NDA Maths · Matrices & Determinants

Matrices: Order, Algebra, Powers & Equations

A matrix is a rectangular array of numbers; you add, scalar-multiply, multiply, transpose, raise to powers, and solve matrix equations — all governed by conformability and the fact that AB ≠ BA.

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Why this matters

Thirty-three PYQs, mostly EASY–MODERATE — the foundation the whole chapter stands on. Questions test matrix order and multiplication conformability, counting matrices, powers and matrix polynomials (A² − kA − I = O), and the algebra traps that catch students who assume matrices behave like numbers. Master the eight concepts below and you bank the easy marks and stop losing the trap ones.

Test yourself — a quick 14-question Matrices & Determinants recall quiz, instant score.

Concept 1 of 8

What a matrix is — order and types

Intuition

A matrix is just a grid of numbers arranged in rows and columns. Its order is (rows × columns), read rows-first. The shape decides almost everything — which matrices can be added, which can be multiplied, and which can have a determinant or inverse.

Definition

A matrix of order $m \times n$ has $m$ rows and $n$ columns. Key types:

Row / column matrix: a single row ( $1\times n$ ) or single column ( $m\times 1$ ).
Square matrix: $m = n$ — only these have a determinant and (possibly) an inverse.
**Null (zero) matrix $O$ :** every entry 0.
Diagonal / scalar / identity: square matrices with entries only on the main diagonal (identity $I$ has 1s there).

Worked example

A matrix has 12 entries. What orders are possible for it?

The order $m \times n$ must satisfy $mn = 12$ .
Factor pairs of 12: $1\times12,\ 2\times6,\ 3\times4,\ 4\times3,\ 6\times2,\ 12\times1$ .
That is 6 possible orders.

Answer:6 possible orders (one per ordered factor pair of 12).

Practice this concept4 quick reps

Concept 2 of 8

Equality, addition, and scalar multiplication

Intuition

Two matrices are equal only if they have the same order AND every corresponding entry matches — which turns a matrix equation into a set of ordinary equations. Addition and scalar multiplication are entry-by-entry, so they need the same order.

Definition

Matrices are equal iff same order and $a_{ij} = b_{ij}$ for all $i, j$ . Addition is defined only for the same order, done entrywise. Scalar multiplication $kA$ multiplies every entry by $k$ . Equating matrices entrywise is the standard way to solve for unknown entries.

Worked example

3A - B = \begin{pmatrix} 4 & 5 \\ 9 & 1 \end{pmatrix}

, where

A = \begin{pmatrix} x & 2 \\ 3 & 1 \end{pmatrix}

and

B = \begin{pmatrix} 2 & 1 \\ 0 & y \end{pmatrix}

, find

x

and

y

Scalar-multiply: $3A = \begin{pmatrix} 3x & 6 \\ 9 & 3 \end{pmatrix}$ .
Subtract entrywise: $3A - B = \begin{pmatrix} 3x - 2 & 5 \\ 9 & 3 - y \end{pmatrix}$ .
Equate with $\begin{pmatrix} 4 & 5 \\ 9 & 1 \end{pmatrix}$ : $3x - 2 = 4$ and $3 - y = 1$ .
Solve: $x = 2,\ y = 2$ . (The off-diagonal entries 5 and 9 already match — a built-in consistency check.)

Answer:

x = 2,\ y = 2

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Matrices & DeterminantsMODERATE

A = \begin{pmatrix}4i-6 & 10i \\ 14i & 6+4i\end{pmatrix}

and

k = \dfrac{1}{2i}

, where

i = \sqrt{-1}

, then

kA

is equal to

[Q11 · Sep · 2017]

Concept 3 of 8

Matrix multiplication and conformability

Intuition

To multiply

A

B

, the inner dimensions must match: an

m\times n

times an

n\times p

gives an

m\times p

. Each entry of the product is a row of

A

dotted with a column of

B

. Crucially, matrix multiplication is not commutative —

AB

and

BA

can differ, or one may not even exist.

Definition

$A_{m\times n}\,B_{n\times p} = (AB)_{m\times p}$ ; the product is defined only when $A$ 's column count equals $B$ 's row count. Entry $(AB)_{ij} = \sum_k a_{ik}b_{kj}$ . In general $AB \neq BA$ . The quadratic form $[x\ y\ z]\,M\,[x\ y\ z]^T$ is $1\times1$ (a scalar).

Order of a product

A_{m\times n}\, B_{n\times p} = (AB)_{m\times p}

Worked example

A

4\times3

and

B

3\times2

, find the orders of

AB

and

BA

(or state if undefined).

$AB$ : inner dims $3 = 3$ match → order = outer dims = $4\times2$ .
$BA$ : inner dims would be 2 ( $B$ 's columns) vs 4 ( $A$ 's rows) — $2 \neq 4$ , so $BA$ is undefined.
$AB$ exists but $BA$ does not — another way $AB \neq BA$ .

Answer:

AB

4\times2

;

BA

is undefined.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Matrices & DeterminantsEASY

A

is a matrix of order

3\times5

and

B

is a matrix of order

5\times3

, then the order of

AB

and

BA

will respectively be

[Q41 · Apr · 2020]

Matrix multiplication is NOT commutative — $AB \neq BA$

Order matters:

AB

and

BA

are generally different matrices (and one may not even exist). Never reorder factors inside a product. If a question gives

AB

, compute

AB

— answering with

BA

is the classic trap.

Drill 12 more on matrix multiplication and conformability

Concept 4 of 8

Counting matrices

Intuition

If each of the

N

entries of a matrix can independently take one of

k

values, there are

k^N

matrices. If instead you're asked how many ORDERS a matrix with

N

entries can have, count the factor pairs of

N

Definition

By entries: an $m\times n$ matrix has $mn$ entries; if each is chosen from a set of $k$ values, there are $k^{mn}$ such matrices. By order: the number of possible orders for a matrix with exactly $N$ entries equals the number of (ordered) factor pairs of $N$ , i.e. the number of divisors of $N$ .

Worked example

How many distinct

2\times2

matrices have every entry from

\{1, 2\}

A $2\times2$ matrix has 4 entries.
Each entry independently is 1 or 2 → 2 choices each.
Total $= 2^4 = 16$ .

Answer:16 matrices.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Matrices & DeterminantsEASY

How many distinct

2\times2

matrices exist with all four entries taken from

\{1,2\}

[Q47 · Sep · 2021]

Drill 2 more on counting matrices

Concept 5 of 8

Transpose and its rules

Intuition

The transpose

A^T

(or

A'

) flips rows into columns. The one rule that trips people up is the reversal law: the transpose of a product reverses the order —

(AB)^T = B^T A^T

Definition

$(A^T)_{ij} = a_{ji}$ — an $m\times n$ matrix transposes to $n\times m$ . Rules:

Self-inverse: $(A^T)^T = A$
Sum: $(A+B)^T = A^T + B^T$
Scalar: $(kA)^T = kA^T$
Reversal: $(AB)^T = B^T A^T$ (order reverses)

Transpose rules

(AB)^T = B^T A^T

Worked example

Which of these are correct for matrices

A, B, C

of the same order: (1)

(A+B+C)^T = A^T+B^T+C^T

; (2)

(AB)^T = A^T B^T

(1) Transpose distributes over addition → TRUE.
(2) Transpose of a product REVERSES order: $(AB)^T = B^T A^T$ , not $A^T B^T$ → FALSE.

Answer:Only (1) is correct; (2) violates the reversal law.

Practice this concept4 quick reps

From the bank · past-year question

Example 5Matrices & DeterminantsMODERATE

Consider the following in respect of matrices A, B and C of same order: 1.

(A+B+C)' = A' + B' + C'

(AB)' = A'B'

(ABC)' = C'B'A'

where A' is the transpose of the matrix A. Which of the above are correct?

[Q28 · Sep · 2018]

$(AB)^T = B^T A^T$ — the order REVERSES

The transpose of a product flips the factors:

(AB)^T = B^T A^T

, not

A^T B^T

. Sum and scalar transposes keep their order (

(A+B)^T = A^T+B^T

); only the PRODUCT reverses. Writing

A^T B^T

is the trap.

Concept 6 of 8

Powers of a matrix

Intuition

A^n

means

A

multiplied by itself

n

times (only square matrices). The exam trick is to compute

A^2

, spot a pattern (it repeats, or equals a multiple of

A

I

), and ride that pattern instead of grinding all

n

products.

Definition

For square $A$ : $A^2 = A\cdot A$ , $A^n = A^{n-1}A$ . Look for cycles: e.g. a swap matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ squares to $I$ , so even powers are $I$ . A rotation by $\theta$ to the $n$ th power is rotation by $n\theta$ .

Powers of a matrix

A^2 = A\cdot A \qquad A^n = A^{n-1}A

Worked example

A = \begin{pmatrix}1&1\\0&1\end{pmatrix}

, find

A^3

$A^2 = \begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}1&2\\0&1\end{pmatrix}$ .
$A^3 = A^2 A = \begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}1&3\\0&1\end{pmatrix}$ .
(Pattern: $A^n = \begin{pmatrix}1&n\\0&1\end{pmatrix}$ .)

Answer:

A^3 = \begin{pmatrix}1&3\\0&1\end{pmatrix}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6Matrices & DeterminantsEASY

A = \begin{pmatrix}0&1\\1&0\end{pmatrix}

, then the value of

A^4

[Q38 · Apr · 2017]

Drill 3 more on powers of a matrix

Concept 7 of 8

Matrix polynomials and equations

Intuition

A matrix can satisfy a polynomial equation like

A^2 - kA - I = O

. Compute

A^2

, write the equation entrywise (or use the Cayley–Hamilton shortcut for

2\times2

: every matrix satisfies

A^2 - (\text{trace})A + (\det)I = O

), and read off the unknown.

Definition

Expressions like $A^2 - 4A$ , $23A^3 - 19A^2 - 4A$ are evaluated by substituting powers of $A$ . For a $2\times2$ matrix, Cayley–Hamilton gives $A^2 - (a+d)A + (ad-bc)I = O$ where $a+d$ is the trace and $ad-bc$ the determinant — the fastest route to the constant $k$ in $A^2 - kA + cI = O$ problems.

Cayley–Hamilton (2×2)

A^2 - (a+d)\,A + (ad - bc)\,I = O

Worked example

A = \begin{pmatrix}3&1\\2&4\end{pmatrix}

and

A^2 - kA + 10I_2 = O

, find

k

Trace $= 3 + 4 = 7$ ; determinant $= 3\cdot4 - 1\cdot2 = 10$ .
Cayley–Hamilton: $A^2 - (\text{trace})A + (\det)I = O$ , i.e. $A^2 - 7A + 10I = O$ .
Compare with $A^2 - kA + 10I = O$ : $k = 7$ .

Answer:

k = 7

(the trace).

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 7Matrices & DeterminantsMODERATE

A = \begin{pmatrix}1 & 2 \\ 2 & 3\end{pmatrix}

and

A^2 - kA - I_2 = O

, then what is k?

[Q26 · Apr · 2018]

Drill 6 more on matrix polynomials and equations

Concept 8 of 8

Matrix algebra — where numbers' rules break

Intuition

Matrices look like numbers but don't always obey number-algebra. Three classics:

AB = O

does NOT force

A

B

to be zero;

(A+B)(A-B) \neq A^2 - B^2

unless

A, B

commute; and

(A+B)^2 \neq A^2 + 2AB + B^2

for the same reason.

Definition

Because $AB \neq BA$ in general:

$(A+B)(A-B) = A^2 - AB + BA - B^2$ , which equals $A^2 - B^2$ only if $AB = BA$ .
$(A+B)^2 = A^2 + AB + BA + B^2$ .
$AB = O$ is possible with both $A \neq O$ and $B \neq O$ (zero divisors exist).

Transpose/adjoint/inverse of a product all reverse order.

Non-commutative expansions

(A+B)^2 = A^2 + AB + BA + B^2 \qquad (A+B)(A-B) = A^2 - AB + BA - B^2

Worked example

For matrices

A, B

of the same order, is

(A+B)(A-B) = A^2 - B^2

always true?

Expand: $(A+B)(A-B) = A^2 - AB + BA - B^2$ .
This equals $A^2 - B^2$ only if $-AB + BA = O$ , i.e. $AB = BA$ .
Since matrices generally don't commute, the identity fails in general.

Answer:No — only when

A

and

B

commute (

AB = BA

Practice this concept4 quick reps

From the bank · past-year question

Example 8Matrices & DeterminantsMODERATE

Consider the following in respect of matrices A and B of same order: 1.

A^2-B^2=(A+B)(A-B)

(A-I)(I+A)=O \Leftrightarrow A^2=I

where I is the identity matrix and O is the null matrix. Which of the above is/are correct?

[Q35 · Sep · 2018]

Don't import $a^2 - b^2 = (a+b)(a-b)$ into matrices

Every number identity that secretly uses commutativity can break for matrices.

(A+B)(A-B)

(A+B)^2

(AB)^2 = A^2B^2

all FAIL unless

AB = BA

. When an option assumes one of these, it's almost always the trap answer.

$AB = O$ does NOT force $A = O$ or $B = O$

Unlike numbers, matrices have zero divisors: two non-zero matrices can multiply to the zero matrix (e.g.

\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix} = O

). So you may NOT 'cancel' a matrix; concluding

A=O

B=O

from

AB=O

is the trap.

Drill 3 more on matrix algebra — where numbers' rules break

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Matrix multiplication and conformability
Order of a product
$A_{m\times n}\, B_{n\times p} = (AB)_{m\times p}$
Transpose and its rules
Transpose rules
$(AB)^T = B^T A^T$
Powers of a matrix
Powers of a matrix
$A^2 = A\cdot A \qquad A^n = A^{n-1}A$
Matrix polynomials and equations
Cayley–Hamilton (2×2)
$A^2 - (a+d)\,A + (ad - bc)\,I = O$
Matrix algebra — where numbers' rules break
Non-commutative expansions
$(A+B)^2 = A^2 + AB + BA + B^2 \qquad (A+B)(A-B) = A^2 - AB + BA - B^2$

Watch out for (4)

Matrix multiplication is NOT commutative — $AB \neq BA$
→ Matrix multiplication and conformability
$(AB)^T = B^T A^T$ — the order REVERSES
→ Transpose and its rules
Don't import $a^2 - b^2 = (a+b)(a-b)$ into matrices
→ Matrix algebra — where numbers' rules break
$AB = O$ does NOT force $A = O$ or $B = O$
→ Matrix algebra — where numbers' rules break

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