Algebra Of Matrices

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The algebra of matrices covers the operations that combine matrices - addition, subtraction, scalar multiplication, and matrix multiplication - together with the special classes of square matrices (symmetric, skew-symmetric, orthogonal, idempotent, nilpotent, and more) and the inverse operation via the adjoint. In JEE Mathematics, this is the operational core of Unit 03: unlike scalar multiplication, matrix multiplication is not commutative, and understanding when $A B = B A$ versus $A B \neq = B A$ is one of the most tested skills.

Key Formulas - Quick Reference

Addition/subtraction requires same order: $[A \pm B]_{ij} = a_{ij} \pm b_{ij}$ .
Scalar multiplication: $[λ A]_{ij} = λ \cdot a_{ij}$ .
Matrix multiplication (row-column rule): $[A B]_{ij} = k = 1 \sum n a_{ik} b_{k j}$ ; requires columns of $A$ to equal rows of $B$ .
Symmetric: $A^{T} = A$ (i.e. $a_{ij} = a_{j i}$ ). Skew-symmetric: $A^{T} = - A$ with diagonal $= 0$ .
Non-singular: $∣ A ∣ \neq = 0$ ; singular: $∣ A ∣ = 0$ . Only non-singular matrices are invertible.
Adjoint of $A$ : $adj (A) = [C_{ij}]^{T}$ , the transpose of the cofactor matrix.
Fundamental identity: $A \cdot adj (A) = ∣ A ∣ I_{n} = adj (A) \cdot A$ .
Inverse: $A^{- 1} = \frac{1}{∣ A ∣} adj (A)$ , defined when $∣ A ∣ \neq = 0$ .
Reversal law: $(A B)^{- 1} = B^{- 1} A^{- 1}$ ; $(A^{- 1})^{- 1} = A$ .
Special matrices - orthogonal: $A A^{T} = A^{T} A = I$ ; unitary: $\overset{ˉ}{A}^{T} A = I$ ; idempotent: $A^{2} = A$ ; involutory: $A^{2} = I$ ; nilpotent: $A^{m} = O$ for some positive integer $m$ .

1. Addition and Subtraction of Matrices

Two matrices can be added (or subtracted) only when they are of the same order, and the resulting matrix has the same order. Matrices of the same order are said to be conformable for addition.

For matrices

A = [a_{ij}]_{m \times n}

and

B = [b_{ij}]_{m \times n}

A + B = [a_{ij} + b_{ij}]_{m \times n}, A - B = [a_{ij} - b_{ij}]_{m \times n}

Example: $a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} + c_{1} c_{2} c_{3} d_{1} d_{2} d_{3} = a_{1} + c_{1} a_{2} + c_{2} a_{3} + c_{3} b_{1} + d_{1} b_{2} + d_{2} b_{3} + d_{3}$ .

Note: Only matrices of the same order can be added or subtracted. A

2 \times 3

and a

3 \times 2

cannot be added, even though both have six entries.

2. Scalar Multiplication

The matrix obtained by multiplying every element of $A$ by a scalar $λ$ is called the scalar multiple of $A$ , denoted $λ A$ .

Example: If $A = [263758]$ , then $2 A = [4126141016]$ .

Properties of Scalar Multiplication

Distributive over matrix addition: $k (A + B) = k A + k B$ .
Distributive over scalar addition: $(k_{1} + k_{2}) A = k_{1} A + k_{2} A$ .
Associative: $k_{1} (k_{2} A) = (k_{1} k_{2}) A = k_{2} (k_{1} A)$ .

3. Matrix Multiplication

Two matrices $A$ and $B$ can be multiplied only when the number of columns of $A$ equals the number of rows of $B$ . Such matrices are said to be conformable for multiplication.

A = [a_{ij}]_{m \times n}

and

B = [b_{ij}]_{n \times p}

, then

C = A B

has order

m \times p

with entries

c_{ij} = a_{i 1} b_{1 j} + a_{i 2} b_{2 j} + \dots + a_{in} b_{nj} = k = 1 \sum n a_{ik} b_{k j}

In words: entry

c_{ij}

is the dot product of row

i

A

with column

j

B

Solved Example 1

Problem: If

A = [213312]

and

B = 121213

, show that

A B \neq = B A

Solution:

$A$ is $2 \times 3$ and $B$ is $3 \times 2$ , so $A B$ is $2 \times 2$ and $B A$ is $3 \times 3$ .

$A B = [2 (1) + 3 (2) + 1 (1) 1 (1) + 3 (2) + 2 (1) 2 (2) + 3 (1) + 1 (3) 1 (2) + 3 (1) + 2 (3)] = [991011]$

$B A = 1 (2) + 2 (1) 2 (2) + 1 (1) 1 (2) + 3 (1) 1 (3) + 2 (3) 2 (3) + 1 (3) 1 (3) + 3 (3) 1 (1) + 2 (2) 2 (1) + 1 (2) 1 (1) + 3 (2) = 4559912547$

Since $A B$ has order $2 \times 2$ and $B A$ has order $3 \times 3$ , they cannot even be equal. So $A B \neq = B A$ .

Key Properties of Matrix Multiplication

Not commutative: in general $A B \neq = B A$ , even when both products exist and have the same order.
Associative: $(A B) C = A (B C)$ (when all products are defined).
Distributive over addition: $A (B + C) = A B + A C$ and $(A + B) C = A C + B C$ .
Identity behaves like $1$ : if $A$ is $n \times n$ and $I_{n}$ is the identity matrix of order $n$ , then $A I_{n} = I_{n} A = A$ .
Idempotence of the identity: $I = I^{2} = I^{3} = \dots = I^{n}$ .

4. Symmetric and Skew-Symmetric Matrices

A square matrix $A = [a_{ij}]$ is symmetric if $a_{ij} = a_{j i}$ for all $i, j$ , i.e. $A^{T} = A$ .

A square matrix $A$ is skew-symmetric if $a_{ij} = - a_{j i}$ for all $i, j$ and all diagonal entries are zero, i.e. $A^{T} = - A$ .

Examples: $a h g h b f g f c$ is symmetric; $0 - h g h 0 - f - g f 0$ is skew-symmetric.

5. Singular and Non-Singular Matrices

A square matrix

A

is non-singular if

∣ A ∣ \neq = 0

, and singular if

∣ A ∣ = 0

. Here

∣ A ∣

(or

det (A)

) is the determinant of

A

Only non-singular matrices have an inverse. Example: $A = [2435] \Rightarrow ∣ A ∣ = 10 - 12 = - 2 \neq = 0$ , so $A$ is non-singular.

6. Special Matrices

Unitary Matrix

A square matrix $A$ is unitary if $A^{θ} A = I$ (where $A^{θ}$ is the transpose-conjugate). Taking determinants: $∣ A^{θ} ∣∣ A ∣ = 1$ , so $∣ A ∣$ has unit modulus. Unitary matrices must be non-singular.

Equivalently, $A^{θ} A = I ⟺ A A^{θ} = I$ .

Orthogonal Matrix

A square matrix $A$ of order $n$ is orthogonal if $A A^{T} = A^{T} A = I_{n}$ . Orthogonality is the real-valued version of unitarity.

Idempotent Matrix

A square matrix $A$ is idempotent if $A^{2} = A$ .

Solved Example 2

Problem: Show that

A = 2 - 1 1 - 2 3 - 2 - 4 4 - 3

is idempotent.

Solution:

We need to verify $A^{2} = A$ . Compute $A^{2} = A \cdot A$ using the row-column rule.

Row 1 of $A^{2}$ : $[2 (2) + (- 2) (- 1) + (- 4) (1), 2 (- 2) + (- 2) (3) + (- 4) (- 2), 2 (- 4) + (- 2) (4) + (- 4) (- 3)]$

$= [4 + 2 - 4, - 4 - 6 + 8, - 8 - 8 + 12] = [2, - 2, - 4]$ . ✓ (matches row 1 of $A$ )

Row 2 of $A^{2}$ : $[(- 1) (2) + 3 (- 1) + 4 (1), (- 1) (- 2) + 3 (3) + 4 (- 2), (- 1) (- 4) + 3 (4) + 4 (- 3)]$

$= [- 2 - 3 + 4, 2 + 9 - 8, 4 + 12 - 12] = [- 1, 3, 4]$ . ✓

Row 3 of $A^{2}$ : $[1 (2) + (- 2) (- 1) + (- 3) (1), 1 (- 2) + (- 2) (3) + (- 3) (- 2), 1 (- 4) + (- 2) (4) + (- 3) (- 3)]$

$= [2 + 2 - 3, - 2 - 6 + 6, - 4 - 8 + 9] = [1, - 2, - 3]$ . ✓

All three rows match $A$ , so $A^{2} = A$ and $A$ is idempotent.

Involutory Matrix

A square matrix $A$ is involutory if $A^{2} = I$ . Every involutory matrix is its own inverse: $A^{- 1} = A$ .

Nilpotent Matrix

A square matrix $A$ is nilpotent if there exists a positive integer $m$ such that $A^{m} = O$ (the null matrix). If $m$ is the smallest such integer, it is called the index of the nilpotent matrix.

Quick Reference: Special Matrices

Type	Defining condition	Key fact
Symmetric	$A^{T} = A$	$a_{ij} = a_{j i}$
Skew-symmetric	$A^{T} = - A$	Diagonal entries are $0$
Orthogonal	$A A^{T} = A^{T} A = I$	$A^{- 1} = A^{T}$ ; $∣ A ∣ = \pm 1$
Unitary	$A^{θ} A = I$	$∣ A ∣$ has unit modulus
Idempotent	$A^{2} = A$	Eigenvalues are $0$ or $1$
Involutory	$A^{2} = I$	$A^{- 1} = A$
Nilpotent	$A^{m} = O$ for some $m \geq 1$	Singular; all eigenvalues $= 0$
Singular	$∣ A ∣ = 0$	Not invertible
Non-singular	$∣ A ∣ \neq = 0$	Invertible

7. Adjoint of a Square Matrix

Let

A = [a_{ij}]

be a square matrix of order

n

, and let

C_{ij}

be the cofactor of

a_{ij}

A

. The adjoint of

A

, denoted

adj (A)

, is the transpose of the cofactor matrix:

adj (A) = [C_{ij}]^{T} \Rightarrow [adj (A)]_{ij} = C_{j i}

For a $3 \times 3$ matrix $A = a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33}$ , the adjoint is:

$adj (A) = C_{11} C_{21} C_{31} C_{12} C_{22} C_{32} C_{13} C_{23} C_{33}^{T} = C_{11} C_{12} C_{13} C_{21} C_{22} C_{23} C_{31} C_{32} C_{33}$

where $C_{ij}$ denotes the cofactor of $a_{ij}$ .

Solved Example 3

Problem: Find the adjoint of

A = [p r q s]

Solution:

Compute the four cofactors:

$C_{11} = (+ 1) \cdot s = s$ ; $C_{12} = (- 1) \cdot r = - r$ ; $C_{21} = (- 1) \cdot q = - q$ ; $C_{22} = (+ 1) \cdot p = p$ .

Cofactor matrix: $[s - q - r p]$ .

Adjoint (transpose of cofactor matrix):

$adj (A) = [s - q - r p]^{T} = [s - r - q p]$

Quick shortcut for $2 \times 2$ : swap the diagonal entries, negate the off-diagonal entries.

Fundamental Adjoint Identity

Theorem: Let

A

be a square matrix of order

n

. Then

A \cdot adj (A) = ∣ A ∣ I_{n} = adj (A) \cdot A

This identity is the key that links the adjoint to the inverse.

8. Inverse of a Matrix

A non-singular square matrix

A

of order

n

is invertible, and there exists a unique square matrix

B

of the same order such that

A B = I_{n} = B A

In this case,

B

is called the inverse of

A

, denoted

A^{- 1} = B

. The inverse is given by

A^{- 1} = \frac{1}{∣ A ∣} adj (A)

The formula follows directly from $A \cdot adj (A) = ∣ A ∣ I_{n}$ : divide both sides by $∣ A ∣$ to get $A \cdot \frac{adj ( A )}{∣ A ∣} = I_{n}$ , so $A^{- 1} = \frac{adj ( A )}{∣ A ∣}$ .

Properties of Inverse

Reversal law: if $A$ and $B$ are invertible matrices of the same order, then $A B$ is invertible and $(A B)^{- 1} = B^{- 1} A^{- 1}$ . More generally, $(A B C \dots)^{- 1} = \dots C^{- 1} B^{- 1} A^{- 1}$ - the order is reversed.
Inverse of inverse: $(A^{- 1})^{- 1} = A$ .
Transpose commutes with inverse: $(A^{T})^{- 1} = (A^{- 1})^{T}$ .
Determinant of inverse: $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ (valid because $∣ A ∣ \neq = 0$ ).

Solved Example 4

Problem: Find the inverse of

A = [2435]

Solution:

Step 1: compute $∣ A ∣ = (2) (5) - (3) (4) = 10 - 12 = - 2 \neq = 0$ , so $A$ is invertible.

Step 2: adjoint (for $2 \times 2$ : swap diagonal, negate off-diagonal):

$adj (A) = [5 - 4 - 3 2]$

Step 3: apply the inverse formula.

$A^{- 1} = \frac{1}{∣ A ∣} adj (A) = \frac{1}{- 2} [5 - 4 - 3 2] = [- 5/2 2 3/2 - 1]$

Verify: $A \cdot A^{- 1} = [2435] [- 5/2 2 3/2 - 1] = [- 5 + 6 - 10 + 10 3 - 3 6 - 5] = [1001] = I$ . ✓

Common Mistakes to Avoid

Watch out

Assuming $A B = B A$ . Matrix multiplication is not commutative in general. Even when $A B$ and $B A$ are both defined and have the same order, they are usually different matrices.
Forgetting the order reversal in the reversal law. $(A B)^{- 1} = B^{- 1} A^{- 1}$ , not $A^{- 1} B^{- 1}$ . Same reversal for transpose: $(A B)^{T} = B^{T} A^{T}$ .
Trying to invert a singular matrix. If $∣ A ∣ = 0$ , then $A^{- 1}$ does not exist. The formula $A^{- 1} = adj (A) /∣ A ∣$ divides by zero when $∣ A ∣ = 0$ .
Multiplying non-conformable matrices. For $A B$ to be defined, the number of columns of $A$ must equal the number of rows of $B$ . Otherwise the product is undefined.
Confusing adjoint with transpose. $adj (A)$ is the transpose of the cofactor matrix, not just the transpose of $A$ . Only for orthogonal matrices do these coincide (up to sign).
Assuming $(A B)^{2} = A^{2} B^{2}$ . This is generally false because $(A B)^{2} = A B A B$ , and unless $A$ and $B$ commute, this does not simplify to $A^{2} B^{2}$ .
Confusing idempotent, involutory, and nilpotent. Idempotent: $A^{2} = A$ . Involutory: $A^{2} = I$ . Nilpotent: $A^{m} = O$ for some $m$ . All three sound similar but are distinct classes.

Frequently Asked Questions

Why is matrix multiplication not commutative?

Because the row-column rule uses the rows of the left matrix and columns of the right matrix. Swapping their order changes which entries get combined, and the resulting products differ. Even when $A B$ and $B A$ have the same order, they represent different combinations of the original entries.

When does the inverse of a matrix exist?

The inverse $A^{- 1}$ exists if and only if $A$ is a non-singular square matrix, i.e. $∣ A ∣ \neq = 0$ . For a non-square matrix, an inverse in the usual sense does not exist (though pseudo-inverses exist, beyond the JEE syllabus).

How is the adjoint different from the transpose?

The transpose $A^{T}$ just swaps rows and columns of $A$ . The adjoint $adj (A)$ is the transpose of the cofactor matrix of $A$ - each entry is a signed sub-determinant, not just a copied entry. For a $2 \times 2$ matrix, $adj (A)$ swaps diagonal entries and negates the off-diagonal entries.

What is a fast trick to compute the adjoint of a $2 \times 2$ matrix?

For $A = [a c b d]$ , the adjoint is $adj (A) = [d - c - b a]$ : swap the diagonal entries and negate the off-diagonal entries.

Are all orthogonal matrices invertible?

Yes. For an orthogonal matrix, $A A^{T} = I$ , which means $A^{T}$ is the inverse of $A$ . In particular, $∣ A ∣^{2} = 1$ , so $∣ A ∣ = \pm 1 \neq = 0$ , confirming that $A$ is non-singular and hence invertible with $A^{- 1} = A^{T}$ .

What is the eigenvalue interpretation of idempotent and involutory matrices?

The eigenvalues of an idempotent matrix are $0$ or $1$ (from $λ^{2} = λ$ ). The eigenvalues of an involutory matrix are $\pm 1$ (from $λ^{2} = 1$ ). This is why projection matrices are idempotent (they project onto a subspace) and reflection matrices are involutory.

Can two non-zero matrices multiply to give the zero matrix?

Yes. For example, $[1111] [1 - 1 - 1 1] = [0000]$ . This shows the matrix algebra is a ring with zero divisors - unlike real numbers, where $ab = 0$ forces $a = 0$ or $b = 0$ .

How is the adjoint used to solve linear equations?

For $A X = B$ with $∣ A ∣ \neq = 0$ , multiplying both sides by $A^{- 1} = adj (A) /∣ A ∣$ gives $X = A^{- 1} B = \frac{adj ( A ) \cdot B}{∣ A ∣}$ . This is the matrix version of Cramer's rule, and the primary application of the adjoint in solving simultaneous equations - covered in detail in the System of Linear Equations concept.

Previous year questions on Algebra Of Matrices

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1. Addition and Subtraction of Matrices

2. Scalar Multiplication

Properties of Scalar Multiplication

3. Matrix Multiplication

Key Properties of Matrix Multiplication

4. Symmetric and Skew-Symmetric Matrices

5. Singular and Non-Singular Matrices

6. Special Matrices

Unitary Matrix

Orthogonal Matrix

Idempotent Matrix

Involutory Matrix

Nilpotent Matrix

Quick Reference: Special Matrices

7. Adjoint of a Square Matrix

Fundamental Adjoint Identity

8. Inverse of a Matrix

Properties of Inverse

Common Mistakes to Avoid

Frequently Asked Questions

Why is matrix multiplication not commutative?

When does the inverse of a matrix exist?

How is the adjoint different from the transpose?

What is a fast trick to compute the adjoint of a 2×2 matrix?

Are all orthogonal matrices invertible?

What is the eigenvalue interpretation of idempotent and involutory matrices?

Can two non-zero matrices multiply to give the zero matrix?

How is the adjoint used to solve linear equations?

Previous year questions on Algebra Of Matrices

Ready to master Matrices?

What is a fast trick to compute the adjoint of a $2 \times 2$ matrix?