NCERT Solutions for Class 12 Maths Chapter 3 Matrices

 

NCERT Solutions for Class 12 Maths Chapter 3 Matrices: Matrix is a Latin word which means "womb", an environment where something grows. In this article, you will find NCERT solutions for class 12 maths chapter 3 matrices. Matrix is an array of numbers or mathematical objects for which some operations like addition and multiplication are defined. Matrices is an important and powerful tool in mathematics and it's basically introduced to solve simultaneous linear equations. Matrices have a lot of applications like solving the system of linear equations, doing transformations of one vector space to another, etc. It has applications in engineering, for example, the image or video that you see are matrices of color intensities. In this chapter, you will learn about matrices and it's properties. In order to get the solutions of NCERT for class 12 maths chapter 3 matrices, you can go through this article. Important topics that are going to be discussed in this chapter are matrices, the order of a matrix, types of matrices, equality of matrices, operations like addition multiplication on matrices, symmetric and skew-symmetric matrices, etc. In CBSE NCERT solutions for class 12 maths chapter 3 matrices article, questions from all topics are covered. The practice is very important to command over any chapter of CBSE maths, so you should try to solve every problem on your own. If you are not able to solve, you can take the help of NCERT solutions for class 12 maths chapter 3 matrices, which are explained in a detailed manner. Check all NCERT solutions from class 6 to 12 at a single place which will be helpful in order to learn CBSE science and maths. 

The topic algebra which contains two topics matrices and determinants which has 13 % weightage in the maths CBSE 12th board final examination, which means you will see 10 marks questions from these two chapters matrices, and determinants in 12th board final exam out of 80 marks. Matrix is a very important chapter from the exam point of view, also from the application point of view, it is very important in further studies like engineering. In this chapter, there are 4 exercises with 62 questions. All these questions are prepared and explained in the NCERT solutions for class 12 maths chapter 3 matrices article. These solutions of NCERT will help you to understand the concept more easily, and perform well in the CBSE 12th  board exam.

What are 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. 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 chapter 3 matrices in a better way total of 28 solved examples are given and also fto practice more, at the end of the chapter, 15 questions are given in the miscellaneous exercise. 

Topics and sub-topics of NCERT Grade 12 Maths Chapter 3 Matrices

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

 

NCERT solutions for class 12 maths chapter 3 matrices- Solved exercise questions

Solutions of NCERT class 12 maths chapter 3 Matrices: 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 a13, a21, a33a24, a23

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. 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.

\therefore m=n

Option (c) is correct.

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.

CBSE NCERT solutions for class 12 maths chapter 3 Matrices: 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(iv) 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:

             AB

Answer:

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

(iv) AB

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

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

AB = \begin{bmatrix} -6 &26 \\ -1 & 19 \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. FindA^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 thatI + 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 = nare 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.

CBSE NCERT Solutions for class 12 maths chapter -3 Matrices: 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

Question 11 Choose the correct answer in the Exercises 11 and 12.

    If A, B are symmetric matrices of same order, then AB – BA is a

            (A) Skew symmetric matrix
            (B) Symmetric matrix
            (C) Zero matrix
            (D) Identity 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.

Option A is correct.

Question 12 Choose the correct answer in the Exercises 11 and 12.

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

       \begin{bmatrix} 1 &3 \\ 2 & 7 \end{bmatrix}

Answer:

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

                A=IA

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

Using elementary transformations

             R_2\rightarrow R_2-2R_1

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

                  

\Rightarrow          R_1\rightarrow R_1-3R_2

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

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

Question 4 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

        \begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}

Answer:

A=\begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}

                A=IA

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

             R_2\rightarrow R_2-2R_1

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