Matrices: Chapter 3 Links
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3.1 | 3.2 | 3.3 | 3.4 | MCQs | QRTs
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This page on Matrices Miscellaneous Exercise of Class 12 Maths Chapter 3 would give you a complete revision of all the important concepts studied in this chapter. In this exercise, you will solve questions based on matrix operations, transpose of matrices, symmetric and skew-symmetric matrices, matrix equations, and invertibility of matrices.
Through these questions, you will also revise important properties related to addition, multiplication, transpose and inverse of matrices. Besides that, you shall also use concepts like equality of matrices, identity matrix, verification of inverse using AB = BA = I and transpose properties such as (AB)' = B'A' . These results form the foundation for advanced applications of matrices and determinants.
This miscellaneous exercise also works as a quick chapter recap before examinations, since it combines a variety of conceptual and computational problems from the complete chapter. By practising these questions, you can strengthen your understanding of matrix properties, improve calculation accuracy and revise the important concepts of Chapter 3 in one place.
Key Concepts
- Matrix: A matrix is an ordered rectangular arrangement of numbers or functions in rows and columns.
- Order of Matrix: If a matrix has m rows and n columns, then its order is m \times n .
- Row Matrix: A matrix having only one row is called a row matrix.
- Column Matrix: A matrix having only one column is called a column matrix.
- Square Matrix: A matrix is called a square matrix if the number of rows and columns are equal.
- Diagonal Matrix: A square matrix is diagonal if all elements except the principal diagonal elements are zero.
- Scalar Matrix: A diagonal matrix in which all diagonal elements are equal is called a scalar matrix.
- Identity Matrix: A matrix I is called identity matrix if diagonal elements are 1 and all other elements are 0, and AI = IA = A .
- Zero Matrix: A matrix whose all elements are zero is called a zero matrix.
- Equality of Matrices: Two matrices are equal if they are of same order and their corresponding elements are equal.
- Addition of Matrices: Two matrices can be added only when they are of the same order.
- Commutative Property: Matrix addition is commutative, i.e. A + B = B + A .
- Associative Property: Matrix addition is associative, i.e. (A + B) + C = A + (B + C) .
- Scalar Multiplication: If k is a scalar, then each element of matrix A gets multiplied by k in kA .
- Matrix Multiplication: The product AB is defined only when the number of columns of A equals the number of rows of B.
💡 Remember that matrix multiplication is not commutative in general.
- Associative Law of Multiplication: Matrix multiplication is associative, i.e. A(BC) = (AB)C .
- Distributive Law: Matrix multiplication is distributive over addition, i.e. A(B + C) = AB + AC .
- Transpose of Matrix: The transpose of a matrix is obtained by interchanging its rows and columns.
- Properties of Transpose: Important properties are (A')' = A , (A + B)' = A' + B' , and (AB)' = B'A' (Reversal Law)
- Symmetric Matrix: A matrix is symmetric if A' = A .
- Skew-Symmetric Matrix: A matrix is skew-symmetric if A' = -A .
- Invertible Matrix: A square matrix is invertible if there exists another matrix such that their product gives identity matrix.
- Inverse Condition: A matrix A has an inverse B if AB = BA = I , where I is the identity matrix.
- Uniqueness of Inverse: If a matrix has an inverse, then it is unique.
- Inverse of Product: If A and B are invertible matrices, then (AB)^{-1} = B^{-1}A^{-1} . This is called the reversal law.
- Verification Method: To verify inverse of matrices, we multiply them and check whether the result is identity matrix.
Now, let’s begin this exercise with Question 1
Question 1: Matrices Miscellaneous Exercise
1. If A and B are symmetric matrices, prove that AB - BA is a skew-symmetric matrix.
Solution
Since A and B are symmetric matrices, we have
A' = A \quad \text{and} \quad B' = B
Let
C = AB - BA
To prove that C is skew-symmetric, we need to show that
C' = -C
Now, taking transpose of both sides,
C' = (AB - BA)'
Using the property (AB)' = B'A' , we get
C' = (BA)' - (AB)'
= A'B' - B'A'
Since A' = A and B' = B ,
So, C' = AB - BA
= -(BA - AB)
= -(AB - BA)
i.e. C = -C
Hence, AB - BA is a skew-symmetric matrix.
You may already be following Maths Better for important concepts and exam-oriented questions, like integration tricks and solving linear equations using the Matrix Method etc. Similarly, this NCERT Solutions series for Class 12 Maths will help strengthen your concepts and improve problem-solving skills.
Now, let’s move on to the next question of Matrices Miscellaneous Exercise.
Question 2: Miscellaneous Exercise Matrices
2. Show that the matrix B'AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.
Solution
Let
C = B'AB
Now, taking transpose of C ,
C' = (B'AB)'
Using the property (ABC)' = C'B'A' , we get
C' = B'A'(B')'
Since (B')' = B ,
C' = B'A'B
Case 1: If A is symmetric, then
A' = A
Therefore,
C' = B'AB = C
Hence, B'AB is symmetric.
Case 2: If A is skew-symmetric, then
A' = -A
Therefore, C' = B'(-A)B
= -B'AB
= -C
Hence, B'AB is skew-symmetric.
Question 3: Miscellaneous Exercise Chapter 3
3. Find the values of x, y, z if the matrix A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} satisfies the equation A'A = I .
Solution
Given,
A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}
Therefore, transpose of A is
A' = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}
Now,
A'A = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}
Multiplying the matrices,
A'A = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}
Since A'A = I , we compare corresponding elements with
I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
Thus,
2x^2 = 1, \quad 6y^2 = 1, \quad 3z^2 = 1
Therefore,
x = \pm \frac{1}{\sqrt{2}}, \quad y = \pm \frac{1}{\sqrt{6}}, \quad z = \pm \frac{1}{\sqrt{3}}
Question 4: Solve for x
4. For what values of x
[1 \quad 2 \quad 1]\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O
Solution
First, multiply
\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix}
We get,
= \begin{bmatrix} (1\times0 + 2\times2 + 0\times x) \\ (2\times0 + 0\times2 + 1\times x) \\ (1\times0 + 0\times2 + 2\times x) \end{bmatrix}
= \begin{bmatrix} 4 \\ x \\ 2x \end{bmatrix}
Now, multiply with [1 \quad 2 \quad 1] .
[1 \quad 2 \quad 1]\begin{bmatrix} 4 \\ x \\ 2x \end{bmatrix} = O
Therefore,
1(4) + 2(x) + 1(2x) = 0
or 4 + 2x + 2x = 0
or 4 + 4x = 0
4x = -4
So, x = -1
Therefore, the required value of x is -1 .
Try solving the next question on your own first, and then compare your approach with the solution given below. We have already solved such questions in Matrices 3.3 also.
Question 5: Matrices Miscellaneous
5. If A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} , show that A^2 - 5A + 7I = O
Solution
Given,
A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}
First, find A^2 .
So, A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}
= \begin{bmatrix} (3\times3 + 1\times-1) & (3\times1 + 1\times2) \\ (-1\times3 + 2\times-1) & (-1\times1 + 2\times2) \end{bmatrix}
= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}
Now,
5A = 5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}
Also,
7I = 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}
Therefore, A^2 - 5A + 7I
= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}
= \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}
So, A^2 - 5A + 7I
= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O
Hence proved.
Question 6: Matrices Miscellaneous
6. Find x , if
[x \quad -5 \quad -1]\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O
Solution
First, multiply \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix}
= \begin{bmatrix} (1\times x + 0\times4 + 2\times1) \\ (0\times x + 2\times4 + 1\times1) \\ (2\times x + 0\times4 + 3\times1) \end{bmatrix}
= \begin{bmatrix} x + 2 \\ 9 \\ 2x + 3 \end{bmatrix}
Now, multiply with [x \quad -5 \quad -1] .
[x \quad -5 \quad -1]\begin{bmatrix} x + 2 \\ 9 \\ 2x + 3 \end{bmatrix} = O
x(x+2) + (-5)(9) + (-1)(2x+3) = 0
or x^2 + 2x - 45 - 2x - 3 = 0
or x^2 - 48 = 0
i.e. x^2 = 48
gives x = \pm 4\sqrt{3}
Therefore, the required values of x are \pm 4\sqrt{3} .
Matrix-based questions are generally direct, but presentation and correct steps play an important role in scoring full marks. You can check that in the following word problem.
Question 7: Matrices Word Problem
7. A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:
\begin{array}{c|ccc} \text{Market} & x & y & z \\ \hline I & 10000 & 2000 & 18000 \\ II & 6000 & 20000 & 8000 \end{array}
(a) If unit sale prices of x, y, z are ₹2.50, ₹1.50 and ₹1.00 respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are ₹2.00, ₹1.00 and 50 paise respectively, find the gross profit.
Solution
The sales matrix is
A = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}
The price matrix is
B = \begin{bmatrix} 2.5 \\ 1.5 \\ 1 \end{bmatrix}
(a) Total Revenue
Total revenue in each market is obtained by AB .
So, AB = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.5 \\ 1.5 \\ 1 \end{bmatrix}
= \begin{bmatrix} (10000\times2.5 + 2000\times1.5 + 18000\times1) \\ (6000\times2.5 + 20000\times1.5 + 8000\times1) \end{bmatrix}
= \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}
Hence, the total revenue in Market I is ₹46,000 and in Market II is ₹53,000.
(b) Gross Profit
The cost matrix is
C = \begin{bmatrix} 2 \\ 1 \\ 0.5 \end{bmatrix}
Total cost in each market therefore,
is AC = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 0.5 \end{bmatrix}
= \begin{bmatrix} (10000\times2 + 2000\times1 + 18000\times0.5) \\ (6000\times2 + 20000\times1 + 8000\times0.5) \end{bmatrix}
= \begin{bmatrix} 31000 \\ 36000 \end{bmatrix}
Therefore, gross profit is (Revenue – Cost)
\begin{bmatrix} 46000 \\ 53000 \end{bmatrix} - \begin{bmatrix} 31000 \\ 36000 \end{bmatrix} = \begin{bmatrix} 15000 \\ 17000 \end{bmatrix}
Hence, the gross profit in Market I is ₹15,000 and in Market II is ₹17,000.
NCERT Class 12 Maths has a total of 61 exercises across Part 1 and Part 2 including the miscellaneous ones. I’ll cover each exercise one by one with clear explanations and exam-oriented solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter.
Here’s the next question.
Question 8: Find a Matrix – Misc. Ex.
8. Find the matrix X such that
X\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}
Solution
Since
\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}
is a matrix of order 2 \times 3 and the product is equal to a matrix of order 2 \times 3 , the matrix X must be a square matrix of order 2 \times 2 .
Let
X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}
Now,
\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}
Multiplying the matrices,
\begin{bmatrix} a + 4b & 2a + 5b & 3a + 6b \\ c + 4d & 2c + 5d & 3c + 6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}
By equality of matrices, corresponding elements are equal.
a + 4b = -7
2a + 5b = -8
Solving these equations,
b = -2, \quad a = 1
Also,
c + 4d = 2
2c + 5d = 4
Solving these equations,
d = 0, \quad c = 2
Therefore,
X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}
Choose the correct answer in the following 3 questions of miscellaneous exercise of chapter 3.
Question 9: Matrices Miscellaneous – MCQ
9. If A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} is such that A^2 = I , then
- (A) 1 + \alpha^2 + \beta\gamma = 0
- (B) 1 - \alpha^2 + \beta\gamma = 0
- (C) 1 - \alpha^2 - \beta\gamma = 0
- (D) 1 + \alpha^2 - \beta\gamma = 0
Solution
Given,
A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}
Now, find A^2 .
Now, A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}
= \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta - \alpha\beta \\ \alpha\gamma - \alpha\gamma & \beta\gamma + \alpha^2 \end{bmatrix}
= \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix}
Since A^2 = I , we compare corresponding elements with
I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
Thus,
\alpha^2 + \beta\gamma = 1
or 1 - \alpha^2 - \beta\gamma = 0
✅️ Hence, the correct answer is (C).
Detailed video explanations of many NCERT questions are also available on the Maths Better YouTube channel, you must check them as well.
Question 10: Miscellaneous Matrices – MCQ
10. If the matrix A is both symmetric and skew-symmetric, then
- (A) A is a diagonal matrix
- (B) A is a zero matrix
- (C) A is a square matrix
- (D) None of these
Solution
If A is symmetric, then
A' = A
If A is skew-symmetric, then
A' = -A
Therefore,
A = -A
2A = O
A = O
Hence, A is a zero matrix.
✅️ Therefore, the correct answer is (B).
Question 11: Matrices Miscellaneous – MCQ
11. If A is a square matrix such that A^2 = A , then (I + A)^3 - 7A is equal to
- (A) A
- (B) I - A
- (C) I
- (D) 3A
Solution
Given,
A^2 = A
Now,
(I + A)^3 = (I + A)(I + A)(I + A)
Using expansion, (we can use it here because AI = A = AI )
(I + A)^3 = I + 3A + 3A^2 + A^3
Since A^2 = A , we get
A^3 = A\cdot A^2 = A\cdot A = A^2 = A
Therefore,
(I + A)^3 = I + 3A + 3A + A
= I + 7A
Hence,
(I + A)^3 - 7A = I
✅️ Therefore, the correct answer is (C).
Common Mistakes to Avoid
- Ignoring matrix order: Students often forget to check whether matrix multiplication is actually possible. Always verify that the number of columns of first matrix equals the number of rows of second matrix.
- Wrong order in multiplication: Matrix multiplication is not commutative, so in general AB \ne BA Changing the order may completely change the result.
- Mistakes in transpose properties: While using transpose, students sometimes write (AB)' = A'B' which is incorrect. The correct property is (AB)' = B'A'
- Errors in equality of matrices: Two matrices are equal only when both their orders and corresponding elements are equal.
- Arithmetic mistakes during multiplication: Small calculation errors in row-column multiplication often lead to incorrect final answers.
- Confusing symmetric and skew-symmetric matrices: In skew-symmetric matrices, diagonal elements are always zero because a_{ii} = -a_{ii} which gives a_{ii} = 0
- Skipping verification in objective questions: In MCQs involving identities like A^2 = I or A^2 = A students sometimes choose options without simplifying properly.
Continue Learning
With this, we have completed the Miscellaneous Exercise of Chapter 3 – Matrices. This exercise revised all the major concepts of matrices including matrix multiplication, transpose, symmetric and skew-symmetric matrices, equality of matrices, and important algebraic properties.
These concepts form the foundation for the next chapter on Determinants, where we will study methods to find inverse of matrices and solve systems of linear equations more efficiently.
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All the best and keep learning 👍
