condition is the one that links

a

and

b

. Without it the diagonal gives one equation in two unknowns and every option looks reachable.

Stopping at cos 2B = 0

The trigonometric determinant is

1 + \cos 2B

, not

\cos 2B

: the two squared terms add to

1

. Zero requires

\cos 2B = -1

, giving

(2n+1)\pi/2

— the option

(2n+1)\pi/4

is the half-finished version.

Inverse of a Matrix — Adjoint Formula, Products and Verification

Learn this subtopic in the notes

Inverse of a 2 × 2 Matrix

2 × 2 inverse

\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \quad ad - bc \ne 0

Invert an Expression: Compute A² − 5A or A + B First, Then Invert

Order of operations

(A^2 - 5A)^{-1} = \left(\text{the matrix } A^2 - 5A\right)^{-1} \ne (A^2)^{-1} - (5A)^{-1}, \qquad (A + B)^{-1} \ne A^{-1} + B^{-1}

Matrices with tan x Entries: |A| = sec²x and adj A = Aᵀ

The tan-entry matrix

A = \begin{pmatrix} 1 & \tan x \\ -\tan x & 1 \end{pmatrix}:\ |A| = \sec^2x,\ \operatorname{adj}A = A^T,\ A^{-1} = \cos^2x\,A^T,\ A^TA^{-1} = \begin{pmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{pmatrix}

(AB)⁻¹ = B⁻¹A⁻¹: Inverting Products and Recovering a Factor

Inverse of a product

(AB)^{-1} = B^{-1}A^{-1} \qquad B^{-1} = (AB)^{-1}A \qquad (A^{-1})^{-1} = A \qquad (A^T)^{-1} = (A^{-1})^T

Unknown Entries and A⁻¹ = A³: Use AA⁻¹ = I

The defining property of the inverse

AB = I \iff B = A^{-1} \qquad A^{-1} = A^{3} \iff A^{4} = I

Common traps

Swapping signs on the diagonal instead of the off-diagonal

The adjoint of

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is

\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

: the diagonal entries SWAP places and keep their signs; the off-diagonal entries stay put and change sign. Every option list contains the other three sign patterns.

Distributing the inverse over a sum

There is no rule for

(A + B)^{-1}

or

(A^2 - 5A)^{-1}

except to form the matrix and invert it. The distractors are exactly what distributing produces.

Sign of the sin 2x entries

A^TA^{-1}

has

-\sin 2x

in the top-right and

+\sin 2x

in the bottom-left. The four sign arrangements are all offered; compute the

(1,2)

entry explicitly —

\dfrac{-\tan x - \tan x}{1 + \tan^2x} = -\sin 2x

.

Keeping the order

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

is false unless the matrices commute. Reversing is the whole rule; with it,

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

and NOT

A\,(AB)^{-1}

.

Solving all nine entries

Three unknowns need three equations; the other six entries are checks. Choosing the entries that isolate one unknown each keeps the arithmetic to a few lines — solving the whole product is where time goes.

Cayley–Hamilton, Matrix Polynomials and Powers

Learn this subtopic in the notes

Cayley–Hamilton for 2 × 2: A² − (tr A)A + |A|·I = 0

Cayley–Hamilton (2 × 2)

A^2 - (\operatorname{tr}A)\,A + |A|\,I = O \qquad A^{-1} = \frac{(\operatorname{tr}A)\,I - A}{|A|}

$\operatorname{tr}A$ sum of the diagonal entries
$O$ the zero matrix

A⁻¹ = αI + βA: Read α and β From the Theorem

Inverse as a combination

A^{-1} = \alpha I + \beta A \ \text{ with } \ \alpha = \frac{\operatorname{tr}A}{|A|},\ \beta = -\frac{1}{|A|}

A Factored Polynomial in A Gives A⁻¹ in One Line

From a polynomial relation to the inverse

A^2 - pA + qI = O \ (q \ne 0) \ \Rightarrow\ A^{-1} = \frac{pI - A}{q} \qquad (A + kI)^{-1}:\ \text{write } A^2 - pA + qI = (A + kI)(A - mI) + cI

Powers of a Matrix: Find the Cycle

Powers modulo a cycle

A^m = cI \ \Rightarrow\ A^{mk + r} = c^k A^r \qquad \begin{pmatrix} i & 1 \\ 1 & 0 \end{pmatrix}^3 = iI,\quad i^4 = 1

Common traps

Sign of the determinant term

It is

A^2 - (\operatorname{tr}A)A + |A|I

: minus the trace, PLUS the determinant. Writing

-|A|I

turns a null-matrix answer into a wrong 'symmetric matrix' option.

α from the wrong entry

Matching the

(1,1)

entry gives

\alpha + \beta

, not

\alpha

— for

\begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix}

that is

\frac23

, and reading it as

\alpha

makes

4(\alpha - \beta) = \frac{10}{3}

, which is offered.

\alpha = \operatorname{tr}A/|A| = \frac56

.

Concluding A = 3I or A = 5I

(A - 3I)(A - 5I) = O

does NOT force either factor to be zero — matrices have zero divisors. The information is the polynomial relation, and only that.

Reducing 2029 modulo 3 instead of 12

A^3 = iI

is a scalar but NOT the identity; the cycle closes at

A^{12} = I

. Using

3

gives

A^{2029} = A^{1}\cdot i^{676} = A

only by luck of the arithmetic — reduce modulo the exponent at which the power returns to

I

.

Systems of Linear Equations and Symmetric, Skew-Symmetric Matrices

Learn this subtopic in the notes

Solving AX = B: Elimination, or X = A⁻¹B

Linear system in matrix form

AX = B,\quad |A| \ne 0 \ \Rightarrow\ X = A^{-1}B \ \text{(unique)}

Homogeneous Systems: Non-Trivial Solutions Need |A| = 0

Homogeneous system

AX = O \text{ has a non-trivial solution} \iff |A| = 0

Symmetric + Skew-Symmetric: The Unique Split, and Why Odd-Order Skew Is Singular

Symmetric and skew-symmetric parts

M = \frac{M + M^T}{2} + \frac{M - M^T}{2} \qquad B^T = -B,\ n \text{ odd} \ \Rightarrow\ |B| = 0

Common traps

Trusting a solution without checking every equation

One sitting's stored key gave

(2, 1, 1)

for a system whose first equation that triple fails. Substitute the solution into ALL three equations before choosing — it takes ten seconds and catches both your slips and the paper's.

Reading |A| = 0 as 'no solution'

For a homogeneous system a zero determinant means infinitely many solutions, never none —

X = O

always works. 'No solution' is only possible for

AX = B

with

B \ne O

.

Expecting a 2 × 2 skew-symmetric matrix to be singular

\begin{pmatrix} 0 & t \\ -t & 0 \end{pmatrix}

has determinant

t^2 \ne 0

. The 'skew ⇒ singular' rule is for ODD order only; in even order the skew part is invertible whenever

t \ne 0

.

Determinants and Matrices formulas

Determinants, Cofactors and the Adjoint Identities

Determinants and Cofactors: Expansion Along a Row

The Adjoint and A·adj(A) = |A|·I

|adj A| = |A|ⁿ⁻¹ and |kA| = kⁿ|A|

A·adj(A) = AAᵀ: Two Equations From the Diagonal and Off-Diagonal

Determinant Equations: When Does |A| Vanish?

Common traps

Inverse of a Matrix — Adjoint Formula, Products and Verification

Inverse of a 2 × 2 Matrix

Invert an Expression: Compute A² − 5A or A + B First, Then Invert

Matrices with tan x Entries: |A| = sec²x and adj A = Aᵀ

(AB)⁻¹ = B⁻¹A⁻¹: Inverting Products and Recovering a Factor

Unknown Entries and A⁻¹ = A³: Use AA⁻¹ = I

Common traps

Cayley–Hamilton, Matrix Polynomials and Powers

Cayley–Hamilton for 2 × 2: A² − (tr A)A + |A|·I = 0

A⁻¹ = αI + βA: Read α and β From the Theorem

A Factored Polynomial in A Gives A⁻¹ in One Line

Powers of a Matrix: Find the Cycle

Common traps

Systems of Linear Equations and Symmetric, Skew-Symmetric Matrices

Solving AX = B: Elimination, or X = A⁻¹B

Homogeneous Systems: Non-Trivial Solutions Need |A| = 0

Symmetric + Skew-Symmetric: The Unique Split, and Why Odd-Order Skew Is Singular

Common traps

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