Class 12Mathematics · AlgebraFull chapter

Matrices

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Matrices: Definition, Order and Types

Quick answer A matrix is a rectangular arrangement of numbers in rows and columns; its order tells us its size, and matrices are classified into special types based on their shape and entries.

A matrix is a rectangular array of numbers (or functions) arranged in horizontal rows and vertical columns, enclosed in brackets. The numbers are called the elements or entries of the matrix. Matrices are usually named with capital letters A, B, C, ... and their elements with lower-case letters.

If a matrix A has m rows and n columns, we say A is a matrix of order m × n (read as "m by n"). The general element in the i-th row and j-th column is written aij, and the whole matrix is written compactly as A = [aij]m×n, where 1 ≤ i ≤ m and 1 ≤ j ≤ n. A matrix of order m × n has exactly m×n elements.

Types of matrices commonly used:

  • Row matrix: exactly one row (order 1×n).
  • Column matrix: exactly one column (order m×1).
  • Square matrix: number of rows = number of columns (order n×n). The elements a11, a22, ..., ann form the principal diagonal.
  • Diagonal matrix: a square matrix in which every non-diagonal element is 0 (diagonal entries may be anything, including 0).
  • Scalar matrix: a diagonal matrix in which all diagonal elements are equal.
  • Identity (unit) matrix In: a scalar matrix in which every diagonal element equals 1.
  • Zero (null) matrix: every element is 0, denoted O.

Worked Example. Construct a 2×2 matrix A = [aij] whose elements are given by aij = (i + 2j)²⁄2.

Here A has order 2×2, so i = 1, 2 and j = 1, 2. Substituting:

  • a11 = (1+2)²/2 = 9/2
  • a12 = (1+4)²/2 = 25/2
  • a21 = (2+2)²/2 = 16/2 = 8
  • a22 = (2+4)²/2 = 36/2 = 18

So A = [[9/2, 25/2], [8, 18]].

Order of a matrix Order = m × n (m rows, n columns) Total number of elements = m × n
General matrix notation A = [aᵢⱼ]_{m×n}, 1 ≤ i ≤ m, 1 ≤ j ≤ n aᵢⱼ is the entry in row i, column j
Identity matrix rule aᵢⱼ = 1 if i = j; aᵢⱼ = 0 if i ≠ j Defines Iₙ, the multiplicative identity
Remember
  • A matrix of order m×n has m rows, n columns and m×n elements, written A = [aᵢⱼ]
  • A square matrix has equal number of rows and columns (order n×n)
  • A diagonal matrix has all off-diagonal entries zero; a scalar matrix additionally has equal diagonal entries
  • The identity matrix Iₙ has 1's on the diagonal and 0's elsewhere; it behaves like '1' in matrix multiplication
  • A zero (null) matrix O has every entry equal to 0

Equality of Matrices and Algebra of Matrices

Quick answer Two matrices are equal only if they have the same order and identical corresponding entries; matrices of the same order can be added, subtracted, and multiplied by a scalar entry-wise.

Two matrices A = [aij] and B = [bij] are said to be equal, written A = B, if and only if (i) they have the same order, and (ii) every corresponding pair of elements is equal, i.e. aij = bij for all i, j. Equality of matrices is often used to set up simultaneous equations to find unknown entries.

Addition and subtraction are defined only when the two matrices have the same order. If A = [aij]m×n and B = [bij]m×n, then A + B = [aij + bij]m×n, i.e. corresponding elements are added. Subtraction A − B is defined as A + (−1)B, subtracting corresponding elements.

Matrix addition satisfies: commutativity (A+B = B+A), associativity ((A+B)+C = A+(B+C)), existence of an additive identity (A + O = A), and existence of an additive inverse (A + (−A) = O).

Scalar multiplication: if k is a number (scalar) and A = [aij]m×n, then kA = [k·aij]m×n — every entry is multiplied by k. This satisfies k(A+B) = kA+kB and (k+l)A = kA+lA for scalars k, l.

Worked Example. If A = [[2, 3], [1, 4]] and B = [[1, −1], [0, 2]], find 2A − 3B.

2A = [[4, 6], [2, 8]] and 3B = [[3, −3], [0, 6]].

2A − 3B = [[4−3, 6−(−3)], [2−0, 8−6]] = [[1, 9], [2, 2]].

Equality of matrices A = B ⇔ same order and aᵢⱼ = bᵢⱼ ∀ i, j Both conditions must hold simultaneously
Matrix addition (A + B)ᵢⱼ = aᵢⱼ + bᵢⱼ Defined only when A and B have the same order
Scalar multiplication (kA)ᵢⱼ = k · aᵢⱼ k is any real scalar
Matrix subtraction A − B = A + (−1)B Entry-wise subtraction
Remember
  • A = B requires both the same order and all corresponding entries equal
  • Addition/subtraction is defined only for matrices of the same order, done entry-wise
  • Matrix addition is commutative and associative, with additive identity O and additive inverse −A
  • Scalar multiplication kA multiplies every entry of A by k
  • Equality of matrices is used to solve for unknown entries by comparing corresponding elements

Multiplication of Matrices and Its Properties

Quick answer Two matrices can be multiplied only when the number of columns of the first equals the number of rows of the second; matrix multiplication is associative and distributive but generally not commutative.

The product AB of two matrices A and B is defined only when the number of columns of A equals the number of rows of B. If A is of order m×n and B is of order n×p, then AB exists and is of order m×p. The element in the i-th row and j-th column of AB is obtained by multiplying corresponding elements of the i-th row of A with the j-th column of B and adding the results: (AB)ij = Σk=1n aik bkj.

A crucial fact for CBSE and competitive exams: matrix multiplication is not commutative in general, i.e. AB ≠ BA even when both products are defined and of the same order. Matrix multiplication is, however, associative — (AB)C = A(BC) — and distributive over addition — A(B+C) = AB+AC and (A+B)C = AC+BC. Multiplication by the identity matrix leaves a matrix unchanged: AI = IA = A. Also, the product of two non-zero matrices can be a zero matrix, unlike ordinary numbers.

Worked Example. Let A = [[1, 2], [3, 4]] and B = [[2, 0], [1, 2]]. Find AB and BA, and verify AB ≠ BA.

AB: row1·col1 = 1(2)+2(1) = 4; row1·col2 = 1(0)+2(2) = 4; row2·col1 = 3(2)+4(1) = 10; row2·col2 = 3(0)+4(2) = 8. So AB = [[4, 4], [10, 8]].

BA: row1·col1 = 2(1)+0(3) = 2; row1·col2 = 2(2)+0(4) = 4; row2·col1 = 1(1)+2(3) = 7; row2·col2 = 1(2)+2(4) = 10. So BA = [[2, 4], [7, 10]].

Since AB = [[4,4],[10,8]] and BA = [[2,4],[7,10]] are different, AB ≠ BA — confirming multiplication is not commutative.

Matrix multiplication rule (AB)ᵢⱼ = Σₖ₌₁ⁿ aᵢₖ · bₖⱼ Defined only when columns of A = rows of B
Order of product A(m×n) · B(n×p) = AB(m×p) Resulting matrix takes outer dimensions
Associativity (AB)C = A(BC) Holds whenever the products are defined
Distributivity A(B + C) = AB + AC Matrix multiplication distributes over addition
Remember
  • AB is defined only if (columns of A) = (rows of B); if A is m×n and B is n×p, AB is m×p
  • (AB)ᵢⱼ = Σ aᵢₖ bₖⱼ, the sum of products of row i of A with column j of B
  • Matrix multiplication is associative and distributive over addition, but not commutative in general (AB ≠ BA)
  • AI = IA = A: the identity matrix acts as the multiplicative identity
  • Unlike ordinary numbers, AB = O does not necessarily mean A = O or B = O

Transpose of a Matrix, Symmetric and Skew-Symmetric Matrices

Quick answer The transpose of a matrix interchanges its rows and columns; symmetric and skew-symmetric matrices are square matrices with special transpose relationships, and every square matrix can be split into these two parts.

The transpose of a matrix A of order m×n, denoted A′ (or AT), is the matrix of order n×m obtained by interchanging the rows and columns of A, so that (A′)ij = aji.

Properties of transpose (for matrices of suitable orders and scalar k):

  • (A′)′ = A
  • (kA)′ = kA′
  • (A + B)′ = A′ + B′
  • (AB)′ = B′A′ (note the reversal of order)

A square matrix A is symmetric if A′ = A, which means aij = aji for all i, j (the matrix is a mirror image of itself about the principal diagonal). A square matrix A is skew-symmetric if A′ = −A, which means aij = −aji for all i, j; setting i = j gives aii = −aii, so every diagonal element of a skew-symmetric matrix must be 0.

An important theorem: for any square matrix A, ½(A + A′) is always symmetric and ½(A − A′) is always skew-symmetric, and their sum reconstructs A. This means every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix.

Worked Example. Express A = [[3, 5], [1, −1]] as the sum of a symmetric and a skew-symmetric matrix.

A′ = [[3, 1], [5, −1]].

A + A′ = [[6, 6], [6, −2]], so P = ½(A + A′) = [[3, 3], [3, −1]] — check: P′ = P, so P is symmetric.

A − A′ = [[0, 4], [−4, 0]], so Q = ½(A − A′) = [[0, 2], [−2, 0]] — check: Q′ = −Q, so Q is skew-symmetric.

P + Q = [[3, 3], [3, −1]] + [[0, 2], [−2, 0]] = [[3, 5], [1, −1]] = A, as required.

Transpose definition (A′)ᵢⱼ = aⱼᵢ Rows of A become columns of A′
Reversal law (AB)′ = B′A′ Order of the factors reverses under transpose
Symmetric condition A′ = A ⇔ aᵢⱼ = aⱼᵢ Square matrix only
Skew-symmetric condition A′ = −A ⇔ aᵢⱼ = −aⱼᵢ, aᵢᵢ = 0 All diagonal entries are zero
Symmetric–skew decomposition A = ½(A + A′) + ½(A − A′) First term symmetric, second term skew-symmetric
Remember
  • Transpose A′ swaps rows and columns: (A′)ᵢⱼ = aⱼᵢ; a matrix of order m×n has transpose of order n×m
  • Key identity: (AB)′ = B′A′ — the order reverses
  • A is symmetric if A′ = A (aᵢⱼ = aⱼᵢ); A is skew-symmetric if A′ = −A (aᵢⱼ = −aⱼᵢ)
  • Every diagonal entry of a skew-symmetric matrix must be zero
  • Any square matrix A = ½(A+A′) + ½(A−A′), a unique sum of a symmetric part and a skew-symmetric part

Elementary Operations and the Inverse of a Matrix

Quick answer Elementary row/column operations transform a matrix step by step and provide a systematic method to find the inverse of an invertible square matrix without using determinants.

An elementary operation (transformation) on a matrix is one of the following, applied to rows (or, similarly, to columns):

  1. Interchange of any two rows: Ri ↔ Rj
  2. Multiplication of the elements of a row by a non-zero scalar: Ri → kRi (k ≠ 0)
  3. Addition to the elements of a row, the corresponding elements of another row multiplied by a non-zero scalar: Ri → Ri + kRj

A square matrix A is called invertible if there exists a square matrix B of the same order such that AB = BA = I. The matrix B is called the inverse of A, written A−1. The inverse, if it exists, is always unique.

Finding the inverse by elementary row operations: Write A = IA. Apply a sequence of elementary row operations to the matrix on the left (A) to reduce it to the identity matrix I, applying the exact same operations simultaneously to the I on the right-hand side. When the left side becomes I, the matrix that has appeared on the right side is A−1. (If, at any stage, all entries of an entire row on the left become zero, A has no inverse.) An analogous method using column operations starts from A = AI.

Worked Example. Find A−1 for A = [[2, 1], [1, 1]] using elementary row operations.

Write A = IA: [[2, 1], [1, 1]] = [[1, 0], [0, 1]] A.

Apply R1 → R1 − R2: [[1, 0], [1, 1]] = [[1, −1], [0, 1]] A.

Apply R2 → R2 − R1: [[1, 0], [0, 1]] = [[1, −1], [−1, 2]] A.

The left side is now I, so A−1 = [[1, −1], [−1, 2]].

Check: A·A−1 = [[2,1],[1,1]]·[[1,−1],[−1,2]] = [[2(1)+1(−1), 2(−1)+1(2)], [1(1)+1(−1), 1(−1)+1(2)]] = [[1, 0], [0, 1]] = I.

Definition of inverse A A⁻¹ = A⁻¹A = I A⁻¹ exists only for square, invertible A
Row-operation starting identity A = I·A Reduce A to I on the left using row operations only
Column-operation starting identity A = A·I Reduce A to I on the right using column operations only
Remember
  • Three elementary row/column operations: interchange, scaling by a non-zero scalar, and adding a scalar multiple of one row/column to another
  • A square matrix A is invertible if there exists B with AB = BA = I; when it exists, A⁻¹ is unique
  • Method: write A = IA and reduce A to I via row operations, applying the same operations to I to obtain A⁻¹
  • If any row on the left becomes entirely zero during reduction, A is not invertible
  • Only square matrices can possess an inverse in this sense

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

Order = m × n (m rows, n columns)
Order of a matrix
A = [aᵢⱼ]_{m×n}, 1 ≤ i ≤ m, 1 ≤ j ≤ n
General matrix notation
aᵢⱼ = 1 if i = j; aᵢⱼ = 0 if i ≠ j
Identity matrix rule
A = B ⇔ same order and aᵢⱼ = bᵢⱼ ∀ i, j
Equality of matrices
(A + B)ᵢⱼ = aᵢⱼ + bᵢⱼ
Matrix addition
(kA)ᵢⱼ = k · aᵢⱼ
Scalar multiplication
A − B = A + (−1)B
Matrix subtraction
(AB)ᵢⱼ = Σₖ₌₁ⁿ aᵢₖ · bₖⱼ
Matrix multiplication rule
A(m×n) · B(n×p) = AB(m×p)
Order of product
(AB)C = A(BC)
Associativity
A(B + C) = AB + AC
Distributivity
(A′)ᵢⱼ = aⱼᵢ
Transpose definition
(AB)′ = B′A′
Reversal law
A′ = A ⇔ aᵢⱼ = aⱼᵢ
Symmetric condition
A′ = −A ⇔ aᵢⱼ = −aⱼᵢ, aᵢᵢ = 0
Skew-symmetric condition
A = ½(A + A′) + ½(A − A′)
Symmetric–skew decomposition
A A⁻¹ = A⁻¹A = I
Definition of inverse
A = I·A
Row-operation starting identity
A = A·I
Column-operation starting identity

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Order of a matrix easy

If a matrix has order 3 × 4, how many elements does it have?

Q2 Types of matrices easy

Which of the following matrices is a diagonal matrix?

Q3 Types of matrices easy

In the identity matrix I of order n, which statement is correct?

Q4 Order of a matrix medium

A matrix has 18 elements. How many different orders (m × n pairs) can this matrix have?

Q5 Equality of matrices medium

If [[x+3, 2y+x],[0,-1]] = [[0,-7],[0,-1]], find the value of x + y.

Q6 Addition of matrices medium

If A = [[1,2],[3,4]] and B = [[4,3],[2,1]], what is A + B?

Q7 Multiplication of matrices medium

A is a matrix of order 3×2 and B is a matrix of order 2×4. What is the order of AB, and is BA defined?

Q8 Multiplication of matrices medium

If A = [[1,2],[3,4]] and B = [[2,0],[1,2]], find AB.

Q9 Transpose properties medium

If A and B are matrices such that AB is defined, then (AB)′ equals:

Q10 Symmetric and skew-symmetric matrices medium

A square matrix A is called skew-symmetric if:

Q11 Inverse using elementary operations hard

Using elementary row operations, find the inverse of A = [[2,1],[1,1]].

Q12 Symmetric–skew decomposition hard

A is symmetric and B is skew-symmetric such that A + B = [[2,3],[5,-1]]. Find B.

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Construct a 3 × 2 matrix whose elements are given by aᵢⱼ = ½|i − 3j|.Construction of a matrix from a rule

The matrix has order 3×2, so i = 1, 2, 3 and j = 1, 2. Compute each entry:

  • a11 = ½|1−3(1)| = ½|1−3| = ½(2) = 1
  • a12 = ½|1−3(2)| = ½|1−6| = ½(5) = 5/2
  • a21 = ½|2−3(1)| = ½|2−3| = ½(1) = 1/2
  • a22 = ½|2−3(2)| = ½|2−6| = ½(4) = 2
  • a31 = ½|3−3(1)| = ½|0| = 0
  • a32 = ½|3−3(2)| = ½|3−6| = ½(3) = 3/2

Therefore the required matrix is A = [[1, 5/2], [1/2, 2], [0, 3/2]].

2 Find X and Y if X + Y = [[7,0],[2,5]] and X − Y = [[3,0],[0,3]].Addition and subtraction of matrices

Adding the two given equations: (X+Y) + (X−Y) = [[7,0],[2,5]] + [[3,0],[0,3]] = [[10,0],[2,8]], so 2X = [[10,0],[2,8]], giving X = [[5,0],[1,4]].

Subtracting the second equation from the first: (X+Y) − (X−Y) = [[7,0],[2,5]] − [[3,0],[0,3]] = [[4,0],[2,2]], so 2Y = [[4,0],[2,2]], giving Y = [[2,0],[1,1]].

Verification: X + Y = [[5,0],[1,4]] + [[2,0],[1,1]] = [[7,0],[2,5]], and X − Y = [[5,0],[1,4]] − [[2,0],[1,1]] = [[3,0],[0,3]].

3 Simplify: cos θ [[cos θ, sin θ], [−sin θ, cos θ]] + sin θ [[sin θ, −cos θ], [cos θ, sin θ]].Scalar multiplication and addition of matrices

Multiply each matrix by its scalar first:

cos θ [[cos θ, sin θ], [−sin θ, cos θ]] = [[cos²θ, sin θ cos θ], [−sin θ cos θ, cos²θ]]

sin θ [[sin θ, −cos θ], [cos θ, sin θ]] = [[sin²θ, −sin θ cos θ], [sin θ cos θ, sin²θ]]

Adding the two matrices entry-wise:

  • (1,1): cos²θ + sin²θ = 1
  • (1,2): sin θ cos θ − sin θ cos θ = 0
  • (2,1): −sin θ cos θ + sin θ cos θ = 0
  • (2,2): cos²θ + sin²θ = 1

So the sum simplifies to [[1,0],[0,1]] = I, the identity matrix.

4 If A′ = [[3,4],[-1,2],[0,1]] and B = [[-1,2,1],[1,2,3]], verify that (A + B)′ = A′ + B′.Transpose of a matrix

Since A′ = [[3,4],[-1,2],[0,1]] (order 3×2), A itself (its transpose) has order 2×3: A = [[3,-1,0],[4,2,1]].

Step 1: Compute A + B. Both A and B have order 2×3, so A + B = [[3-1, -1+2, 0+1],[4+1, 2+2, 1+3]] = [[2,1,1],[5,4,4]].

Step 2: Compute (A+B)′ by interchanging rows and columns: (A+B)′ = [[2,5],[1,4],[1,4]].

Step 3: Compute B′. B = [[-1,2,1],[1,2,3]], so B′ = [[-1,1],[2,2],[1,3]].

Step 4: Compute A′ + B′ = [[3,4],[-1,2],[0,1]] + [[-1,1],[2,2],[1,3]] = [[2,5],[1,4],[1,4]].

Since (A+B)′ = [[2,5],[1,4],[1,4]] = A′ + B′, the property (A+B)′ = A′ + B′ is verified.

5 Using elementary row transformations, find the inverse of A = [[1,2],[2,-1]].Inverse of a matrix using elementary operations

Write A = IA: [[1,2],[2,-1]] = [[1,0],[0,1]] A.

Step 1: Apply R₂ → R₂ − 2R₁. Left side: row2 becomes (2−2(1), −1−2(2)) = (0,−5). Right side: row2 becomes (0,1) − 2(1,0) = (−2,1).

Now: [[1,2],[0,-5]] = [[1,0],[-2,1]] A.

Step 2: Apply R₂ → R₂ ÷ (−5). Left row2 becomes (0,1). Right row2 becomes (−2/−5, 1/−5) = (2/5, −1/5).

Now: [[1,2],[0,1]] = [[1,0],[2/5,-1/5]] A.

Step 3: Apply R₁ → R₁ − 2R₂. Left row1 becomes (1−0, 2−2) = (1,0). Right row1 becomes (1,0) − 2(2/5,−1/5) = (1−4/5, 0+2/5) = (1/5, 2/5).

Now: [[1,0],[0,1]] = [[1/5,2/5],[2/5,-1/5]] A.

The left side is now I, so A⁻¹ = [[1/5, 2/5], [2/5, -1/5]].

Check: A·A⁻¹ = [[1,2],[2,-1]]·[[1/5,2/5],[2/5,-1/5]] = [[1/5+4/5, 2/5-2/5],[2/5-2/5, 4/5+1/5]] = [[1,0],[0,1]] = I.

6 If A = [[3,-2],[4,-2]], find the value of k such that A² = kA − 2I.Matrix multiplication and matrix equations

Step 1: Compute A². A² = A·A = [[3,-2],[4,-2]]·[[3,-2],[4,-2]].

Row1·Col1 = 3(3)+(-2)(4) = 9-8 = 1. Row1·Col2 = 3(-2)+(-2)(-2) = -6+4 = -2.

Row2·Col1 = 4(3)+(-2)(4) = 12-8 = 4. Row2·Col2 = 4(-2)+(-2)(-2) = -8+4 = -4.

So A² = [[1,-2],[4,-4]].

Step 2: Write kA − 2I. kA = [[3k,-2k],[4k,-2k]], and 2I = [[2,0],[0,2]], so kA − 2I = [[3k-2, -2k],[4k, -2k-2]].

Step 3: Equate A² = kA − 2I entry by entry:

  • 1 = 3k − 2 ⇒ 3k = 3 ⇒ k = 1
  • −2 = −2k ⇒ k = 1
  • 4 = 4k ⇒ k = 1
  • −4 = −2k − 2 ⇒ −2k = −2 ⇒ k = 1

All four equations give the same value, so k = 1.

Previous-year board questions 4

Q1 If A is a square matrix such that A² = A, find the value of (I + A)² − 3A. 2023 2 marks

Expand (I + A)² using distributivity (I and A being square matrices of the same order commute with themselves and with I):

(I + A)² = (I + A)(I + A) = I² + IA + AI + A² = I + A + A + A² = I + 2A + A².

Since A² = A (given), substitute:

(I + A)² = I + 2A + A = I + 3A.

Therefore, (I + A)² − 3A = (I + 3A) − 3A = I.

The required value is the identity matrix I.

Q2 Using elementary row transformations, find the inverse of the matrix A = [[2,-3],[-1,2]]. 2024 3 marks

Write A = IA: [[2,-3],[-1,2]] = [[1,0],[0,1]] A.

Step 1: Apply R₁ → R₁ + 2R₂. Left row1 becomes (2+2(-1), -3+2(2)) = (0,1). Right row1 becomes (1,0) + 2(0,1) = (1,2).

Now: [[0,1],[-1,2]] = [[1,2],[0,1]] A.

Step 2: Interchange R₁ ↔ R₂: [[-1,2],[0,1]] = [[0,1],[1,2]] A.

Step 3: Apply R₁ → R₁ − 2R₂. Left row1 becomes (-1-0, 2-2) = (-1,0). Right row1 becomes (0,1) − 2(1,2) = (-2,-3).

Now: [[-1,0],[0,1]] = [[-2,-3],[1,2]] A.

Step 4: Apply R₁ → R₁ × (−1). Left row1 becomes (1,0). Right row1 becomes (2,3).

Now: [[1,0],[0,1]] = [[2,3],[1,2]] A.

So A⁻¹ = [[2,3],[1,2]].

Check: A·A⁻¹ = [[2,-3],[-1,2]]·[[2,3],[1,2]] = [[4-3, 6-6],[-2+2,-3+4]] = [[1,0],[0,1]] = I.

Q3 A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are: Market I — 10,000 units of x, 2,000 units of y, 18,000 units of z; Market II — 6,000 units of x, 20,000 units of y, 8,000 units of z. If the unit sale prices of x, y, z are ₹2.50, ₹1.50, and ₹1.00 respectively, find the total revenue in each market using matrix multiplication. 2022 5 marks

Represent the sales data as a 2×3 matrix S (rows = markets, columns = x, y, z) and the unit prices as a 3×1 column matrix P:

S = [[10000, 2000, 18000], [6000, 20000, 8000]], P = [[2.50], [1.50], [1.00]].

Since S is 2×3 and P is 3×1, the product SP is defined and gives a 2×1 matrix of total revenues:

Revenue in Market I = 10000(2.50) + 2000(1.50) + 18000(1.00) = 25000 + 3000 + 18000 = ₹46,000.

Revenue in Market II = 6000(2.50) + 20000(1.50) + 8000(1.00) = 15000 + 30000 + 8000 = ₹53,000.

So SP = [[46000],[53000]]. The total revenue is ₹46,000 in Market I and ₹53,000 in Market II.

Q4 If A = [[1,2],[2,3]] satisfies the relation A² − kA − I = O, where I is the identity matrix and O is the zero matrix, find the value of k. 2023 4 marks

Step 1: Compute A². A² = [[1,2],[2,3]]·[[1,2],[2,3]].

Row1·Col1 = 1(1)+2(2) = 5. Row1·Col2 = 1(2)+2(3) = 8.

Row2·Col1 = 2(1)+3(2) = 8. Row2·Col2 = 2(2)+3(3) = 13.

So A² = [[5,8],[8,13]].

Step 2: Write kA + I. kA = [[k,2k],[2k,3k]], I = [[1,0],[0,1]], so kA + I = [[k+1, 2k],[2k, 3k+1]].

Step 3: Use A² − kA − I = O, i.e., A² = kA + I, and equate entries:

  • 5 = k + 1 ⇒ k = 4
  • 8 = 2k ⇒ k = 4
  • 8 = 2k ⇒ k = 4
  • 13 = 3k + 1 ⇒ 3k = 12 ⇒ k = 4

All four entries are consistent, so k = 4.

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