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About Matrices — Class 10 ICSE

Perform matrix operations including addition, subtraction, and multiplication of 2x2 matrices. This topic is part of the ICSE Class 10 mathematics syllabus (chapter: Chapter 10). On this page you can practice 60 questions across three difficulty levels — 20 easy, 20 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 25-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

What you'll learn in Matrices

  • Introduction to Matrices: Organizing Data Systematically
  • Matrix Addition and Subtraction: Combining Data
  • Scalar Multiplication: Scaling Matrix Values
  • Matrix Multiplication: The 'Row by Column' Rule
  • Summary, Special Matrices & Exam Preparation

Interactive lesson · about 15 minutes · checkpoint question after every unit

Matrices — solved examples for Class 10 ICSE

Example 1easy

What is the order of the matrix A=(3542)A = \begin{pmatrix} 3 & 5 \\ 4 & -2 \end{pmatrix}?
  1. A)1×21 \times 2
  2. B)2×12 \times 1
  3. C)2×22 \times 2
  4. D)4×14 \times 1

Step-by-step solution

  1. Count rows: 2 rows (horizontal lines of numbers).
  2. Count columns: 2 columns (vertical lines of numbers).
  3. Order = rows x columns
    =2×2= 2 \times 2

Answer: 2×22 \times 2

Example 2medium

If A=(1221)A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} and B=(2112)B = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}, find A(BA)A(BA).
  1. A)(22212122)\begin{pmatrix} 22 & 21 \\ 21 & 22 \end{pmatrix}
  2. B)(21222221)\begin{pmatrix} 21 & 22 \\ 22 & 21 \end{pmatrix}
  3. C)(20232320)\begin{pmatrix} 20 & 23 \\ 23 & 20 \end{pmatrix}
  4. D)(22222121)\begin{pmatrix} 22 & 22 \\ 21 & 21 \end{pmatrix}

Step-by-step solution

  1. Find BABA:
    BA=(2(1)+1(2)2(2)+1(1)1(1)+2(2)1(2)+2(1))=(4554)BA = \begin{pmatrix} 2(1)+1(2) & 2(2)+1(1) \\ 1(1)+2(2) & 1(2)+2(1) \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 5 & 4 \end{pmatrix}
  2. Find A(BA)A(BA):
    A(4554)=(1(4)+2(5)1(5)+2(4)2(4)+1(5)2(5)+1(4))A \begin{pmatrix} 4 & 5 \\ 5 & 4 \end{pmatrix} = \begin{pmatrix} 1(4)+2(5) & 1(5)+2(4) \\ 2(4)+1(5) & 2(5)+1(4) \end{pmatrix}
  3. Result:
    =(14131314)= \begin{pmatrix} 14 & 13 \\ 13 & 14 \end{pmatrix}

Answer: (22212122)\begin{pmatrix} 22 & 21 \\ 21 & 22 \end{pmatrix}

Example 3hard

If X=(4112)X = \begin{pmatrix} 4 & 1 \\ -1 & 2 \end{pmatrix}, show that 6XX2=9I6X - X^2 = 9I. What is X2X^2?
  1. A)(15663)\begin{pmatrix} 15 & 6 \\ -6 & 3 \end{pmatrix}
  2. B)(16664)\begin{pmatrix} 16 & 6 \\ -6 & 4 \end{pmatrix}
  3. C)(15663)\begin{pmatrix} 15 & 6 \\ 6 & 3 \end{pmatrix}
  4. D)(17663)\begin{pmatrix} 17 & 6 \\ -6 & 3 \end{pmatrix}

Step-by-step solution

  1. X2=X×XX^2 = X \times X:
  2. (1,1):
    4(4)+1(1)=154(4) + 1(-1) = 15
  3. (1,2):
    4(1)+1(2)=64(1) + 1(2) = 6
  4. (2,1):
    1(4)+2(1)=6-1(4) + 2(-1) = -6
  5. (2,2):
    1(1)+2(2)=3-1(1) + 2(2) = 3
  6. So X2=(15663)X^2 = \begin{pmatrix} 15 & 6 \\ -6 & 3 \end{pmatrix}
  7. Verify:
    6XX2=(2415666+6123)=(9009)=9I  6X - X^2 = \begin{pmatrix} 24-15 & 6-6 \\ -6+6 & 12-3 \end{pmatrix} = \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix} = 9I \; \checkmark

Answer: (15663)\begin{pmatrix} 15 & 6 \\ -6 & 3 \end{pmatrix}

Practice questions on Matrices

  1. Q1.easy

    If A=(2513)A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} and B=(4126)B = \begin{pmatrix} 4 & 1 \\ 2 & 6 \end{pmatrix}, find A+BA + B.
    1. A)(6639)\begin{pmatrix} 6 & 6 \\ 3 & 9 \end{pmatrix}
    2. B)(6439)\begin{pmatrix} 6 & 4 \\ 3 & 9 \end{pmatrix}
    3. C)(6629)\begin{pmatrix} 6 & 6 \\ 2 & 9 \end{pmatrix}
    4. D)(8639)\begin{pmatrix} 8 & 6 \\ 3 & 9 \end{pmatrix}
    Show answer

    Answer: (6639)\begin{pmatrix} 6 & 6 \\ 3 & 9 \end{pmatrix}

    Hint: Add corresponding elements: top-left + top-left, top-right + top-right, etc.

  2. Q2.easy

    If A=(7238)A = \begin{pmatrix} 7 & 2 \\ 3 & 8 \end{pmatrix} and B=(4125)B = \begin{pmatrix} 4 & 1 \\ 2 & 5 \end{pmatrix}, find ABA - B.
    1. A)(3113)\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}
    2. B)(3313)\begin{pmatrix} 3 & 3 \\ 1 & 3 \end{pmatrix}
    3. C)(113513)\begin{pmatrix} 11 & 3 \\ 5 & 13 \end{pmatrix}
    4. D)(3153)\begin{pmatrix} 3 & 1 \\ 5 & 3 \end{pmatrix}
    Show answer

    Answer: (3113)\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}

    Hint: Subtract corresponding elements of B from A.

  3. Q3.easy

    Find 3A3A if A=(2145)A = \begin{pmatrix} 2 & -1 \\ 4 & 5 \end{pmatrix}.
    1. A)(631215)\begin{pmatrix} 6 & -3 \\ 12 & 15 \end{pmatrix}
    2. B)(611215)\begin{pmatrix} 6 & -1 \\ 12 & 15 \end{pmatrix}
    3. C)(63415)\begin{pmatrix} 6 & -3 \\ 4 & 15 \end{pmatrix}
    4. D)(5278)\begin{pmatrix} 5 & 2 \\ 7 & 8 \end{pmatrix}
    Show answer

    Answer: (631215)\begin{pmatrix} 6 & -3 \\ 12 & 15 \end{pmatrix}

    Hint: Multiply every element of A by 3.

  4. Q4.medium

    If A=(1221)A = \begin{pmatrix} 1 & -2 \\ 2 & -1 \end{pmatrix} and B=(3221)B = \begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}, find 2BA22B - A^2.
    1. A)(3441)\begin{pmatrix} 3 & 4 \\ -4 & -1 \end{pmatrix}
    2. B)(9445)\begin{pmatrix} 9 & 4 \\ -4 & 5 \end{pmatrix}
    3. C)(3441)\begin{pmatrix} 3 & 4 \\ 4 & -1 \end{pmatrix}
    4. D)(3441)\begin{pmatrix} 3 & -4 \\ -4 & -1 \end{pmatrix}
    Show answer

    Answer: (3441)\begin{pmatrix} 3 & 4 \\ -4 & -1 \end{pmatrix}

    Hint: Find A2A^2 first, then 2B2B, then subtract.

  5. Q5.medium

    If A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, B=(2142)B = \begin{pmatrix} 2 & 1 \\ 4 & 2 \end{pmatrix} and C=(5174)C = \begin{pmatrix} 5 & 1 \\ 7 & 4 \end{pmatrix}, compute A(B+C)A(B + C).
    1. A)(29156533)\begin{pmatrix} 29 & 15 \\ 65 & 33 \end{pmatrix}
    2. B)(29336515)\begin{pmatrix} 29 & 33 \\ 65 & 15 \end{pmatrix}
    3. C)(15293365)\begin{pmatrix} 15 & 29 \\ 33 & 65 \end{pmatrix}
    4. D)(29153365)\begin{pmatrix} 29 & 15 \\ 33 & 65 \end{pmatrix}
    Show answer

    Answer: (29156533)\begin{pmatrix} 29 & 15 \\ 65 & 33 \end{pmatrix}

    Hint: First compute B+CB + C, then multiply AA by the result.

  6. Q6.medium

    If A=(2102)A = \begin{pmatrix} 2 & 1 \\ 0 & -2 \end{pmatrix}, B=(4132)B = \begin{pmatrix} 4 & 1 \\ -3 & -2 \end{pmatrix} and C=(3214)C = \begin{pmatrix} -3 & 2 \\ -1 & 4 \end{pmatrix}, find A2+AC5BA^2 + AC - 5B.
    1. A)(142176)\begin{pmatrix} -14 & 2 \\ 17 & 6 \end{pmatrix}
    2. B)(142176)\begin{pmatrix} -14 & -2 \\ 17 & 6 \end{pmatrix}
    3. C)(142176)\begin{pmatrix} 14 & 2 \\ -17 & 6 \end{pmatrix}
    4. D)(142176)\begin{pmatrix} -14 & 2 \\ 17 & -6 \end{pmatrix}
    Show answer

    Answer: (142176)\begin{pmatrix} -14 & 2 \\ 17 & 6 \end{pmatrix}

    Hint: Calculate A2A^2, ACAC, and 5B5B separately, then combine.

  7. Q7.hard

    Show that (1221)\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} is a solution of X22X3I=OX^2 - 2X - 3I = O. What is X22XX^2 - 2X?
    1. A)(3003)\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}
    2. B)(3223)\begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix}
    3. C)(3003)\begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}
    4. D)(0000)\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
    Show answer

    Answer: (3003)\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}

    Hint: If X22X3I=OX^2 - 2X - 3I = O, then X22X=3IX^2 - 2X = 3I.

  8. Q8.hard

    If A=(4444)A = \begin{pmatrix} 4 & -4 \\ -4 & 4 \end{pmatrix}, find A2A^2. If A2=pAA^2 = pA, find pp.
    1. A)p=8p = 8
    2. B)p=4p = 4
    3. C)p=16p = 16
    4. D)p=2p = 2
    Show answer

    Answer: p=8p = 8

    Hint: Compute A2A^2, then compare with pApA element by element.

  9. Q9.hard

    If A=(3412)A = \begin{pmatrix} 3 & -4 \\ -1 & 2 \end{pmatrix}, find matrix BB such that BA=IBA = I (identity matrix). What is BB?
    1. A)(121232)\begin{pmatrix} 1 & 2 \\ \frac{1}{2} & \frac{3}{2} \end{pmatrix}
    2. B)(2413)\begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix}
    3. C)(2413)\begin{pmatrix} 2 & -4 \\ 1 & -3 \end{pmatrix}
    4. D)(2413)\begin{pmatrix} -2 & -4 \\ -1 & -3 \end{pmatrix}
    Show answer

    Answer: (121232)\begin{pmatrix} 1 & 2 \\ \frac{1}{2} & \frac{3}{2} \end{pmatrix}

    Hint: Let B=(abcd)B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}. Then BA=IBA = I gives 4 equations.

These are 9 of the 60 questions available for Matrices. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.