Careers360 Logo
NCERT Solutions for Class 12 Maths Chapter 3 Matrices

NCERT Solutions for Class 12 Maths Chapter 3 Matrices

Edited By Ramraj Saini | Updated on Sep 13, 2023 08:58 PM IST | #CBSE Class 12th

Matrices Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 3 Matrices are provided here. Matrices is an important and powerful tool in mathematics and it's basically introduced to solve simultaneous linear equations. maths chapter 3 class 12 includes the type of matrices, operations of matrices, and many more concepts. These concepts are used to solve challenging problems of 3d geometry, and determinants Therefore matrices class 12 solutions become important to get command of concepts. The NCERT Solutions are prepared by expert team according the latest syllabus of CBSE 2023. Also, Students can refer matrices class 12 NCERT solutions which covers all the questions of NCERT Books for Class 12 Maths.

The practice from NCERT solutions for class 12 maths Chapter 3 Matrices is very important to score well in academics as well as in competitive exams. Matrices class 12 also includes exercises you can also solve to build hold on the concepts. All the questions in NCERT maths class 12 chapter 3 are explained in a detailed manner. You can also refer to class 12 maths ncert solutions chapter 3 matrices by clicking on the link NCERT solutions for Class 12.

Apply to Aakash iACST Scholarship Test 2024

Applications for Admissions are open.

Also read:

Aakash iACST Scholarship Test 2024

Get up to 90% scholarship on NEET, JEE & Foundation courses

ALLEN Digital Scholarship Admission Test (ADSAT)

Register FREE for ALLEN Digital Scholarship Admission Test (ADSAT)

Matrices Class 12 Questions And Answers PDF Free Download

Download PDF

NCERT Solutions for Class 12 Maths Chapter 3 Matrices - Important Formulae

Matrix Definition and Properties:

A matrix is an ordered rectangular array of numbers or functions.

A matrix of order m × n consists of m rows and n columns.

The order of a matrix is written as m × n, where m is the number of rows and n is the number of columns.

A matrix is called a square matrix when m = n.

A diagonal matrix A = [aij]m×m has aij = 0 when i ≠ j.

A scalar matrix A = [aij]n×n has aij = 0 when i ≠ j, aij = k (where k is a constant)

when i = j.

An identity matrix A = [aij]n×n has aij = 1 when i = j and aij = 0 when i ≠ j.

A zero matrix contains all its elements as zero.

A column matrix is of the form [A]n × 1.

A row matrix is of the form [A]1 × n.

Equality of Matrices:

Two matrices A and B are equal (A = B) if they have the same order and aij = bij for all the corresponding values of i and j.

Operations on Matrices:

Matrix Addition:

  • If A = [aij]m × n and B = [bij]m × n, then A + B = [aij + bij]m × n.

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Matrix Subtraction:

  • If A = [aij]m × n and B = [bij]m × n, then A - B = [aij - bij]m × n.

TOEFL ® Registrations 2024

Accepted by more than 11,000 universities in over 150 countries worldwide

PTE Exam 2024 Registrations

Register now for PTE & Save 5% on English Proficiency Tests with ApplyShop Gift Cards

Multiplication of a Matrix by Scalar:

  • Let A = [aij]m × n be a matrix and k is a scalar, then kA is obtained by multiplying each element of A by the scalar k, i.e., kA = [kaij]m × n.

Multiplication of Matrices:

  • Let A be an m × p matrix, and B be a p × n matrix. Their product AB is defined if the number of columns in A is equal to the number of rows in B. The resulting matrix is an m × n matrix, and the elements are calculated as follows: (AB)ij = Σ(ai * bj), where the sum is taken over all values of p.

Transpose of a Matrix:

The transpose of a matrix A, denoted as AT, is obtained by interchanging its rows and columns.

Symmetric and Skew-Symmetric Matrices:

A matrix A is symmetric if A = AT (i.e., it is equal to its transpose).

A matrix A is skew-symmetric if AT = -A (i.e., the transpose of A is equal to the negative of A).

Elementary Operation or Transformation of a Matrix:

Elementary row operations include:

  • Interchanging any two rows.

  • Multiplying a row by a non-zero scalar.

  • Adding or subtracting a multiple of one row from another row.

Inverse of a Matrix by Elementary Operations:

You can find the inverse of a matrix using elementary row operations. If the matrix A is invertible, you can transform it into the identity matrix I through row operations on an augmented matrix [A | I], where I is the identity matrix of the same order as A. If this process is successful, the resulting matrix on the left will be I, and the matrix on the right will be the inverse of A.

Free download Matrices Class 12 Questions And Answers for CBSE Exam.

Class 12 Maths Chapter 3 Question Answer (Intext Questions and Exercise)

Matrices Class 12 Questions And Answers: Exercise 3.1

Question:1(i). In the matrix A = \begin{bmatrix} 2& 5 &19 &-7 \\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt3& 1 &-5 &17 \end{bmatrix} , write:

The order of the matrix

Answer:

A = \begin{bmatrix} 2& 5 &19 &-7 \\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt3& 1 &-5 &17 \end{bmatrix}

(i) The order of the matrix = number of row \times number of columns = 3\times 4 .

Question 1(ii). In the matrix A = \begin{bmatrix}2&5&19&-7&\\ 35& -2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix} , write:

The number of elements

Answer:

A = \begin{bmatrix}2&5&19&-7&\\ 35& -2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}

(ii) The number of elements 3\times 4=12 .

Question 1(iii). In the matrix A = \begin{bmatrix}2&5&19&-7&\\35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix} , write:

Write the elements a 13 , a 21 , a 33 , a 24 , a 23

Answer:

A = \begin{bmatrix}2&5&19&-7&\\35&-2&\frac{5}{2}&12\\\sqrt3&1&-5&17 \end{bmatrix}

(iii) An element a_{ij} implies the element in raw number i and column number j.

a_1_3= 19 a_2_1= 35

a_3_3= -5 a_2_4= 12

a_2_3= \frac{5}{2}

Question 2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Answer:

A matrix has 24 elements.

The possible orders are :

1\times 24,24\times 1,2\times 12,12\times 2,3\times 8,8\times 3,4\times 6 \, \, and\, \, 6\times 4 .

If it has 13 elements, then possible orders are :

1\times 13\, \, \, and \, \, \, \, 13\times 1 .

Question 3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Answer:

A matrix has 18 elements.

The possible orders are as given below

1\times 18,18\times 1,2\times 9,9\times 2,3\times 6\, \, \, and\, \, \, \, 6\times 3

If it has 5 elements, then possible orders are :

1\times 5\, \, \, and \, \, \, \, 5\times 1 .

Question 4(i). Construct a 2 × 2 matrix, A = [a_{ij} ] whose elements are given by:

a_{ij} = \frac{(i + j)^2}{2}

Answer:

A = [a_{ij} ]

(i) a_{ij} = \frac{(i + j)^2}{2}

Each element of this matrix is calculated as follows

a_1_1 = \frac{(1+1)^{2}}{2} =\frac{2^{2}}{2}=\frac{4}{2}=2 a_2_2 = \frac{(2+2)^{2}}{2} =\frac{4^{2}}{2}=\frac{16}{2}=8

a_1_2 = \frac{(1+2)^{2}}{2} =\frac{3^{2}}{2}=\frac{9}{2}=4.5 a_2_1 = \frac{(2+1)^{2}}{2} =\frac{3^{2}}{2}=\frac{9}{2}=4.5

Matrix A is given by

A = \begin{bmatrix} 2&4.5 \\4.5 & 8 \end{bmatrix}

Question 4(ii). Construct a 2 × 2 matrix, A = [a_{ij} ] , whose elements are given by:

a_{ij} = \frac{i}{j}

Answer:

A 2 × 2 matrix, A = [a_{ij} ]

(ii) a_{ij} = \frac{i}{j}

a_1_1 = \frac{1}{1}=1 a_2_2 = \frac{2}{2}=1

a_1_2 = \frac{1}{2} a_2_1 = \frac{2}{1}=2

Hence, the matrix is

A = \begin{bmatrix} 1& \frac{1}{2} \\ 2 & 1 \end{bmatrix}

Question 4(iii). Construct a 2 × 2 matrix, A = [a_{ij} ] , whose elements are given by:

a_{ij} = \frac{(i+2j)^2}{2}

Answer:

(iii)

a_{ij} = \frac{(i+2j)^2}{2}

a_1_1 = \frac{(1+(2\times 1))^{2}}{2}= \frac{(1+2)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2} a_2_2 = \frac{(2+(2\times 2))^{2}}{2}= \frac{(2+4)^{2}}{2}=\frac{6^{2}}{2}=\frac{36}{2}=18

a_2_1 = \frac{(2+(2\times 1))^{2}}{2}= \frac{(2+2)^{2}}{2}=\frac{4^{2}}{2}=\frac{16}{2}=8 a_1_2 = \frac{(1+(2\times 2))^{2}}{2}= \frac{(1+4)^{2}}{2}=\frac{5^{2}}{2}=\frac{25}{2}

Hence, the matrix is given by

A = \begin{bmatrix} \frac{9}{2}& \frac{25}{2} \\ 8 & 18 \end{bmatrix}

Question 5(i). Construct a 3 × 4 matrix, whose elements are given by:

a_{ij} = \frac{1}{2}|-3i + j|

Answer:

(i)

a_{ij} = \frac{1}{2}|-3i + j|

a_1_1 = \frac{\left | -3+1 \right |}{2}=\frac{2}{2}=1 a_1_2 = \frac{\left | (-3\times 1)+2 \right |}{2}=\frac{1}{2} a_1_3 = \frac{\left | (-3\times 1)+3 \right |}{2}=0

a_2_1 = \frac{\left | (-3\times 2)+1 \right |}{2}=\frac{5}{2} a_2_2 = \frac{\left | (-3\times 2)+2 \right |}{2}=\frac{4}{2}=2 a_2_3 = \frac{\left | (-3\times 2)+3 \right |}{2}=\frac{\left | -6+3 \right |}{2}=\frac{\left | -3 \right |}{2} =\frac{3}{2}

a_3_1 = \frac{\left | (-3\times 3)+1 \right |}{2}=\frac{8}{2}=4 a_3_2 = \frac{\left | (-3\times 3)+2 \right |}{2}=\frac{7}{2} a_3_3 = \frac{\left | (-3\times 3)+3 \right |}{2}=\frac{\left | -9+3 \right |}{2}=\frac{\left | -6 \right |}{2} =\frac{6}{2}=3

a_1_4 = \frac{\left | (-3\times 1)+4 \right |}{2}=\frac{\left | -3+4 \right |}{2}=\frac{\left | 1 \right |}{2} =\frac{1}{2} a_2_4 = \frac{\left | (-3\times 2)+4 \right |}{2}=\frac{\left | -6+4 \right |}{2}=\frac{\left | -2 \right |}{2} =\frac{2}{2}=1

a_3_4 = \frac{\left | (-3\times 3)+4 \right |}{2}=\frac{\left | -9+4 \right |}{2}=\frac{\left | -5 \right |}{2} =\frac{5}{2}

Hence, the required matrix of the given order is

A = \begin{bmatrix} 1& \frac{1}{2} & 0&\frac{1}{2} \\ \frac{5}{2} & 2&\frac{3}{2}&1 \\4&\frac{7}{2}&3&\frac{5}{2}\end{bmatrix}

Question 5(ii) Construct a 3 × 4 matrix, whose elements are given by:

a_{ij} = 2i - j

Answer:

A 3 × 4 matrix,

(ii) a_{ij} = 2i - j

a_1_1 = 2\times 1-1 =2-1=1 a_1_2 = 2\times 1-2 =2-2=0 a_1_3 = 2\times 1-3 =2-3=-1

a_2_1 = 2\times 2-1 =4-1=3 a_2_2= 2\times 2-2 =4-2=2 a_2_3 = 2\times 2-3 =4-3=1 a_3_1 = 2\times 3-1 =6-1=5 a_3_2 = 2\times 3-2 =6-2=4 a_3_3 = 2\times 3-3 =6-3=3

a_1_4 = 2\times 1-4 =2-4=-2 a_2_4= 2\times 2-4 =4-4=0 a_3_4= 2\times 3-4 =6-4=2

Hence, the matrix is

A = \begin{bmatrix} 1 & 0& -1& -2 \\ \ 3 & 2&1& 0 \\5&4&3&2\end{bmatrix}

Question 6(i). Find the values of x, y and z from the following equations:

\begin{bmatrix}4&3\\x&5 \end{bmatrix} = \begin{bmatrix}y&z\\1&5 \end{bmatrix}

Answer:

(i) \begin{bmatrix}4&3\\x&5 \end{bmatrix} = \begin{bmatrix}y&z\\1&5 \end{bmatrix}

If two matrices are equal, then their corresponding elements are also equal.

\therefore x=1\, \, \, ,\, \, \, y=4\, \, \, \, and\, \, \, \, z=3

Question 6(ii) Find the values of x, y and z from the following equations:

\begin{bmatrix} x +y & 2\\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 &2 \\ 5 & 8 \end{bmatrix}

Answer:

(ii)

\begin{bmatrix} x +y & 2\\ 5 + z & xy \end{bmatrix} = \begin{bmatrix} 6 &2 \\ 5 & 8 \end{bmatrix}

If two matrices are equal, then their corresponding elements are also equal.

\therefore x+y=6 \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)

x=6-y

xy=8 \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)

Solving equation (i) and (ii) ,

(6-y)y =8

6y-y^{2}=8

y^{2}-6y+8=0

solving this equation we get,

y=4 \, \, and\, \, y=2

Putting the values of y, we get

x=2 \, \, and\, \, x=4

And also equating the first element of the second raw

5+z = 5 , z=0

Hence,

x=2,y=4,z=0\, \, \, \, \, and\, \, \, \, \, \, x=4,y=2,z=0

Question 6(iii) Find the values of x, y and z from the following equations

\begin{bmatrix} x + y + z\\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9\\5 \\7 \end{bmatrix}

Answer:

(iii)

\begin{bmatrix} x + y + z\\ x + z \\ y + z \end{bmatrix} = \begin{bmatrix} 9\\5 \\7 \end{bmatrix}

If two matrices are equal, then their corresponding elements are also equal

x+y+z=9........(1)

x+z=5..............(2)

y+z=7..............(3)

subtracting (2) from (1) we will get y=4

substituting the value of y in equation (3) we will get z=3

now substituting the value of z in equation (2) we will get x=2

therefore,

x=2 , y=4 and z=3

Question 7. Find the value of a, b, c and d from the equation:

\begin{bmatrix} a -b & 2a + c\\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5\\ 0 & 13 \end{bmatrix}

Answer:

\begin{bmatrix} a -b & 2a + c\\ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5\\ 0 & 13 \end{bmatrix}

If two matrices are equal, then their corresponding elements are also equal

a-b=-1 .............................1

2a+c=5 .............................2

2a-b=0 .............................3

3c+d=13 .............................4

Solving equation 1 and 3 , we get

a=1 \, \, \, \, and \, \, \, \, b=2

Putting the value of a in equation 2, we get

c=3

Putting the value of c in equation 4 , we get

d=4

Question 8. A = [a_{ij}]_{m\times n} is a square matrix, if

(A) m <n

(B) m >n

(C) m =n

(D) None of these

Answer:

A square matrix has the number of rows and columns equal.

Thus, for A = [a_{ij}]_{m\times n} to be a square matrix m and n should be equal.

Option (c) is .

Question 9. Which of the given values of x and y make the following pair of matrices equal

\begin{bmatrix} 3x + 7 &5 \\ y + 1 & 2 -3x \end{bmatrix} , \begin{bmatrix} 0 & y - 2 \\ 8 & 4 \end{bmatrix}

(A) x = \frac{-1}{3}, y = 7

(B) Not possible to find

(C) y =7, x = \frac{-2}{3}

(D) x = \frac{-1}{3}, y = \frac{-2}{3}

Answer:

Given, \begin{bmatrix} 3x + 7 &5 \\ y + 1 & 2 -3x \end{bmatrix} =\begin{bmatrix} 0 & y - 2 \\ 8 & 4 \end{bmatrix}

If two matrices are equal, then their corresponding elements are also equal

3x+7=0\Rightarrow x=\frac{-7}{3}

y-2=5 \Rightarrow y=5+2=7

y+1=8\Rightarrow y=8-1=7

2-3x=4\Rightarrow 3x=2-4\Rightarrow 3x=-2\Rightarrow x=\frac{-2}{3}

Here, the value of x is not unique, so option B is correct.

Question 10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27
(B) 18
(C) 81
(D) 512

Answer:

Total number of elements in a 3 × 3 matrix

=3\times 3=9

If each entry is 0 or 1 then for every entry there are 2 permutations.

The total permutations for 9 elements

=2^{9}=512

Thus, option (D) is correct.


Class 12 Maths Chapter 3 Question Answer: Exercise 3.2

Question 1(i) Let A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} , B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix} , C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}

Find each of the following:

A + B

Answer:

A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}

(i) A + B

The addition of matrix can be done as follows

A+B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}

A+B = \begin{bmatrix} 2+1 &4+3 \\ 3+(-2) & 2+5 \end{bmatrix}

A+B = \begin{bmatrix} 3 &7 \\ 1 & 7 \end{bmatrix}

Question 1(ii) Let A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} , B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix} , C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}

Find each of the following:

A - B

Answer:

A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}

(ii) A - B

A-B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}

A-B = \begin{bmatrix} 2-1 &4-3 \\ 3-(-2) & 2-5 \end{bmatrix}

A-B = \begin{bmatrix} 1 &1 \\ 5 & -3 \end{bmatrix}

Question 1(iii) Let A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} , B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix} , C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}

Find each of the following:

3A - C

Answer:

A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}

(iii) 3A - C

First multiply each element of A with 3 and then subtract C

3A -C = 3\begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}

3A -C = \begin{bmatrix} 6 &12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}

3A -C = \begin{bmatrix} 6-(-2) &12-5 \\ 9-3 & 6-4 \end{bmatrix}

3A -C = \begin{bmatrix} 8 &7 \\ 6 & 2 \end{bmatrix}

Question 1(v) Let A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} , B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix} , C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}

Find each of the following:

BA

Answer:

The multiplication is performed as follows

A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix} , B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}

BA = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix} \times \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}

BA = \begin{bmatrix} 1\times 2+3\times 3 &1\times 4+3\times 2 \\ -2\times 2+5\times 3& -2\times 4+2\times 5 \end{bmatrix}

BA = \begin{bmatrix} 11 &10 \\ 11& 2 \end{bmatrix}

Question 2(i). Compute the following:

\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}

Answer:

(i) \begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}

= \begin{bmatrix} a+a &b+b \\ -b+b & a+a \end{bmatrix}

= \begin{bmatrix} 2a &2b \\ 0 & 2a \end{bmatrix}

Question 2(ii). Compute the following:

\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}

Answer:

(ii) The addition operation can be performed as follows

\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}

=\begin{bmatrix} a^2 + b^2+2ab& b^2+c^2+2bc\\ a^2 + c^2-2ac& a^2 + b^2-2ab \end{bmatrix}

=\begin{bmatrix} (a+b)^2 & (b+c)^2\\ (a-c)^2 & (a-b)^2 \end{bmatrix}

Question 2(iii). Compute the following:

\begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 &5 \\ 3 & 2 & 4 \end{bmatrix}

Answer:

(iii) The addition of given three by three matrix is performed as follows

\begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 &5 \\ 3 & 2 & 4 \end{bmatrix}

=\begin{bmatrix} -1+12 & 4+7 & -6+6\\ 8+8 & 5+0 & 16+5\\ 2+3 & 8+2 & 5+4 \end{bmatrix}

=\begin{bmatrix} 11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix}

Question 2(iv). Compute the following:

\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}

Answer:

(iv) the addition is done as follows

\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}

=\begin{bmatrix} \cos^2+ \sin^2 x &\sin^2 x+\cos^2 x\\ \sin^2 x+\cos^2 x & \cos^2x+ \sin^2 x \end{bmatrix} since sin^2x+cos^2x=1

=\begin{bmatrix} 1 &1\\ 1 & 1 \end{bmatrix}

Question 3(i). Compute the indicated products.

\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}

Answer:

(i) The multiplication is performed as follows

\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}

=\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \times \begin{bmatrix} a & -b \\ b &a \end{bmatrix}

=\begin{bmatrix} a\times a+b\times b &a\times -b+b\times a \\ -b\times a+a\times b &-b\times -b+a\times a \end{bmatrix}

=\begin{bmatrix} a^{2}+b^{2} & 0 \\ 0 & b^{2}+a^{2} \end{bmatrix}

Question 3(ii). Compute the indicated products.

\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}

Answer:

(ii) the multiplication can be performed as follows

\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}

=\begin{bmatrix} 1\times 2 &1\times 3&1\times 4\\ 2\times 2&2\times 3&2\times 4\\3\times 2&3\times 3&3\times 4 \end{bmatrix}

=\begin{bmatrix} 2 &3& 4\\ 4&6&8\\6&9&12 \end{bmatrix}

Question 3(iii). Compute the indicated products.

\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}

Answer:

(iii) The multiplication can be performed as follows

\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}

=\begin{bmatrix} 1\times 1+(-2)\times 2 & 1\times 2+(-2)\times 3&1\times 3+(-2)\times 1\\ 2\times 1+3\times 2 & 2\times 2+3\times 3&2\times 3+3\times 1 \end{bmatrix}

Question 3(iv). Compute the indicated products.

\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}

Answer:

(iv) The multiplication is performed as follows

\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}

=\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix}\times \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}

=\begin{bmatrix} 2\times 1+3\times 0+4\times 3 \, \, & 2\times (-3)+3\times 2+4\times 0 \, \, & 2\times 5+3\times 4+4\times 5 \\ 3\times 1+4\times 0+5\times 3 \, \, & 3\times (-3)+4\times 2+5\times 0 & 3\times 5+4\times 4+5\times 5 \\ 4\times 1+5\times 0+6\times 3 \, \, & 4\times (-3)+5\times 2+6\times 0\, \, & 4\times 5+5\times 4+6\times 5 \end{bmatrix}

= \begin{bmatrix} 14 & 0 & 42\\ 18 & -1 & 56\\ 22 & -2 & 70 \end{bmatrix}

Question 3(v). Compute the indicated products.

\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}

Answer:

(v) The product can be computed as follows

\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}

=\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\times \begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}

=\begin{bmatrix} 2\times 1+1\times (-1) &2\times 0+1\times (2) & 2\times 1+1\times (1) \\ 3\times 1+2\times (-1) & 3\times 0+2\times (2) &3\times 1+2\times (1) \\ (-1)\times 1+1\times (-1) & (-1)\times 0+1\times (2) & (-1)\times 1+1\times (1) \end{bmatrix}

=\begin{bmatrix} 1 &2&3 \\ 1 & 4&5\\ -2 & 2&0 \end{bmatrix}

Question 3(vi). Compute the indicated products.

\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}

Answer:

(vi) The given product can be computed as follows

\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}

=\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}

=\begin{bmatrix} 3 \times 2+(-1)\times 1+3\times 3\, \, \, & 3 \times (-3)+(-1)\times 0+3\times 1 \\ (-1) \times 2+ 0 \times 1+2\times 3 \, \, \, & (-1) \times -3+0\times 0+2\times 1 \end{bmatrix}

=\begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}

Question 4. If A = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix} , B = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix} and C = \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix} , then compute (A+B) and (B-C). Also verify that A + (B - C) = (A + B) - C

Answer:

A = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix} , B = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix} and C = \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}

A+B = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix} + \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}

A+B = \begin{bmatrix} 1+3 &2+(-1) &-3+2 \\ 5+4 &0+2 &2+5 \\ 1+2 & -1+0 &1+3 \end{bmatrix}

A+B = \begin{bmatrix} 4 &1 &-1 \\ 9 &2 &7 \\ 3 & -1 &4 \end{bmatrix}

B-C = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix} -\begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}

B-C = \begin{bmatrix} 3-4 &-1-1 &2-2 \\ 4-0 &2-3 &5-2 \\ 2-1 & 0-(-2) &3-3 \end{bmatrix}

B-C = \begin{bmatrix} -1 &-2 &0 \\ 4 &-1 &3 \\ 1 & 2 &0 \end{bmatrix}

Now, to prove A + (B - C) = (A + B) - C

L.H.S\, \, :\, A+(B-C)

A+(B-C)=\begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix} + \begin{bmatrix} -1 &-2 &0 \\ 4 &-1 &3 \\ 1 & 2 &0 \end{bmatrix} (Puting value of B-C from above)

A+(B-C)=\begin{bmatrix} 1-1 &2-2 &-3+0 \\ 5+4 &0+(-1) &2+3 \\ 1+1 & -1+2 &1+0 \end{bmatrix}

A+(B-C)=\begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}

R.H.S\, \, :\, (A+B)-C

(A+B)-C = \begin{bmatrix} 4 &1 &-1 \\ 9 &2 &7 \\ 3 & -1 &4 \end{bmatrix} - \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}

(A+B)-C = \begin{bmatrix} 4-4 &1-1 &-1-2 \\ 9-0 &2-3 &7-2 \\ 3-1 & -1-(-2) &4-3 \end{bmatrix}

(A+B)-C = \begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}

Hence, we can see L.H.S = R.H.S = \begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}

Question 5. If A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} and B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix} , then compute 3A - 5B

Answer:

A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} and B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}

3A-5B = 3\times \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} -5\times \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}

3A-5B = \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}

3A-5B = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}

3A-5B = 0

Question 6. Simplify \cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix} .

Answer:

The simplification is explained in the following step

\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}

= \begin{bmatrix} \cos^{2}\theta & \sin\theta \cos\theta \\ -\sin\theta \cos\theta & \cos^{2}\theta \end{bmatrix} +\begin{bmatrix} \sin^{2}\theta & - \sin\theta \cos\theta\\ \sin\theta\cos\theta & \sin^{2}\theta \end{bmatrix}

= \begin{bmatrix} \cos^{2}\theta+\sin^{2}\theta & \sin\theta \cos\theta - \sin\theta \cos\theta \\ -\sin\theta \cos\theta + \sin\theta \cos\theta & \cos^{2}\theta + \sin^{2}\theta\end{bmatrix}

= \begin{bmatrix} 1&0 \\ 0 & 1\end{bmatrix} =I

the final answer is an identity matrix of order 2

Question 7(i). Find X and Y, if

X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix} and X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}

Answer:

(i) The given matrices are

X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix} and X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}

X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}.............................1

X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}.............................2

Adding equation 1 and 2, we get

2 X = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix} + \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}

2 X = \begin{bmatrix} 7+3 &0+0 \\ 2+0 &5+3 \end{bmatrix}

2 X = \begin{bmatrix} 10 &0 \\ 2 &8 \end{bmatrix}

X = \begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}

Putting the value of X in equation 1, we get

\begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix} +Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}

Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix} - \begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}

Y = \begin{bmatrix} 7-5 &0-0 \\ 2-1 &5-4 \end{bmatrix}

Y = \begin{bmatrix} 2 &0 \\ 1 &1 \end{bmatrix}

Question 7(ii). Find X and Y, if

2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} and 3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}

Answer:

(ii) 2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} and 3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}

2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}..........................1

3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}......................2

Multiply equation 1 by 3 and equation 2 by 2 and subtract them,

3(2X + 3Y)-2(3X+2Y) = 3 \times \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} - \, \, \, 2\times \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}

6X + 9Y-6X-4Y= \begin{bmatrix} 6 &9 \\ 12 & 0 \end{bmatrix} - \begin{bmatrix} 4 &-4 \\ -2 & 10 \end{bmatrix}

9Y-4Y= \begin{bmatrix} 6-4 &9-(-4) \\ 12-(-2) & 0-10 \end{bmatrix}

5Y= \begin{bmatrix} 2 &13 \\ 14 & -10 \end{bmatrix}

Y= \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}

Putting value of Y in equation 1 , we get

2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}

2X + 3 \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}

2X + \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}

2X = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix}

2X = \begin{bmatrix} 2-\frac{6}{5} &3-\frac{39}{5} \\ 4-\frac{42}{5} & 0 -(-6)\end{bmatrix}

2X = \begin{bmatrix} \frac{4}{5} &-\frac{24}{5} \\ -\frac{22}{5} & 6\end{bmatrix}

X = \begin{bmatrix} \frac{2}{5} &-\frac{12}{5} \\ -\frac{11}{5} & 3\end{bmatrix}

Question 8. Find X, if Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix} and 2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}

Answer:

Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}

2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}

Substituting the value of Y in the above equation

2X+ \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}

2X = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}- \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}

2X = \begin{bmatrix} 1-3 &0-2 \\ -3-1 & 2-4 \end{bmatrix}

2X = \begin{bmatrix} -2 &-2 \\ -4 & -2 \end{bmatrix}

X = \begin{bmatrix} -1 &-1 \\ -2 & -1 \end{bmatrix}

Question 9. Find x and y, if 2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}

Answer:

2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}

\begin{bmatrix} 2 & 6\\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}

\begin{bmatrix} 2+y & 6+0\\ 0+1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}

\begin{bmatrix} 2+y & 6\\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}

Now equating LHS and RHS we can write the following equations

2+y=5 2x+2=8

y=5-2 2x=8-2

y=3 2x=6

x=3

Question 10. Solve the equation for x, y, z and t, if 2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}

Answer:

2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}

Multiplying with constant terms and rearranging we can rewrite the matrix as

\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9 &15\\ 12 & 18 \end{bmatrix} - 3\begin{bmatrix} 1& -1\\ 0 & 2 \end{bmatrix}

\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9 &15\\ 12 & 18 \end{bmatrix} - \begin{bmatrix} 3& -3\\ 0 & 6 \end{bmatrix}

\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9-3 &15-(-3)\\ 12-0 & 18-6 \end{bmatrix}

\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 6 &18\\ 12 & 12 \end{bmatrix}

Dividing by 2 on both sides

\begin{bmatrix}x & z \\ y &t \end{bmatrix} = \begin{bmatrix} 3 &9\\ 6 & 6 \end{bmatrix}

x=3,y=6,z=9\, \, and\, \, t=6

Question 11. If x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix} , find the values of x and y.

Answer:

x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}

\begin{bmatrix}2x\\3x \end{bmatrix} + \begin{bmatrix} -y\\y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}

Adding both the matrix in LHS and rewriting

\begin{bmatrix}2x-y\\3x+y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}

2x-y=10........................1

3x+y=5........................2

Adding equation 1 and 2, we get

5x=15

x=3

Put the value of x in equation 2, we have

3x+y=5

3\times 3+y=5

9+y=5

y=5-9

y=-4

Question 12. Given 3\begin{bmatrix}x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 &x + y \\ z + w & 3 \end{bmatrix} , find the values of x, y, z and w.

Answer:

3\begin{bmatrix}x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 &x + y \\ z + w & 3 \end{bmatrix}

\begin{bmatrix}3x &3 y \\3 z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}

If two matrices are equal than corresponding elements are also equal.

Thus, we have

3x=x+4

3x-x=4

2x=4

x=2

3y=6+x+y

Put the value of x

3y-y=6+2

2y=8

y=4

3w=2w+3

3w-2w=3

w=3

3z=-1+z+w

3z-z=-1+3

2z=2

z=1

Hence, we have x=2,y=4,z=1\, \, and\, \, w=3.

Question 13. If F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix} , show that F(x) F(y) = F(x + y) .

Answer:

F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}

To prove : F(x) F(y) = F(x + y)

R.H.S : F(x + y)

F(x+y) = \begin{bmatrix} \cos (x+y) & -\sin (x+y)& 0\\\sin (x+y) &\cos (x+y) & 0 \\ 0 &0&1\end{bmatrix}

L.H.S : F(x) F(y)

F(x)F(y) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}\times \begin{bmatrix} \cos y & -\sin y& 0\\\sin y &\cos y & 0 \\ 0 &0&1\end{bmatrix}

F(x)F(y) = \begin{bmatrix} \cos x \cos y- \sin x\sin y+0 & -\cos x \sin y-\sin x\cos y+0& 0+0+0\\\ sin x\cos y+\cos x \sin y+0 & - \sin x\sin y+\cos x \cos y+0 &0+0+0 \\ 0+0+0 &0+0+0&0+0+1\end{bmatrix}


F(x) F(y)= \begin{bmatrix} \cos (x+y) & -\sin (x+y)& 0\\\sin (x+y) &\cos (x+y) & 0 \\ 0 &0&1\end{bmatrix}

Hence, we have L.H.S. = R.H.S i.e. F(x) F(y) = F(x + y) .

Question 14(i). Show that

\begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}

Answer:

To prove:

\begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}

L.H.S : \begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}

= \begin{bmatrix}5\times 2+(-1)\times 3 &5\times 1+(-1)\times 4\\6\times 2+7\times 3&6\times 1+7\times 4 \end{bmatrix}

= \begin{bmatrix}7 &1\\33&34 \end{bmatrix}

R.H.S : \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}

= \begin{bmatrix} 2\times 5+1\times 6 & 2\times (-1)+1\times 7\\ 3\times 5+4\times 6 & 3\times (-1)+4\times 7 \end{bmatrix}

= \begin{bmatrix} 16 & 5\\ 39 & 25 \end{bmatrix}

Hence, the right-hand side not equal to the left-hand side, that is

Question 14(ii). Show that

\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}

Answer:

To prove the following multiplication of three by three matrices are not equal

\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}

L.H.S: \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix}

= \begin{bmatrix}1\times(-1)+2\times 0+3\times 2 \, \, \, & 1\times(1)+2\times (-1)+3\times 3\, \, \, &1\times(0)+2\times 1+3\times 4\\0\times(-1)+1\times 0+0\times 2\, \, \, &0\times(1)+1\times (-1)+0\times 3\, \, \, &0\times(0)+1\times 1+0\times 4\\1\times(-1)+1\times 0+0\times 2\, \, \, &1\times(1)+1\times (-1)+0\times 3\, \, \, &1\times(0)+1\times 1+0\times 4 \end{bmatrix}

= \begin{bmatrix}5& 8&14\\0&-1&1\\-1&0&1\end{bmatrix}


R.H.S : \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}

= \begin{bmatrix}-1\times(1)+1\times 0+0\times 1 \, \, \, & -1\times(2)+1\times (1)+0\times 1\, \, \, &-1\times(3)+1\times 0+0\times 0\\0\times(1)+-(1)\times 0+1\times 1\, \, \, &0\times(2)+(-1)\times (1)+1\times 1\, \, \, &0\times(3)+(-1)\times 0+1\times 0\\2\times(1)+3\times 0+4\times 1\, \, \, &2\times(2)+3\times (1)+4\times 1\, \, \, &2\times(3)+3\times 0+4\times 0 \end{bmatrix}

= \begin{bmatrix}-1& -1&-3\\1&0&0\\6&11&6\end{bmatrix}

Hence, L.H.S \neq R.H.S i.e. \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} .

Question 15. Find A^2 -5A + 6I , if

A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}

Answer:

A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}

First, we will find ou the value of the square of matrix A

A\times A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}\times \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}

A^{2} = \begin{bmatrix} 2\times 2+0\times 2+1\times 1 & 2\times 0+0\times 1+1\times -1 & 2\times 1+0\times 3+1\times 0\\ 2\times 2+1\times 2+3\times 1& 2\times 0+1\times 1+3\times -1 &2\times 1+1\times 3+3\times 0 \\ 1\times 2+(-1)\times 2+0\times 1 & 1\times 0+(-1)\times 1+0\times -1 & 1\times 1+(-1)\times 3+0\times 0 \end{bmatrix}

A^{2} = \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}

I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}

\therefore A^2 -5A + 6I

= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix} -5 \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix} +6 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}

= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5\\ 10 & 5 &15 \\ 5 & -5 & 0 \end{bmatrix} +\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 &0 \\ 0 & 0 & 6 \end{bmatrix}

= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0\\ 9-10+0 & -2-5+6 &5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix}

= \begin{bmatrix} 1 & -1 & -3\\ -1 & -1 &-10 \\ -5 & 4 & 4 \end{bmatrix}

Question 16. If A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix} prove that A^3 - 6A^2 + 7A + 2I = 0 .

Answer:

A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}

First, find the square of matrix A and then multiply it with A to get the cube of matrix A

A\times A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix} \times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}

A^{2} = \begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9 \end{bmatrix}

A^{2} = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}

A^{3}=A^{2}\times A

A^{2}\times A = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix} \times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}

A^{3} = \begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39 \end{bmatrix}

A^{3} = \begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}

I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}

\therefore A^3 - 6A^2 + 7A + 2I = 0

L.H.S :

\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix} - 6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix} +7 \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix} +2 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}

=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix} - \begin{bmatrix}30&0&48\\12&24&30\\48&0&78 \end{bmatrix} + \begin{bmatrix}7&0&14\\0&14&7\\14&0&21 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 &0 \\ 0 & 0 & 2 \end{bmatrix}

=\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2 \end{bmatrix}

=\begin{bmatrix}30-30&0&48-48\\12-12&24-24&30-30\\48-48&0&78-78 \end{bmatrix}

= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 &0 \\ 0 & 0 & 0 \end{bmatrix}=0

Hence, L.H.S = R.H.S

i.e. A^3 - 6A^2 + 7A + 2I = 0 .

Question 17. If A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} and I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix} , find k so that A^{2} = kA - 2I .

Answer:

A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}

I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}

A \times A= \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} \times \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}

A^{2} = \begin{bmatrix}9-8 &-6+4\\12-8&-8+4 \end{bmatrix}

A^{2} = \begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}

A^{2} = kA - 2I

\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}= k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} - 2 \begin{bmatrix}1 &0\\0&1 \end{bmatrix}

\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}= k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} - \begin{bmatrix}2 &0\\0&2 \end{bmatrix}

\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}+ \begin{bmatrix}2 &0\\0&2 \end{bmatrix} =k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}

\begin{bmatrix}1+2 &-2+0\\4+0&-4+2 \end{bmatrix} =\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}

\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} =\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}

We have, 3=3k

k=\frac{3}{3}=1

Hence, the value of k is 1.

Question 18. If A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix} and I is the identity matrix of order 2, show that I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}

Answer:

A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}

I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}

To prove : I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}

L.H.S : I+A

I+A = \begin{bmatrix}1 &0\\0&1 \end{bmatrix} + \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}

I+A = \begin{bmatrix} 1+0&0-\tan\frac{\alpha}{2}\\0+\tan\frac{\alpha}{2}&1+ 0\end{bmatrix}

I+A = \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}

R.H.S : (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}

(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix} = (\begin{bmatrix}1 &0\\0&1 \end{bmatrix}- \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}) \times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}

(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix} =\begin{bmatrix} 1-0&0-(-\tan\frac{\alpha}{2})\\0-\tan\frac{\alpha}{2}&1- 0\end{bmatrix} \times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}

(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix} =\begin{bmatrix} 1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix} \times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}

=\begin{bmatrix} \cos\alpha + \sin \alpha\tan\frac{\alpha}{2} &- \sin \alpha+ \cos \alpha \tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} \cos\alpha + \sin \alpha &\tan\frac{\alpha}{2} \sin\alpha + \cos \alpha \end{bmatrix}

=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}\tan\frac{\alpha}{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ (2\cos^{2} \frac{\alpha }{2} -1)\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} (2\cos^{2} \frac{\alpha }{2} -1) + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} &\tan\frac{\alpha}{2} 2\sin\frac{\alpha } {2} \ cos \frac{\alpha }{2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}

=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin^{2}\frac{\alpha }{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} -\tan\frac{\alpha}{2}\\-2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+\tan\frac{\alpha}{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} & 2\sin^{2}\frac{\alpha } {2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}

= \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}

Hence, we can see L.H.S = R.H.S

i.e. I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix} .

Question 19(i). A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 1800

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

\begin{bmatrix}x &(30000-x) \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} =1800 (simple interest for 1 year =\frac{pricipal\times rate}{100} )

\frac{5}{100}x+\frac{7}{100}(30000-x) = 1800

5x+210000-7x=180000

210000-180000=7x-5x

30000=2x

x=15000

Thus, to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs 15000 in the first bond and Rs 15000 in the second bond.

Question 19(ii). A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 2000

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

\begin{bmatrix}x &(30000-x) \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} =2000 (simple interest for 1 year =\frac{pricipal\times rate}{100} )

\frac{5}{100}x+\frac{7}{100}(30000-x) = 2000

5x+210000-7x=200000

210000-200000=7x-5x

10000=2x

x=5000

Thus, to obtain an annual total interest of Rs. 2000, the trust fund should invest Rs 5000 in the first bond and Rs 25000 in the second bond.

Question 20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.

Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively.

The total amount the bookshop will receive from selling all the books:

12 \begin{bmatrix}10 &8&10 \end{bmatrix} \begin{bmatrix}80\\60\\40 \end{bmatrix}

=12(10\times 80+8\times 60+10\times 40)

= 12(800+480+ 400)

= 12(1680)

=20160

The total amount the bookshop will receive from selling all the books is 20160.

Question 21 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k , respectively. Choose the correct answer in Exercises 21 and 22.

Q21. The restriction on n, k and p so that PY + WY will be defined are:
(A) k = 3, p = n

(B) k is arbitrary, p = 2

(C) p is arbitrary, k = 3

(D) k = 2, p = 3

Answer:

P and Y are of order p*k and 3*k respectivly.

\therefore PY will be defined only if k=3, i.e. order of PY is p*k .

W and Y are of order n*3 and 3*k respectivly.

\therefore WY is defined because the number of columns of W is equal to the number of rows of Y which is 3, i.e. the order of WY is n*k

Matrices PY and WY can only be added if they both have same order i.e = n*k implies p=n.

Thus, k = 3, p = n are restrictions on n, k and p so that PY + WY will be defined.

Option (A) is correct.

Question 22 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k ,
respectively. Choose the correct answer in Exercises 21 and 22.
If n = p , then the order of the matrix 7X - 5Z is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

Answer:

X has of order 2*n .

\therefore 7X also has of order 2*n .

Z has of order 2*p .

\therefore 5Z also has of order 2*p .

Mtarices 7X and 5Z can only be subtracted if they both have same order i.e 2*n = 2*p and it is given that p=n.

We can say that both matrices have order of 2*n .

Thus, order of 7X - 5Z is 2*n .

Option (B) is correct.

Class 12 Maths Chapter 3 Question Answer: Exercise 3.3

Question 1(i). Find the transpose of each of the following matrices:

\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}

Answer:

A=\begin{bmatrix} 5\\ \frac{1}{2} \\-1 \end{bmatrix}

The transpose of the given matrix is

A^{T}=\begin{bmatrix} 5& \frac{1}{2} &-1 \end{bmatrix}

Question 1(ii). Find the transpose of each of the following matrices:

\begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}

Answer:

A=\begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}

interchanging the rows and columns of the matrix A we get

A^{T}=\begin{bmatrix} 1 & 2\\ -1 & 3 \end{bmatrix}

Question 1(iii) Find the transpose of each of the following matrices:

\begin{bmatrix} -1 & 5 & 6\\ \sqrt3& 5 &6 \\ 2 &3 &-1 \end{bmatrix}

Answer:

A = \begin{bmatrix} -1 & 5 & 6\\ \sqrt3& 5 &6 \\ 2 &3 &-1 \end{bmatrix}

Transpose is obtained by interchanging the rows and columns of matrix

A^{T} = \begin{bmatrix} -1 & \sqrt3 & 2\\ 5& 5 &3 \\ 6 &6 &-1 \end{bmatrix}

Question 2(i). If A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix} and B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix} , then verify

(A + B)' = A' + B'

Answer:

A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix} and B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}

(A + B)' = A' + B'

L.H.S : (A + B)'

A+B = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix} + \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}

A+B = \begin{bmatrix} -1+(-4) & 2+1 & 3+(-5)\\ 5+1 &7+2 &9+0 \\ -2+1 & 1+3 & 1+1 \end{bmatrix}

A+B = \begin{bmatrix} -5 & 3 & -2\\ 6 &9 &9 \\ -1 & 4 & 2 \end{bmatrix}

(A+B)' = \begin{bmatrix} -5 & 6 & -1\\ 3 &9 &4 \\ -2 & 9 & 2 \end{bmatrix}

R.H.S : A' + B'

A'+B' = \begin{bmatrix} -1 & 5 & -2\\ 2 &7 &1 \\ 3 & 9 & 1 \end{bmatrix} + \begin{bmatrix} -4 & 1 & 1\\ 1 &2 &3\\ -5 & 0 & 1 \end{bmatrix}

A'+B' = \begin{bmatrix} -1+(-4) & 5+1 & -2+1\\ 2+1 &7+2 &1+3 \\ 3+(-5) & 9+0 & 1+1 \end{bmatrix}

A'+B' = \begin{bmatrix} -5 & 6 & -1\\ 3 &9 &4 \\ -2 & 9 & 2 \end{bmatrix}

Thus we find that the LHS is equal to RHS and hence verified.

Question 2(ii). If A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix} and B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix} , then verify

(A - B)' = A' - B'

Answer:

A = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix} and B = \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}

(A - B)' = A' - B'

L.H.S : (A - B)'

A-B = \begin{bmatrix} -1 & 2 & 3\\ 5 &7 &9 \\ -2 & 1 & 1 \end{bmatrix} - \begin{bmatrix} -4 & 1 & -5\\ 1 &2 &0 \\ 1 & 3 & 1 \end{bmatrix}

A-B = \begin{bmatrix} -1-(-4) & 2-1 & 3-(-5)\\ 5-1 &7-2 &9-0 \\ -2-1 & 1-3 & 1-1 \end{bmatrix}

A-B = \begin{bmatrix} 3 & 1 & 8\\ 4 &5 &9 \\ -3 & -2& 0 \end{bmatrix}

(A-B)' = \begin{bmatrix} 3 & 4 & -3\\ 1 &5 &-2 \\ 8 & 9& 0 \end{bmatrix}

R.H.S : A' - B'

A'-B' = \begin{bmatrix} -1 & 5 & -2\\ 2 &7 &1 \\ 3 & 9 & 1 \end{bmatrix} - \begin{bmatrix} -4 & 1 & 1\\ 1 &2 &3\\ -5 & 0 & 1 \end{bmatrix}

A'-B' = \begin{bmatrix} -1-(-4) & 5-1 & -2-1\\ 2-1 &7-2 &1-3 \\ 3-(-5) & 9-0 & 1-1 \end{bmatrix}

A'-B' = \begin{bmatrix} 3 & 4 & -3\\ 1 &5 &-2 \\ 8 & 9& 0 \end{bmatrix}

Hence, L.H.S = R.H.S. so verified that

(A - B)' = A' - B' .

Question 3(i). If A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix} and B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix} , then verify

(A + B)' = A' + B'

Answer:

A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix} B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}

A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}

To prove: (A + B)' = A' + B'

L.H.S : (A + B)' =

A+B = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}

A+B = \begin{bmatrix} 3+(-1) & -1+(-1)&0+1\\ 4+1 &2+2 & 1+3 \end{bmatrix}

A+B = \begin{bmatrix} 2 & -2&1\\ 5 &4 & 4 \end{bmatrix}

\therefore \, \, \, (A+B)' = \begin{bmatrix} 2 & 5\\ 1 &4\\1 & 4 \end{bmatrix}

R.H.S: A' + B'

A'+B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}

A'+B' = \begin{bmatrix} 2 & 5\\ 1 &4 \\ 1 & 4 \end{bmatrix}

Hence, L.H.S = R.H.S i.e. (A + B)' = A' + B' .

Question 3(ii). If A = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix} and B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix} , then verify

(A - B)' = A' - B'

Answer:

A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix} B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}

A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}

To prove: (A - B)' = A' - B'

L.H.S : (A - B)' =

A-B = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}

A-B = \begin{bmatrix} 3-(-1) & -1-(2)&0-1\\ 4-1 &2-2 & 1-3 \end{bmatrix}

A-B = \begin{bmatrix} 4 & -3&-1\\ 3 &0 & -2 \end{bmatrix}

\therefore \, \, \, (A-B)' = \begin{bmatrix} 4 & 3\\ -3 &0\\-1 & -2 \end{bmatrix}

R.H.S: A' - B'

A'-B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}

A'-B' = \begin{bmatrix} 4 & 3\\ -3 &0 \\ -1 & -2 \end{bmatrix}

Hence, L.H.S = R.H.S i.e. (A - B)' = A' - B' .

Question 4. If A' = \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix} and B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix} , then find (A + 2B)'

Answer:

B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}

A' = \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}

A=(A')' = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}

(A + 2B)' :

A+2B = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix} +2 \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}

A+2B = \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0\\ 2 & 4 \end{bmatrix}

A+2B = \begin{bmatrix} -2+(-2) & 1+0\\ 3+2 & 2+4 \end{bmatrix}

A+2B = \begin{bmatrix} -4 & 1\\ 5 & 6 \end{bmatrix}

Transpose is obtained by interchanging rows and columns and the transpose of A+2B is

(A+2B)' = \begin{bmatrix} -4 & 5\\ 1 & 6 \end{bmatrix}

Question 5(i) For the matrices A and B, verify that (AB)' = B'A' , where

A = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix} , B = \begin{bmatrix} -1& 2 &1 \end{bmatrix}

Answer:

A = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix} , B = \begin{bmatrix} -1& 2 &1 \end{bmatrix}

To prove : (AB)' = B'A'

L.H.S : (AB)'

AB = \begin{bmatrix} 1\\-4 \\3 \end{bmatrix} \begin{bmatrix} -1& 2 &1 \end{bmatrix}

AB = \begin{bmatrix} -1&2&1\\4&-8&-4 \\-3 &6&3\end{bmatrix}

(AB)' = \begin{bmatrix} -1&4&-3\\2&-8&6 \\1 &-4&3\end{bmatrix}

R.H.S : B'A'

B' = \begin{bmatrix} -1\\2 \\1 \end{bmatrix}

A' = \begin{bmatrix} 1& -4 &3 \end{bmatrix}

B'A' = \begin{bmatrix} -1\\2 \\1 \end{bmatrix} \begin{bmatrix} 1& -4 &3 \end{bmatrix}

B'A' = \begin{bmatrix} -1&4&-3\\2&-8&6 \\1&-4&3 \end{bmatrix}

Hence, L.H.S =R.H.S

so it is verified that (AB)' = B'A' .

Question 5(ii) For the matrices A and B, verify that (AB)' = B'A' , where

A = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix} , B = \begin{bmatrix} 1 & 5&7 \end{bmatrix}

Answer:

A = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix} , B = \begin{bmatrix} 1 & 5&7 \end{bmatrix}

To prove : (AB)' = B'A'

L.H.S : (AB)'

AB = \begin{bmatrix} 0\\1 \\2 \end{bmatrix} \begin{bmatrix} 1& 5 &7 \end{bmatrix}

AB = \begin{bmatrix} 0&0&0\\1&5&7 \\2 &10&14\end{bmatrix}

(AB)' = \begin{bmatrix} 0&1&2\\0&5&10 \\0 &7&14\end{bmatrix}

R.H.S : B'A'

B' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix}

A' = \begin{bmatrix} 0& 1 &2 \end{bmatrix}

B'A' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix} \begin{bmatrix} 0& 1 &2 \end{bmatrix}

B'A' = \begin{bmatrix} 0&1&2\\0&5&10 \\0&7&14 \end{bmatrix}

Heence, L.H.S =R.H.S i.e. (AB)' = B'A' .

Question 6(i). If A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} , then verify that A'A =I

Answer:

A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}

By interchanging rows and columns we get transpose of A

A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}

To prove: A'A =I

L.H.S : A'A

A'A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}

A'A = \begin{bmatrix} \cos^{2}\alpha + \sin ^{2}\alpha & \sin\alpha \cos \alpha - \sin \alpha \ cos \alpha \\ \sin\alpha \cos \alpha - \sin \alpha \cos \alpha & \ sin^{2}\alpha +\cos^{2}\alpha \end{bmatrix}

A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S

Question 6(ii). If A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix} , then verify that A'A = I

Answer:

A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}

By interchanging columns and rows of the matrix A we get the transpose of A

A' = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}

To prove: A'A =I

L.H.S : A'A

A'A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix} \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}

A'A = \begin{bmatrix} \cos^{2}\alpha + \sin ^{2}\alpha & \sin\alpha \cos \alpha - \sin \alpha \ cos \alpha \\ \sin\alpha \cos \alpha - \sin \alpha \cos \alpha & \ sin^{2}\alpha +\cos^{2}\alpha \end{bmatrix}

A'A = \begin{bmatrix} 1& 0 \\ 0& 1 \end{bmatrix}=I=R.H.S

Question 7(i). Show that the matrix A = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix} is a symmetric matrix.

Answer:

A = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}

the transpose of A is

A' = \begin{bmatrix} 1 &- 1& 5\\ -1 & 2 & 1\\ 5 & 1 & 3 \end{bmatrix}

Since, A' = A so given matrix is a symmetric matrix.

Question 7(ii) Show that the matrix A = \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix} is a skew-symmetric matrix.

Answer:

A = \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}

The transpose of A is

A' = \begin{bmatrix} 0 & -1 & 1\\ 1 & 0 &-1 \\- 1 & 1 &0 \end{bmatrix}

A' =- \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 &1 \\ 1 & -1 &0 \end{bmatrix}

A' =- A

Since, A' =- A so given matrix is a skew-symmetric matrix.

Question 8(i). For the matrix A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix} , verify that

(A + A') is a symmetric matrix.

Answer:

A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}

A' = \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}

A + A'= \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}

A + A'= \begin{bmatrix} 1+1 & 5+6\\ 6+5 & 7+7 \end{bmatrix}

A + A'= \begin{bmatrix}2 & 11\\ 11& 14 \end{bmatrix}

(A + A')'= \begin{bmatrix}2 & 11\\ 11& 14 \end{bmatrix}

We have A+A'=(A + A')'

Hence , (A + A') is a symmetric matrix.

Question 8(ii) For the matrix A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix} , verify that

(A - A') is a skew symmetric matrix.

Answer:

A = \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix}

A' = \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}

A - A'= \begin{bmatrix} 1 & 5\\ 6 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 6\\ 5 & 7 \end{bmatrix}

A - A'= \begin{bmatrix} 1-1 & 5-6\\ 6-5 & 7-7 \end{bmatrix}

A - A'= \begin{bmatrix}0 & -1\\ 1& 0 \end{bmatrix}

(A - A')'= \begin{bmatrix}0 & 1\\ -1& 0 \end{bmatrix}=-(A-A')

We have A-A'=-(A - A')'

Hence , (A - A') is a skew-symmetric matrix.

Question 9. Find \frac{1}{2}(A+A') and \frac{1}{2}(A-A') , when A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}

Answer:

A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}

the transpose of the matrix is obtained by interchanging rows and columns

A' = \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix}

\frac{1}{2}(A+A') = \frac{1}{2}(\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix} +\begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix})

\frac{1}{2}(A+A') = \frac{1}{2}(\begin{bmatrix} 0+0 & a+(-a) & b+(-b)\\ -a+a & 0+0 & c+(-c)\\ -b+b & -c+c & 0+0 \end{bmatrix})

\frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}

\frac{1}{2}(A+A') = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}

\frac{1}{2}(A+A') = 0


\frac{1}{2}(A-A') = \frac{1}{2}(\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix} - \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix})

\frac{1}{2}(A-A') = \frac{1}{2}(\begin{bmatrix} 0-0 & a-(-a) & b-(-b)\\ -a-a & 0-0 & c-(-c)\\ -b-b & -c-c & 0-0 \end{bmatrix})

\frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 2a &2 b\\ -2a & 0 & 2c\\ -2b & -2c & 0 \end{bmatrix}

\frac{1}{2}(A-A') = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}

Question 10(i). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}

Answer:

A=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix}

A'=\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}

A+A'=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix} +\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}

A+A'=\begin{bmatrix} 6 & 6\\ 6 & -2 \end{bmatrix}

Let

B=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 6 & 6\\ 6 & -2 \end{bmatrix} =\begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}

B'=\begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix}=B

Thus, \frac{1}{2}(A+A') is a symmetric matrix.


A-A'=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix} -\begin{bmatrix} 3 & 1\\ 5 & -1 \end{bmatrix}

A-A'=\begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix}

Let

C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix}

C'= \begin{bmatrix} 0 & -2\\ 2 & 0 \end{bmatrix}

C=-C'

Thus, \frac{1}{2}(A-A') is a skew symmetric matrix.

Represent A as sum of B and C.

B+C = \begin{bmatrix} 3 & 3\\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2\\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 5\\ 1 & -1\end{bmatrix}=A

Question:10(ii). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}

Answer:

A=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}

A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}

A+A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix} + \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}

A+A'=\begin{bmatrix} 12 & -4 & 4\\ -4 & 6 & -2\\ 4 & -2 & 6 \end{bmatrix}

Let

B= \frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 12 & -4 & 4\\ -4 & 6 & -2\\ 4 & -2 & 6 \end{bmatrix} = \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}

B'= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}=B

Thus, \frac{1}{2}(A+A') is a symmetric matrix.


A-A'=\begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix} - \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}

A-A'=\begin{bmatrix} 0 & 0&0\\ 0 & 0&0 \\0&0&0\end{bmatrix}

Let

C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix} =\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}

C'=\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix}

C=-C'

Thus, \frac{1}{2}(A-A') is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C= \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix} +\begin{bmatrix} 0 & 0&0\\ 0&0 & 0\\0&0&0 \end{bmatrix} = \begin{bmatrix} 6 & -2 & 2\\ -2 & 3 & -1\\ 2 & -1 & 3 \end{bmatrix}=A

Question 10(iii). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}

Answer:

A=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}

A'=\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}

A+A'=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix} +\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}

A+A'=\begin{bmatrix} 6 & 1 & -5\\ 1& -4 & -4\\ -5 & -4 & 4 \end{bmatrix}

Let

B= \frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 6 & 1 & -5\\ 1 & -4 & -4\\ -5 & -4 & 4 \end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}

B'= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix}=B

Thus, \frac{1}{2}(A+A') is a symmetric matrix.


A-A'=\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix} -\begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}

A-A'=\begin{bmatrix} 0 & 5&3\\ -5 & 0&6 \\-3&-6&0\end{bmatrix}

Let

C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 5&3\\ -5&0 & 6\\-3&-6&0 \end{bmatrix} =\begin{bmatrix} 0 & \frac{5}{2}&\frac{3}{2}\\ -\frac{5}{2}&0 & 3\\\frac{-3}{2}&-3&0 \end{bmatrix}

C'=\begin{bmatrix} 0 &- \frac{5}{2}&-\frac{3}{2}\\ \frac{5}{2}&0 &- 3\\\frac{3}{2}&3&0 \end{bmatrix}

C=-C'

Thus, \frac{1}{2}(A-A') is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C= \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2}\\ \frac{1}{2} & -2 & -2\\ \frac{-5}{2} & -2 & 2 \end{bmatrix} +\begin{bmatrix} 0 & \frac{5}{2}&\frac{3}{2}\\ -\frac{5}{2}&0 & 3\\\frac{-3}{2}&-3&0 \end{bmatrix} =\begin{bmatrix} 3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2 \end{bmatrix}=A

Question 10(iv). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}

Answer:

A =\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix}

A'=\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}

A+A'=\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix} +\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}

A+A'=\begin{bmatrix} 2 & 4\\ 4 & 4 \end{bmatrix}

Let

B=\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 2 & 4\\ 4 & 4 \end{bmatrix} =\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}

B'=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix}=B

Thus, \frac{1}{2}(A+A') is a symmetric matrix.

A-A'=\begin{bmatrix} 1 & 5\\ -1 & 2 \end{bmatrix} -\begin{bmatrix} 1 & -1\\ 5 & 2 \end{bmatrix}

A-A'=\begin{bmatrix} 0 & 6\\ -6 & 0 \end{bmatrix}

Let

C= \frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix} 0 & 6\\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3\\ -3 & 0 \end{bmatrix}

C'= \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix}

C=-C'

Thus, \frac{1}{2}(A-A') is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C=\begin{bmatrix} 1 & 2\\ 2 & 2 \end{bmatrix} - \begin{bmatrix} 0 & -3\\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 5\\ -1 & 2\end{bmatrix}=A


NCERT solutions for class 12 maths chapter 3 Matrices: Exercise 3.4

Question 1 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix}1&-1\\2&3 \end{bmatrix}

Answer:

Use the elementary transformation we can find the inverse as follows

A=\begin{bmatrix}1&-1\\2&3 \end{bmatrix}

A=IA

\Rightarrow \begin{bmatrix}1&-1\\2&3 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_2\rightarrow R_2-2R_1

\Rightarrow \begin{bmatrix}1&-1\\0&5 \end{bmatrix} = \begin{bmatrix}1&0\\-2&1 \end{bmatrix}A

R_2\rightarrow \frac{R_2}{5}

\Rightarrow \begin{bmatrix}1&-1\\0&1 \end{bmatrix} = \begin{bmatrix}1&0\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}A

R_1\rightarrow R_1+R_2

\Rightarrow \begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}A

\therefore A^{-1}= \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}

Question 2 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 2&1\\1&1\end{bmatrix}

Answer:

A=\begin{bmatrix} 2&1\\1&1\end{bmatrix}

A=IA

\Rightarrow \begin{bmatrix} 2&1\\1&1\end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_1\rightarrow R_1-R_2

\Rightarrow \begin{bmatrix}1&0\\1&1 \end{bmatrix} = \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A

R_2\rightarrow R_2-R_1

\Rightarrow \begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}1&-1\\-1& 2 \end{bmatrix}A

A^{-1}= \begin{bmatrix}1&-1\\-1& 2 \end{bmatrix}

Thus we have obtained the inverse of the given matrix through elementary transformation

Question 7 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix}

Answer:

A=\begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix}

A=IA

\Rightarrow \begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_2\rightarrow R_2-R_1

\Rightarrow \begin{bmatrix} 3 & 1\\ 2 & 1 \end{bmatrix} = \begin{bmatrix}1&0\\-1&1 \end{bmatrix}A

\Rightarrow R_1\rightarrow R_1-R_2

\Rightarrow \begin{bmatrix} 1 & 0\\ 2 & 1 \end{bmatrix} = \begin{bmatrix}2&-1\\-1&1 \end{bmatrix}A

R_2\rightarrow R_2-2R_1

\Rightarrow \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix}2&-1\\-5&3 \end{bmatrix}A

\therefore A^{-1}= \begin{bmatrix}2&-1\\-5&3 \end{bmatrix} .

Thus the inverse of matrix A is obtained using elementary transformation.

Question 8 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}

Answer:

A=\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}

A=IA

\Rightarrow\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_1\rightarrow R_1-R_2

\Rightarrow\begin{bmatrix} 1 &1 \\ 3 & 4 \end{bmatrix}= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A

\Rightarrow R_2\rightarrow R_2-3R_1

\Rightarrow \begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}1&-1\\-3&4 \end{bmatrix}A

R_1\rightarrow R_1-R_2

\Rightarrow \begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}4&-5\\-3&4 \end{bmatrix}A

Thus using elementary transformation inverse of A is obtained as

\therefore A^{-1}= \begin{bmatrix}4&-5\\-3&4 \end{bmatrix} .

Question 9 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}

Answer:

A=\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}

A=IA

\Rightarrow\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_1\rightarrow R_1-R_2

\Rightarrow\begin{bmatrix} 1& 3\\ 2 & 7 \end{bmatrix}= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A

\Rightarrow R_2\rightarrow R_2-2R_1

\Rightarrow\begin{bmatrix} 1& 3\\ 0 & 1 \end{bmatrix}= \begin{bmatrix}1&-1\\-2&3 \end{bmatrix}A

R_1\rightarrow R_1-3R_2

\Rightarrow\begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}= \begin{bmatrix}7&-10\\-2&3 \end{bmatrix}A

Thus using elementary transformation the inverse of A is obtained as

\therefore A^{-1}= \begin{bmatrix}7&-10\\-2&3 \end{bmatrix} .

Question 10 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}

Answer:

A=\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}

A=IA

\Rightarrow\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_2\rightarrow R_2+R_1

\Rightarrow\begin{bmatrix} 3 & -1\\ -1 & 1 \end{bmatrix}= \begin{bmatrix}1&0\\1&1 \end{bmatrix}A

\Rightarrow R_1\rightarrow R_1+2R_2

\Rightarrow\begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix}= \begin{bmatrix}3&2\\1&1 \end{bmatrix}A

R_2\rightarrow R_2+R_1

\begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix}= \begin{bmatrix}3&2\\4&3 \end{bmatrix}A

R_2\rightarrow \frac{R_2}{2}

\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}= \begin{bmatrix}3&2\\2&\frac{3}{2} \end{bmatrix}A

R_1\rightarrow R_1-R_2

\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}= \begin{bmatrix}1&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix}A

\therefore A^{-1}= \begin{bmatrix}1&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix} .

Thus the inverse of A is obtained using elementary transformation.

Question 11 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}

Answer:

A=\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}

A=IA

\Rightarrow\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_1\rightarrow R_1-3R_1

\Rightarrow\begin{bmatrix} -1 & 0\\ 1 & -2 \end{bmatrix}= \begin{bmatrix}1&-3\\0&1 \end{bmatrix}A

\Rightarrow R_2\rightarrow R_2+R_1

\Rightarrow\begin{bmatrix} -1 & 0\\ 0 & -2 \end{bmatrix}= \begin{bmatrix}1&-3\\1&-2 \end{bmatrix}A

R_1\rightarrow -R_1

\Rightarrow\begin{bmatrix} 1 & 0\\ 0 & -2 \end{bmatrix}= \begin{bmatrix}-1&3\\1&-2 \end{bmatrix}A

R_2\rightarrow \frac{R_2}{-2}

\Rightarrow\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}= \begin{bmatrix}-1&3\\\frac{1}{-2}&1 \end{bmatrix}A

thus the inverse of matrix A is

\therefore A^{-1}= \begin{bmatrix}-1&3\\\frac{1}{-2}&1 \end{bmatrix} .

Question 12 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}

Answer:

A=\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}

A=IA

\Rightarrow \begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_1\rightarrow \frac{R_1}{6}

\Rightarrow \begin{bmatrix} 1& -\frac{1}{2}\\ -2 & 1 \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\0&1 \end{bmatrix}A

\Rightarrow R_2\rightarrow R_2+2R_1

\Rightarrow \begin{bmatrix} 1& -\frac{1}{2}\\ 0 & 0 \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\\frac{1}{3}&1 \end{bmatrix}A

Hence, we can see all the zeros in the second row of the matrix in L.H.S so A^{-1} does not exist.

Question 14 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}

Answer:

A=\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}

A=IA

\Rightarrow\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

R_2\rightarrow \frac{R_2}{2}

\Rightarrow\begin{bmatrix} 2 & 1\\ 2 & 1 \end{bmatrix}= \begin{bmatrix}1&0\\0&\frac{1}{2} \end{bmatrix}A

\Rightarrow R_1\rightarrow R_1-R_2

\Rightarrow\begin{bmatrix} 0 & 0\\ 2 & 1 \end{bmatrix}= \begin{bmatrix}1&\frac{-1}{2}\\0&\frac{1}{2} \end{bmatrix}A

Hence, we can see all upper values of matirix are zeros in L.H.S so A^{-1} does not exists.

Question 16 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}

Answer:

A=\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}

A=IA

\Rightarrow\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}= \begin{bmatrix}1&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A

R_2\rightarrow R_2+3R_1 and R_3\rightarrow R_3-2R_1

\Rightarrow\begin{bmatrix} 1 &3 & -2\\ 0& 9 &-11 \\ 0 & -1 & 4 \end{bmatrix}= \begin{bmatrix}1&0&0\\3&1&0 \\-2&0&1 \end{bmatrix}A

R_1\rightarrow R_1+3R_3 and R_2\rightarrow R_2+8R_3

\Rightarrow\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & -1 & 4 \end{bmatrix}= \begin{bmatrix}-5&0&3\\-13&1&8 \\-2&0&1 \end{bmatrix}A

\Rightarrow R_3\rightarrow R_3+R_2

\Rightarrow\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & 0 & 25 \end{bmatrix}= \begin{bmatrix}-5&0&3\\-13&1&8 \\-15&1&9 \end{bmatrix}A

R_3\rightarrow \frac{R_3}{25}

\Rightarrow\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & 0 & 1 \end{bmatrix}= \begin{bmatrix}-5&0&3\\-13&1&8 \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A

R_1\rightarrow R_1-10R_3 and R_2\rightarrow R_2-21R_3

\Rightarrow\begin{bmatrix} 1 &0 & 0\\ 0& 1 &0 \\ 0 & 0 & 1 \end{bmatrix}= \begin{bmatrix}1&\frac{-2}{5}&\frac{-3}{5}\\\frac{-2}{5}&\frac{4}{25}&\frac{11}{25} \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A

Thus the inverse of three by three matrix A is

\therefore A^{-1}= . \begin{bmatrix}1&\frac{-2}{5}&\frac{-3}{5}\\\frac{-2}{5}&\frac{4}{25}&\frac{11}{25} \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix} .

Question:18 Matrices A and B will be inverse of each other only if

(A) AB = BA

(B) AB = BA = 0

(C) AB = 0, BA = I

(D) AB = BA = I

Answer:

We know that if A is a square matrix of order n and there is another matrix B of same order n, such that AB=BA=I , then B is inverse of matrix A.

In this case, it is clear that A is inverse of B.

Hence, m atrices A and B will be inverse of each other only if AB=BA=I .

Option D is correct .


NCERT solutions for class 12 maths chapter - 3 Matrices: Miscellaneous exercise

Question:1 Let A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix} , show that (aI + bA)^n = a^n I + na^{n-1} bA , where I is the identity matrix of order 2 and n \in N .

Answer:

Given :

A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}

To prove : (aI + bA)^n = a^n I + na^{n-1} bA

For n=1, aI + bA = a I + a^{0} bA =a I + bA

The result is true for n=1.

Let result be true for n=k,

(aI + bA)^k = a^k I + ka^{k-1} bA

Now, we prove that the result is true for n=k+1,

(aI + bA)^k^+^1 = (aI + bA)^k^(aI + bA)

= (a^k I + ka^{k-1} bA) (aI + bA)

=a^{k+1}I+Ka^{k}bAI+a^{k}bAI+ka^{k-1}b^{2}A^{2}

=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}

A^{2} = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}

A^{2} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}=0

Put the value of A^{2} in above equation,

(aI + bA)^k^+^1 =a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}

(aI + bA)^k^+^1 =a^{k+1}I+(k+1)a^{k}bAI+0

=a^{k+1}I+(k+1)a^{k}bAI

Hence, the result is true for n=k+1.

Thus, we have (aI + bA)^n = a^n I + na^{n-1} bA where A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix} , n \in N .

Question 2. If A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix} then show that A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix} , n\in N .

Answer:

Given :

A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}

To prove:

A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}

For n=1, we have

A^1 =\begin{bmatrix} 3^{1-1} & 3^{1-1} &3^{1-1} \\ 3^{1-1}& 3^{1-1} & 3^{1-1}\\ 3^{1-1} & 3^{1-1}& 3^{1-1} \end{bmatrix} =\begin{bmatrix} 3^{0} & 3^{0} &3^{0} \\ 3^{0}& 3^{0} & 3^{0}\\ 3^{0} & 3^{0}& 3^{0} \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}=A

Thus, the result is true for n=1.

Now, take n=k,

A^k =\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}

For, n=k+1,

A^{K+1}=A.A^K

= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix} \begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}

=\begin{bmatrix}3. 3^{k-1} & 3.3^{k-1} &3.3^{k-1} \\3. 3^{k-1}& 3.3^{k-1} & 3.3^{k-1}\\3. 3^{k-1} & 3.3^{k-1}&3. 3^{k-1} \end{bmatrix}

=\begin{bmatrix} 3^{(K+1)-1} &3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1}&3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1} & 3^{(K+1)-1}& 3^{(K+1)-1}\end{bmatrix}

Thus, the result is true for n=k+1.

Hence, we have A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix} , n\in N where A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix} .

Question 3. If A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix} , then prove that A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix} , where n is any positive integer.

Answer:

Given :

A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}

To prove:

A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix}

For n=1, we have

A^1 = \begin{bmatrix} 1+2\times 1 & -4\times 1\\ 1 & 1-2\times 1 \end{bmatrix} = \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}=A

Thus, result is true for n=1.

Now, take result is true for n=k,

A^k = \begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}

For, n=k+1,

A^{K+1}=A.A^K

= \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix} \begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}

=\begin{bmatrix} 3(1+2k)-4k & -12k-4(1-2k)\\ (1+2k)-k &-4k-(1-2k) \end{bmatrix}

=\begin{bmatrix} 3+6k-4k & -12k-4k+8k\\ 1+k &-4k-1+2k \end{bmatrix}

=\begin{bmatrix} 3+2k & -4k-4k\\ 1+k &-2k-1 \end{bmatrix}

=\begin{bmatrix} 1+2(k+1)& -4(k+1)\\ 1+k &1-2(k+1) \end{bmatrix}

Thus, the result is true for n=k+1.

Hence, we have A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix} , where A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix} .

Question 4. If A and B are symmetric matrices, prove that AB - BA is a skew symmetric matrix.

Answer:

If A, B are symmetric matrices then

A'=A and B' = B

we have, \left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'

=BA-AB

= -(AB-BA)

Hence, we have (AB-BA) = -(AB-BA)'

Thus,( AB-BA)' is skew symmetric.

Question 5. Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

Let be a A is symmetric matrix , then A'=A

Consider, (B'AB)' ={B'(AB)}'

={(AB)}'(B')'

= B'A'(B)

= B'(A'B)

Replace A' by A

=B'(AB)

i.e. (B'AB)' =B'(AB)

Thus, if A is a symmetric matrix than B'(AB) is a symmetric matrix.


Now, let A be a skew-symmetric matrix, then A'=-A .

(B'AB)' ={B'(AB)}'

={(AB)}'(B')'

= B'A'(B)

= B'(A'B)

Replace A' by - A ,

=B'(-AB)

= - B'AB

i.e. (B'AB)' = - B'AB .

Thus, if A is a skew-symmetric matrix then - B'AB is a skew-symmetric matrix.

Hence, the matrix B′AB is symmetric or skew-symmetric according to as A is symmetric or skew-symmetric.

Question 6. Find the values of x , y , z if the matrix A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix} satisfy the equation A'A = I

Answer:

A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}

A' = \begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}

A'A = I

\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} = \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}

\begin{bmatrix} x^{2}+x^{2} & xy-xy& -xz+xz\\ xy-xy& 4y^{2}+y^{2}+y^{2} & 2yz-yz-yz\\ -zx+zx & 2yz-yz-yz &z^{2}+z^{2}+z^{2}\end{bmatrix} = \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}

\begin{bmatrix} 2x^{2} & 0& 0\\ 0& 6y^{2} & 0\\ 0 & 0 &3z^{2}\end{bmatrix} = \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}

Thus equating the terms elementwise

2x^{2} = 1 6y^{2} = 1 3z^{2} = 1

x^{2} = \frac{1}{2} y^{2} = \frac{1}{6} z^{2}=\frac{1}{3}

x = \pm \frac{1}{\sqrt{2}} y= \pm \frac{1}{\sqrt{6}} z=\pm \frac{1}{\sqrt{3}}

Question 7. For what values of x: \begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O ?

Answer:

\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O

\begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O

\begin{bmatrix} 6& 2& 4 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O

\begin{bmatrix} 0+4+4x \end{bmatrix} = O

4+4x=0

4x=-4

x=-1

Thus, value of x is -1.

Question 8. If A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix} , show that A^2 -5A + 7I= 0 .

Answer:

A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}

A^{2} = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}

A^{2} = \begin{bmatrix} 9-1 &3+2 \\ -3-2 & -1+4 \end{bmatrix}

A^{2} = \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}

I= \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}

To prove: A^2 -5A + 7I= 0

L.H.S : A^2 -5A + 7I

= \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix} -5 \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}

=\begin{bmatrix} 8-15+7 &5-5+0 \\ -5+5+0& 3-10+7 \end{bmatrix}

=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix} =0=R.H.S

Hence, we proved that

A^2 -5A + 7I= 0 .

Question 9. Find x, if \begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0 .

Answer:

\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0

\begin{bmatrix} x +0-2& 0-10+0 & 2x-5-3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0

\begin{bmatrix} x -2& -10 & 2x-8 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0

\begin{bmatrix}x (x -2)-40+(2x-8) \end{bmatrix} = 0

\begin{bmatrix}x ^{2}-2x-40+2x-8\end{bmatrix} = 0

\therefore \, \, x ^{2}-48= 0

x ^{2}=48

thus the value of x is

x =\pm 4\sqrt{3}

Question 10(a) A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

Answer:

The unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively.

The total revenue in the market I with the help of matrix algebra can be represented as :

\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}

= 10000\times 2.50+2000\times 1.50+18000\times 1.00

= 25000+3000+18000

= 46000

The total revenue in market II with the help of matrix algebra can be represented as :

\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}

= 6000\times 2.50+20000\times 1.50+8000\times 1.00

= 15000+30000+8000

= 53000

Hence, total revenue in the market I is 46000 and total revenue in market II is 53000.

Question 10(b). A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit.

Answer:

The unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively.

The total cost price in market I with the help of matrix algebra can be represented as :

\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}

= 10000\times 2.00+2000\times 1.00+18000\times 0.50

= 20000+2000+9000

= 31000

Total revenue in the market I is 46000 , gross profit in the market is = 46000-31000 =Rs. 15000

The total cost price in market II with the help of matrix algebra can be represented as :

\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}

= 6000\times 2.0+20000\times 1.0+8000\times 0.50

= 12000+20000+4000

= 36000

Total revenue in market II is 53000, gross profit in the market is = 53000-36000= Rs. 17000

Question 11. Find the matrix X so that X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}

Answer:

X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}

The matrix given on R.H.S is 2\times 3 matrix and on LH.S is 2\times 3 matrix.Therefore, X has to be 2\times 2 matrix.

Let X be \begin{bmatrix} a & c\\ b & d \end{bmatrix}

\begin{bmatrix} a & c\\ b & d \end{bmatrix} \begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}

\begin{bmatrix} a+4c & 2a+5c &3a+6c \\ b+4d & 2b+5d & 3b+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}

a+4c=-7 2a+5c=-8 3a+6c=-9

b+4d=2 2b+5d=4 3b+6d=6

Taking, a+4c=-7

a=-4c-7

2a+5c=-8

-8c-14+5c=-8

-3c=6

c=-2

a=-4\times -2-7

a=8-7=1

b+4d=2

b=-4d+2

2b+5d=4

\Rightarrow -8d+4+5d=4

\Rightarrow -3d=0

\Rightarrow d=0

b=-4d+2

\Rightarrow b=-4\times 0+2=2

Hence, we have a=1, b=2,c=-2,d=0

Matrix X is \begin{bmatrix} 1 & -2\\ 2 & 0 \end{bmatrix} .

Question 12. If A and B are square matrices of the same order such that AB = BA , then prove by induction that AB^n = B^n A . Further, prove that (AB)^n = A^n B^n for all n \in N .

Answer:

A and B are square matrices of the same order such that AB = BA ,

To prove : AB^n = B^n A , n \in N

For n=1, we have AB^1 = B^1 A

Thus, the result is true for n=1.

Let the result be true for n=k,then we have AB^k = B^k A

Now, taking n=k+1 , we have AB^{k+1} = AB^k .B

AB^{k+1} = (B^kA) .B

AB^{k+1} = (B^k) .AB

AB^{k+1} = (B^k) .BA

AB^{k+1} = (B^k.B) .A

AB^{k+1} = (B^k^+^1) .A

Thus, the result is true for n=k+1.

Hence, we have AB^n = B^n A , n \in N .

To prove: (AB)^n = A^n B^n

For n=1, we have (AB)^1 = A^1 B^1

Thus, the result is true for n=1.

Let the result be true for n=k,then we have (AB)^k = A^k B^k

Now, taking n=k+1 , we have (AB)^{k+1} = (A B)^k.(AB)

(AB)^{k+1} = A^k B^k.(AB)

(AB)^{k+1} = A^{K}( B^kA)B

(AB)^{k+1} = A^{K}( AB^k)B

(AB)^{k+1} = (A^{K}A)(B^kB)

(AB)^{k+1} = (A^{k+1})(B^{k+1})

Thus, the result is true for n=k+1.

Hence, we have AB^n = B^n A and (AB)^n = A^n B^n for all n \in N .

Question 14. 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

Answer:

If the matrix A is both symmetric and skew-symmetric, then

A'=A and A'=-A

A'=A'

\Rightarrow \, \, \, \, \, \, \, A=-A

\Rightarrow \, \, \, \, \, \, \, A+A=0

\Rightarrow \, \, \, \, \, \, \, 2A=0

\Rightarrow \, \, \, \, \, \, \, A=0

Hence, A is a zero matrix.

Option B is correct.

Question 15. If A is square matrix such that A^{2}=A , then (I + A)^3 - 7 A is equal to

(A) A
(B) I – A
(C) I
(D) 3A

Answer:

A is a square matrix such that A^{2}=A

(I + A)^3 - 7 A

=I^{3}+A^{3}+3I^{2}A+3IA^{2}-7A

=I+A^{2}.A+3A+3A^{2}-7A

=I+A.A+3A+3A-7A (Replace A^{2} by A )

=I+A^{2}+6A-7A

=I+A-A

=I

Hence, we have (I + A)^3 - 7 A=I

Option C is correct.

If you are interested in Matrices Class 12 NCERT Solutions exercises then these are listed below.

Matrices Class 12 NCERT Solutions Exercise 3.1

Matrices Class 12 NCERT Solutions Exercise 3.2

Matrices Class 12 NCERT Solutions Exercise 3.3

Matrices Class 12 NCERT Solutions Exercise 3.4

Matrices Class 12 NCERT Solutions Miscellaneous Exercise

More about NCERT Solutions for Class 12 Maths Chapter 3 Matrices

Matrix is an array of numbers. Matrix is a mode of representing data to ease calculation and it is one of the most important tools of mathematics because matrices simplify our work to a great extent when compared with other straight forward methods. Exercises are given in chapter 3 maths class 12 students should refer for practice. Matrices class 12 solutions also provided by careers360’s expert team if you are facing problems then you can refer to it as well.

Matrices are used as a representation of the coefficients in the system of linear equations, electronic spreadsheet programs, also used in business and science. For the students to understand NCERT class 12 maths solutions chapter 3 in a better way, a total of 28 solved examples are given to practice more, at the end of the chapter, 15 questions are given in the miscellaneous exercise. For command on concepts you can uses these practice problems of matrices class 12.

Matrices Class 12 NCERT solutions - Topics

3.1 Introduction

3.2 Matrix

3.2.1 Order of a matrix

3.3 Types of Matrices

3.3.1 Equality of matrices

3.4 Operations on Matrices

3.4.1 Addition of matrices

3.4.2 Multiplication of a matrix by a scalar

3.4.3 Properties of matrix addition

3.4.4 Properties of scalar multiplication of a matrix

3.4.5 Multiplication of matrices

3.4.6 Properties of multiplication of matrices

3.5. Transpose of a Matrix

3.5.1 Properties of the transpose of the matrices

3.6 Symmetric and Skew-Symmetric Matrices

3.7 Elementary Operation (Transformation) of a Matrix

3.8 Invertible Matrices

3.8.1 Inverse of a matrix by elementary operations

Topics of NCERT Class 12 Maths Chapter Matrices

The main topics covered in maths chapter 3 class 12 are:

  • Matrix

It is an ordered representation of numbers and functions, known as elements in a rectangular array. In this class 12 NCERT topics discuss concepts related to the order of a matrix, elements in raw, and columns of a matrix. there are good quality questions in matrix class 12 solutions.

  • Type of matrices

This ch 1 maths class 12 concerns different types of matrices like column matrix, raw matrix, square matrix, diagonal matrix, scalar matrix, identity matrix, zero matrix, etc. In this ch 3 maths class 12 also discuss comprehensively conditions of equality of matrix. To get command on these concepts you can refer to NCERT solutions for class 12 maths chapter 3.

  • Operations on matrices

This ch 3 maths class 12 also includes concepts of addition of matrices, multiplication of a matrix by a scalar, negative matrix, difference of matrices. maths class 12 chapter 3 also contains properties of matrix addition that include commutative leas, associative laws, the existence of an additive identity, the existence of additive inverse. Properties of scalar multiplication of matrix elaborated in this chapter. You can refer to class 12 NCERT solutions for questions about these concepts.

  • multiplication of matrix

Properties of multiplication of matrix that includes associative law, distributive laws, the existence of multiplicative identity. to get command of these concepts you can go through the NCERT solution for class 12 maths chapter 3.

  • transpose of a matrix

properties of transpose of the matrix are discussed in maths class 12 chapter 3. matrix class 12 solutions include quality questions to understand the concepts.

ch 3 maths class 12 also discuss in detail the concepts of symmetric and skew symmetric matrix, elementary operations (transformation) of a matrix. for questions on these concepts, you can browse NCERT solutions for class 12 chapter 3.

  • Invertible matrix

In class 12 NCERT chapter you get an idea of the inverse of a matrix by elementary operation. class 12 NCERT solutions include problems related to these concepts

Topics enumerated in class 12 NCERT are very important and students are suggested to go through all the concepts discussed in the topics. Questions related to all the above topics are covered in the NCERT solutions for class 12 maths chapter 3

Also read,

NCERT Solutions for Class 12 - Subject Wise

NCERT solutions for class 12 maths - Chapter wise

How to use NCERT Solutions for Class 12 Maths Chapter 3 Matrices

  • NCERT Class 12 Maths solutions chapter 3 covers all the questions given in the textbook. Make sure you have gone through all the important concepts of NCERT class 12th maths Matricesbefore solving the questions.

  • Firstly try to solve the questions on your own and then check the NCERT class 12 maths chapter 3 solutions.

  • Along with the NCERT Solutions for Class 12 Maths Chapter 3 Matrices, solve the previous year questions as well.

  • NCERT Class 12 maths chapter 3 solutions pdf download will be made available soon. Till then you can save this webpage instead of NCERT class 12 maths chapter 3 pdf to practice questions offline.

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. How are the NCERT solutions helpful in the board exam?

NCERT solutions are not only important when you are stuck while solving the problems but you will get to know how to answer in the board exam in order to get good marks in the board exam. these class 12 maths ch 3 question answer also make you understand new concepts or in depth understanding of topics and can solve you doubts. in this way ncert solutions becomes important so you can refer ncert textbooks, ncert exercises and ncert solutions.

2. What are the important topics in chapter matrices?

Matrices, order of a matrix, types of matrices, equality of matrices, operations like addition multiplication on matrices, symmetric and skew-symmetric matrices are the important topics in this chapter. these topics are very important because of use in other supplementary topics like 3d geometry.

3. Can using class 12 maths chapter 3 ncert solutions improve your performance on the board exam?

Yes, Class 12 Matrices solutions definitely improve the performance on the board exam. It's recommended that students begin by tackling simpler problems before moving on to more challenging ones. Class 12 matrices NCERT solutions will enable them to identify areas where they need improvement. Through repeated practice on their weaker concepts, students can strengthen their skills and perform well on the board exams. Additionally, short-cut tips are provided to assist students in finding easier ways to solve complex problems.

4. What is the weightage of the chapter matrices for CBSE board exam?

The topic algebra which contains two topics matrices and determinants which has 13 % weightage in the maths CBSE 12th board final examination. students can list out topics according to respective weightage and channelise their energy according to priority of the topics.

5. What are the primary themes covered in NCERT Solutions for class 12 chapter 3 maths?

Mathematics often has simple chapters that can be enjoyable to study once understood, and matrices are one example. NCERT Class 12 maths matrices focus on the following main topics: matrices, types of matrices, operations on matrices, transpose of a matrix, symmetric and skew symmetric matrices, elementary operations on matrices, and invertible matrices. The material is presented in a straightforward manner to help students achieve good grades in the board exams, regardless of their intelligence. For ease, students can study matrices class 12 ncert pdf both online and offline.

Articles

Explore Top Universities Across Globe

University of Essex, Colchester
 Wivenhoe Park Colchester CO4 3SQ
University College London, London
 Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
 Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
 Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
 University Park, Nottingham NG7 2RD
Lancaster University, Lancaster
 Bailrigg, Lancaster LA1 4YW

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


Best CBSE schools in Delhi: Click Here


In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top