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State Board · Class 12 · Mathematics Part 1

Lesson 2: Matrices

5 concepts · 40 practice questions with answers · Free to read

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Types of matrices, addition and scalar multiplication

From daily life

A shop register with three items and two days of sales is a 3 × 2 matrix. Building on class 11 : a determinant has a value; a matrix is only an arrangement and has no single value.

Simple definition

A matrix is an arrangement of numbers in m rows and n columns, of order m × n. Types: square (m = n), diagonal, identity (I), zero and triangular. Only matrices of the same order can be added, element by element.

Rule 1: Addition / subtraction — For A ± B the orders must be equal; operate on matching elements.

Rule 2: Scalar multiplication — In kA multiply every element by k.

Rule 3: Equality — Two matrices are equal if their orders are equal and every matching element is equal.

  • Easy : A = [[2, 1], [0, 3]], B = [[1, 4], [2, −1]]: A + B = [[3, 5], [2, 2]].
  • Medium : 3A − B = [[6, 3], [0, 9]] − [[1, 4], [2, −1]] = [[5, −1], [−2, 10]].
  • Exam-type : [[x + y, 2], [3, x − y]] = [[5, 2], [3, 1]] ⇒ x + y = 5, x − y = 1 ⇒ x = 3, y = 2.
Remember

If the orders differ, addition is impossible.

Common mistake

Adding a 3 × 2 and a 2 × 3 matrix.

Shortcut

In a matrix equality write a separate equation for every position. Memory clue : "Check the order first, then operate".

CET trap

A matrix has no single value like a determinant.

CET-style example : How many elements are there in a 2 × 3 matrix? (A) 5 (B) 6 (C) 8 (D) 9

Answer : (B). 2 × 3 = 6. Time-saving method : Rows × columns.

Spot the difference

वैशिष्ट्यFeature
कर्णाबाहेर सर्व 0all 0 off the diagonal
कर्णावर 1, बाकी 01 on the diagonal, 0 elsewhere
सर्व घटक 0every element 0

Practice questions

Easy
  1. 1What is the order of a matrix with 3 rows and 4 columns?
  2. 2How many elements are in a 3 × 4 matrix?
  3. 3Write the 2 × 2 identity matrix.
Medium
  1. 4For A = [[2, 1], [0, 3]], B = [[1, 4], [2, −1]] find A + B.
  2. 5For the same A, B find 3A − B.
  3. 6If [[x + y, 2], [3, x − y]] = [[5, 2], [3, 1]], find x and y.
Challenging
  1. 7If A = [[1, 2], [3, 4]] and A + X = [[5, 5], [5, 5]], find X.
  2. 8If 2X + [[1, 2], [0, 3]] = [[5, 4], [2, 9]], find X.
Show answers (try them yourself first)
  1. 13 × 4.
  2. 212.
  3. 3[[1, 0], [0, 1]].
  4. 4[[3, 5], [2, 2]].
  5. 5[[5, −1], [−2, 10]].
  6. 6x = 3, y = 2.
  7. 7X = [[4, 3], [2, 1]].
  8. 82X = [[4, 2], [2, 6]]; X = [[2, 1], [1, 3]].

Matrix multiplication and transpose

From daily life

If a table gives the numbers of items bought by three friends and another gives the item prices, "row × column" multiplication gives each friend's total bill.

Simple definition

AB exists when the columns of A equal the rows of B; for A (m × n) and B (n × p) the product is m × p. The (i, j) element of AB is row i of A times column j of B (sum of products). The transpose Aᵀ swaps rows and columns.

Rule 1: Row times column — A = [[1, 2], [3, 4]], B = [[2, 0], [1, 5]]: AB = [[1·2 + 2·1, 1·0 + 2·5], [3·2 + 4·1, 3·0 + 4·5]] = [[4, 10], [10, 20]].

Rule 2: Properties — In general AB ≠ BA; (AB)ᵀ = BᵀAᵀ; (A + B)ᵀ = Aᵀ + Bᵀ; AI = IA = A.

Rule 3: Symmetric / skew-symmetric — If Aᵀ = A the matrix is symmetric; if Aᵀ = −A it is skew-symmetric (zeros on the diagonal).

How is the formula obtained? : BA = [[2·1 + 0·3, 2·2 + 0·4], [1·1 + 5·3, 1·2 + 5·4]] = [[2, 4], [16, 22]] ≠ AB, so multiplication is not commutative.

  • Easy : [[1, 2], [3, 4]] × [[1], [1]] = [[3], [7]] (2×2 times 2×1 gives 2×1).
  • Medium : (2×3)(3×2): [[1, 2, 0], [3, 1, 4]] × [[2, 1], [0, 3], [1, 1]] = [[2, 7], [10, 10]].
  • Exam-type : [[0, 3], [−3, 0]] is skew-symmetric; [[1, 5], [5, 2]] is symmetric.
Remember

(AB)ᵀ = BᵀAᵀ — the order reverses.

Common mistake

Assuming AB = BA.

Shortcut

Before multiplying check: inner numbers equal, outer numbers give the order. Memory clue : "Inner same, outer is the answer" (m×n · n×p = m×p).

CET trap

AB = 0 does not force A = 0 or B = 0.

CET-style example : If A is 2 × 3 and B is 3 × 4, the order of AB is: (A) 2 × 4 (B) 3 × 3 (C) 4 × 2 (D) 2 × 3

Answer : (A). Time-saving method : Read the two outer numbers.

Spot the difference

खरे आहे का ?True?
सहसा नाहीnot in general
होयyes
होयyes

Practice questions

Easy
  1. 1A is 3 × 2, B is 2 × 5: order of AB?
  2. 2A is 2 × 3, B is 2 × 3: does AB exist?
  3. 3Write the transpose of [[1, 2], [3, 4]].
Medium
  1. 4For A = [[1, 2], [3, 4]], B = [[2, 0], [1, 5]] find AB.
  2. 5For the same A, B find BA.
  3. 6Verify (AB)ᵀ = BᵀAᵀ for the same AB.
Challenging
  1. 7Find [[1, 2, 0], [3, 1, 4]] × [[2, 1], [0, 3], [1, 1]].
  2. 8If [[0, k], [−3, 0]] is skew-symmetric, find k.
Show answers (try them yourself first)
  1. 13 × 5.
  2. 2No (3 ≠ 2).
  3. 3[[1, 3], [2, 4]].
  4. 4[[4, 10], [10, 20]].
  5. 5[[2, 4], [16, 22]].
  6. 6(AB)ᵀ = [[4, 10], [10, 20]]; BᵀAᵀ is the same.
  7. 7[[2, 7], [10, 10]].
  8. 8Aᵀ = −A ⇒ −3 = −k ⇒ k = 3.

Elementary transformations and inverse (transformation method)

From daily life

While solving equations we swap two lines, multiply a line by a number or add a multiple of one line to another — these are the three elementary transformations.

Simple definition

The three row transformations are Rᵢ ↔ Rⱼ; Rᵢ → kRᵢ (k ≠ 0); Rᵢ → Rᵢ + kRⱼ. For columns use Cᵢ. The inverse A⁻¹ satisfies AA⁻¹ = A⁻¹A = I; it exists only for a square matrix with |A| ≠ 0.

Rule 1: Set-up — Write A = IA and use only row transformations to turn the left A into I; the I on the right then becomes A⁻¹. (Do not mix row and column transformations.)

How is the formula obtained? : A = [[1, 2], [3, 4]]: R₂ → R₂ − 3R₁ gives [[1, 2], [0, −2]]; R₂ → (−½)R₂ gives [[1, 2], [0, 1]]; R₁ → R₁ − 2R₂ gives I. The right side becomes A⁻¹ = [[−2, 1], [3/2, −1/2]].

  • Easy : A = [[1, 3], [2, 7]]: R₂ → R₂ − 2R₁, R₁ → R₁ − 3R₂ gives A⁻¹ = [[7, −3], [−2, 1]].
  • Medium : A = [[2, 1], [5, 3]]: A⁻¹ = [[3, −1], [−5, 2]]; check AA⁻¹ = I.
  • Exam-type : A = [[2, 4], [1, 2]]: |A| = 0, so no inverse (a row of zeros appears).
Remember

If a row of zeros appears, the inverse does not exist.

Common mistake

Changing a row and a column in one step.

Shortcut

Finally check by multiplying AA⁻¹ = I. Memory clue : "Left A → I, right I → A⁻¹".

CET trap

When the determinant is 0 there is no inverse; do not force the transformation method.

CET-style example : The inverse of [[1, 2], [3, 4]] is (A) [[−2, 1], [3/2, −1/2]] (B) [[4, −2], [−3, 1]] (C) [[1, 2], [3, 4]] (D) [[−2, 1], [1, 3/2]]

Answer : (A). (B) is the adjoint; divide by |A| = −2. Time-saving method : Multiply the option by A to test.

Spot the difference

चिन्हSymbol
Rᵢ ↔ RⱼRᵢ ↔ Rⱼ
Rᵢ → kRᵢRᵢ → kRᵢ
Rᵢ → Rᵢ + kRⱼRᵢ → Rᵢ + kRⱼ

Practice questions

Easy
  1. 1Which matrices have an inverse?
  2. 2What is the inverse of [[1, 0], [0, 1]]?
  3. 3Does [[2, 4], [1, 2]] have an inverse?
Medium
  1. 4Find the inverse of [[1, 3], [2, 7]] by transformations.
  2. 5Find the inverse of [[2, 1], [5, 3]].
  3. 6Find the inverse of [[3, 1], [5, 2]].
Challenging
  1. 7Find the inverse of [[1, 2], [3, 4]] and verify AA⁻¹ = I.
  2. 8If A = [[1, 2], [3, 4]], what is (A⁻¹)⁻¹?
Show answers (try them yourself first)
  1. 1Square matrices with determinant ≠ 0.
  2. 2I itself.
  3. 3No; |A| = 4 − 4 = 0.
  4. 4[[7, −3], [−2, 1]].
  5. 5[[3, −1], [−5, 2]].
  6. 6[[2, −1], [−5, 3]].
  7. 7[[−2, 1], [3/2, −1/2]]; the product is I.
  8. 8A itself.

Inverse by the adjoint method

From daily life

Undoing a copy to get the original is what an inverse does. The adjoint method gives a direct formula, with no transformations needed.

Simple definition

The cofactor Cᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ, where Mᵢⱼ is the determinant left after deleting row i and column j. The adjoint adj A is the transpose of the cofactor matrix. A⁻¹ = adj A / |A| with |A| ≠ 0.

Rule 1: 2 × 2 shortcut — adj of [[a, b], [c, d]] is [[d, −b], [−c, a]]: swap the main diagonal, change signs on the other.

Rule 2: Key identity — A (adj A) = (adj A) A = |A| I.

How is the formula obtained? : A = [[2, 1, 0], [1, 3, 1], [0, 1, 2]]: |A| = 2(6 − 1) − 1(2 − 0) = 8; the cofactor matrix is [[5, −2, 1], [−2, 4, −2], [1, −2, 5]]; it is symmetric so adj A is the same. A⁻¹ = (1/8) [[5, −2, 1], [−2, 4, −2], [1, −2, 5]].

  • Easy : [[3, 1], [5, 2]]: |A| = 6 − 5 = 1; adj = [[2, −1], [−5, 3]]; A⁻¹ = [[2, −1], [−5, 3]].
  • Medium : [[2, 1], [5, 3]]: |A| = 1; A⁻¹ = [[3, −1], [−5, 2]].
  • Exam-type : [[1, 2, 3], [0, 1, 4], [1, 0, 6]]: |A| = 11 ≠ 0, so the inverse exists; adj A = [[6, −12, 5], [4, 3, −4], [−1, 2, 1]].
Remember

Do not forget to transpose the cofactor matrix to get adj A.

Common mistake

Forgetting the sign (−1)ⁱ⁺ʲ of a cofactor.

Shortcut

Sign pattern: [+ − +; − + −; + − +]. Memory clue : "Cofactor → transpose → divide by |A|".

CET trap

|adj A| = |A|ⁿ⁻¹ (|A|² for a 3×3).

CET-style example : adj of [[a, b], [c, d]] is: (A) [[d, −b], [−c, a]] (B) [[a, −b], [−c, d]] (C) [[d, b], [c, a]] (D) [[−d, b], [c, −a]]

Answer : (A). Time-saving method : Swap one diagonal, negate the other.

Spot the difference

कधी सोपीBest when
संख्या सोप्या असतानाnumbers are simple
2 × 2 किंवा सूत्र हवे असतानाa 2 × 2 or a formula is needed

Practice questions

Easy
  1. 1Write adj of [[a, b], [c, d]].
  2. 2Find the determinant of [[3, 1], [5, 2]].
  3. 3Does [[4, 2], [2, 1]] have an inverse?
Medium
  1. 4Find the inverse of [[3, 1], [5, 2]] by the adjoint method.
  2. 5Find the inverse of [[2, 1], [5, 3]].
  3. 6Find |A| for [[2, 1, 0], [1, 3, 1], [0, 1, 2]].
Challenging
  1. 7Find A⁻¹ for the 3 × 3 matrix above.
  2. 8For [[1, 2, 3], [0, 1, 4], [1, 0, 6]] give |A| and the first row of adj A.
Show answers (try them yourself first)
  1. 1[[d, −b], [−c, a]].
  2. 26 − 5 = 1.
  3. 3No; |A| = 4 − 4 = 0.
  4. 4[[2, −1], [−5, 3]].
  5. 5[[3, −1], [−5, 2]].
  6. 62(6 − 1) − 1(2 − 0) + 0 = 8.
  7. 7(1/8)[[5, −2, 1], [−2, 4, −2], [1, −2, 5]].
  8. 8|A| = 11; the first row of adj A is (6, −12, 5).

Solving linear equations with matrices

From daily life

To find the prices of two items from two purchase bills, each bill is one equation. A matrix writes all of them in a single line as AX = B.

Simple definition

A system is written AX = B (A = coefficient matrix, X = column of unknowns, B = constants). If |A| ≠ 0 the unique solution is X = A⁻¹B. If |A| = 0 there is no solution or infinitely many.

Rule 1: Inverse method — Find A⁻¹ and compute X = A⁻¹B.

Rule 2: Reduction method — Apply row transformations to [A | B] until A becomes triangular or I; then read the answers from bottom to top.

How is the formula obtained? : x + 2y = 5, 3x + 4y = 11: A = [[1, 2], [3, 4]], B = [[5], [11]]. R₂ → R₂ − 3R₁ gives [[1, 2 | 5], [0, −2 | −4]], so y = 2 and x = 5 − 4 = 1.

  • Easy : 3x + y = 7, 5x + 2y = 11: A⁻¹ = [[2, −1], [−5, 3]] gives X = A⁻¹B = [[3], [−2]], so x = 3, y = −2.
  • Medium : 2x + y = 4, x + 3y + z = 10, y + 2z = 8: |A| = 8; X = (1/8)·adj A·B = (1/8)[[8], [16], [24]] = [[1], [2], [3]].
  • Exam-type : x + 2y = 3, 2x + 4y = 7: |A| = 0; doubling the first gives 6 but the second says 7, so there is no solution.
Remember

In X = A⁻¹B the order is A⁻¹ first, then B.

Common mistake

Writing X = BA⁻¹ (wrong order).

Shortcut

Check the answer by putting it back in the equations. Memory clue : "A⁻¹ hits B from the left".

CET trap

|A| = 0 means no solution or infinitely many, not always "none".

CET-style example : The solution of x + y = 3, x − y = 1 is: (A) (2, 1) (B) (1, 2) (C) (3, 0) (D) (0, 3)

Answer : (A). Add: 2x = 4. Time-saving method : Substitute the options.

Spot the difference

निष्कर्षConclusion
एकमेव उत्तर X = A⁻¹Bunique solution X = A⁻¹B
उत्तर नाही किंवा अनंतno solution or infinitely many

Practice questions

Easy
  1. 1Solve x + y = 3, x − y = 1.
  2. 2Solve 2x + y = 5, x − y = 1.
  3. 3What is the formula for X in AX = B?
Medium
  1. 4Solve x + 2y = 5, 3x + 4y = 11.
  2. 5Solve 3x + y = 7, 5x + 2y = 11 by the inverse.
  3. 6What can you say about x + 2y = 3, 2x + 4y = 6?
Challenging
  1. 7Solve 2x + y = 4, x + 3y + z = 10, y + 2z = 8.
  2. 8Verify the above three values in all three equations.
Show answers (try them yourself first)
  1. 1x = 2, y = 1.
  2. 2Add: 3x = 6; x = 2, y = 1.
  3. 3X = A⁻¹B (|A| ≠ 0).
  4. 4x = 1, y = 2.
  5. 5x = 3, y = −2.
  6. 6|A| = 0 and both equations are the same line ⇒ infinitely many solutions.
  7. 7x = 1, y = 2, z = 3.
  8. 82+2 = 4; 1+6+3 = 10; 2+6 = 8. Correct.