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Do you know how search engines like Google list websites one by one? Or how congested traffic flow is managed on busy streets? Matrices, a powerful mathematical tool, help provide answers to such questions by solving complex systems of equations and efficiently managing data.
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Class 12 Maths NCERT Chapter 3: Matrices covers several important topics such as the Definition of a Matrix, Order of a Matrix, Types of Matrices, Operations on Matrices, Transpose of a Matrix, and more. Understanding these concepts enables students to solve matrix-related problems with ease and enhances their problem-solving ability in real-world applications.
The main purpose of the NCERT Solutions for Class 12 Chapter 3: Matrices is to give students a clear understanding of the fundamental concepts of matrices and their operations. These solutions also aim to strengthen their problem-solving skills in real-life applications, such as solving systems of linear equations and performing transformations in geometry.
This article on NCERT solutions for Class 12 Maths Chapter 3: Matrices offers clear, step-by-step solutions for the exercise problems given in the NCERT textbook. Following the latest syllabus, these solutions have been prepared by Careers360’s experienced subject matter experts, ensuring that students can effectively grasp the fundamental concepts.
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Students who wish to access the Class 12 Maths Chapter 3 Solutions PDF can click on the given below link to download the complete solution in PDF.
Class 12 Maths chapter 3 solutions Exercise: 3.1 Page number: 42-43 Total questions: 10 |
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:
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 row number i and column number j.
$a_{13}= 19$ $a_{21}= 35$
$a_{33}= -5$ $a_{24}= 12$
$a_{23}= \frac{5}{2}$
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 the 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
$a_{11} = \frac{(1+1)^{2}}{2} =\frac{2^{2}}{2}=\frac{4}{2}=2$
$a_{22} = \frac{(2+2)^{2}}{2} =\frac{4^{2}}{2}=\frac{16}{2}=8$
$a_{12} = \frac{(1+2)^{2}}{2} =\frac{3^{2}}{2}=\frac{9}{2}=4.5$
$a_{21} = \frac{(2+1)^{2}}{2} =\frac{3^{2}}{2}=\frac{9}{2}=4.5$
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:
Answer:
A 2 × 2 matrix, $A = [a_{ij} ]$
(ii) $a_{ij} = \frac{i}{j}$
$a_{11} = \frac{1}{1}=1$
$a_{22} = \frac{2}{2}=1$
$a_{12} = \frac{1}{2}$
$a_{21} = \frac{2}{1}=2$
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:
Answer:
(iii)
$a_{ij} = \frac{(i+2j)^2}{2}$
$a_{11} = \frac{(1+(2\times 1))^{2}}{2}= \frac{(1+2)^{2}}{2}=\frac{3^{2}}{2}=\frac{9}{2}$
$a_{22} = \frac{(2+(2\times 2))^{2}}{2}= \frac{(2+4)^{2}}{2}=\frac{6^{2}}{2}=\frac{36}{2}=18$
$a_{21} = \frac{(2+(2\times 1))^{2}}{2}= \frac{(2+2)^{2}}{2}=\frac{4^{2}}{2}=\frac{16}{2}=8$
$a_{12} = \frac{(1+(2\times 2))^{2}}{2}= \frac{(1+4)^{2}}{2}=\frac{5^{2}}{2}=\frac{25}{2}$
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)
$a_{ij} = \frac{1}{2}|-3i + j|$
$a_{11} = \frac{\left | -3+1 \right |}{2}=\frac{2}{2}=1$
$a_{12} = \frac{\left | (-3\times 1)+2 \right |}{2}=\frac{1}{2}$
$a_{13} = \frac{\left | (-3\times 1)+3 \right |}{2}=0$
$a_{21} = \frac{\left | (-3\times 2)+1 \right |}{2}=\frac{5}{2}$
$a_{22} = \frac{\left | (-3\times 2)+2 \right |}{2}=\frac{4}{2}=2$
$a_{23} = \frac{\left | (-3\times 2)+3 \right |}{2}=\frac{\left | -6+3 \right |}{2}=\frac{\left | -3 \right |}{2} =\frac{3}{2}$
$a_{31} = \frac{\left | (-3\times 3)+1 \right |}{2}=\frac{8}{2}=4$
$a_{32} = \frac{\left | (-3\times 3)+2 \right |}{2}=\frac{7}{2}$
$a_{33} = \frac{\left | (-3\times 3)+3 \right |}{2}=\frac{\left | -9+3 \right |}{2}=\frac{\left | -6 \right |}{2} =\frac{6}{2}=3$
$a_{14} = \frac{\left | (-3\times 1)+4 \right |}{2}=\frac{\left | -3+4 \right |}{2}=\frac{\left | 1 \right |}{2} =\frac{1}{2}$
$a_{24} = \frac{\left | (-3\times 2)+4 \right |}{2}=\frac{\left | -6+4 \right |}{2}=\frac{\left | -2 \right |}{2} =\frac{2}{2}=1$
$a_{34} = \frac{\left | (-3\times 3)+4 \right |}{2}=\frac{\left | -9+4 \right |}{2}=\frac{\left | -5 \right |}{2} =\frac{5}{2}$
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:
Answer:
A 3 × 4 matrix,
(ii) $a_{ij} = 2i - j$
$a_{11} = 2\times 1-1 =2-1=1$
$a_{12} = 2\times 1-2 =2-2=0$
$a_{13} = 2\times 1-3 =2-3=-1$
$a_{21} = 2\times 2-1 =4-1=3$
$a_{22}= 2\times 2-2 =4-2=2$
$a_{23} = 2\times 2-3 =4-3=1$
$a_{31} = 2\times 3-1 =6-1=5$
$a_{32} = 2\times 3-2 =6-2=4$
$a_{33} = 2\times 3-3 =6-3=3$
$a_{14} = 2\times 1-4 =2-4=-2$
$a_{24}= 2\times 2-4 =4-4=0$
$a_{34}= 2\times 3-4 =6-4=2$
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:
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
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 correct.
Question 9: Which of the given values of x and y make the following pair of matrices equal
(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:
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 solutions Exercise: 3.2 Page number: 58-61 Total questions: 22 |
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
$A+B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $+ \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$
$A+B = \begin{bmatrix} 2+1 &4+3 \\ 3+(-2) & 2+5 \end{bmatrix}$
$A+B = \begin{bmatrix} 3 &7 \\ 1 & 7 \end{bmatrix}$
Answer:
$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$
(ii) A - B
$A-B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$
$A-B = \begin{bmatrix} 2-1 &4-3 \\ 3-(-2) & 2-5 \end{bmatrix}$
$A-B = \begin{bmatrix} 1 &1 \\ 5 & -3 \end{bmatrix}$
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
$3A -C = 3\begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$
$3A -C = \begin{bmatrix} 6 &12 \\ 9 & 6 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$
$3A -C = \begin{bmatrix} 6-(-2) &12-5 \\ 9-3 & 6-4 \end{bmatrix}$
$3A -C = \begin{bmatrix} 8 &7 \\ 6 & 2 \end{bmatrix}$
Answer:
$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$
(iv) AB
$AB = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $\times \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$
$AB = \begin{bmatrix} 2\times 1+4\times -2 & \, \, \, 2\times 3+4\times 5 \\ 3\times 1+2\times -2 & \, \, \, 3\times 3+2 \times 5 \end{bmatrix}$
$AB = \begin{bmatrix} -6 &26 \\ -1 & 19 \end{bmatrix}$
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:
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:
Answer:
(iii) The addition of the 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:
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.
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.
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.
Answer:
(vi) The given product can be computed as follows
$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$
$=\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$
$=\begin{bmatrix} 3 \times 2+(-1)\times 1+3\times 3\, \, \, & 3 \times (-3)+(-1)\times 0+3\times 1 \\ (-1) \times 2+ 0 \times 1+2\times 3 \, \, \, & (-1) \times -3+0\times 0+2\times 1 \end{bmatrix}$
$=\begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$
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 = $\begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$
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 $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$
$3A-5B = 3\times \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ $-5\times \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$
$3A-5B = \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}$ $- \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}$
$3A-5B = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
$3A-5B = 0$
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
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
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
$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$
$2X + 3 \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$
$2X + \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$
$2X = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix}$
$2X = \begin{bmatrix} 2-\frac{6}{5} &3-\frac{39}{5} \\ 4-\frac{42}{5} & 0 -(-6)\end{bmatrix}$
$2X = \begin{bmatrix} \frac{4}{5} &-\frac{24}{5} \\ -\frac{22}{5} & 6\end{bmatrix}$
$X = \begin{bmatrix} \frac{2}{5} &-\frac{12}{5} \\ -\frac{11}{5} & 3\end{bmatrix}$
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
$2X+ \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$
$2X = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}- \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$
$2X = \begin{bmatrix} 1-3 &0-2 \\ -3-1 & 2-4 \end{bmatrix}$
$2X = \begin{bmatrix} -2 &-2 \\ -4 & -2 \end{bmatrix}$
$X = \begin{bmatrix} -1 &-1 \\ -2 & -1 \end{bmatrix}$
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
$2+y=5$ $2x+2=8$
$y=5-2$ $2x=8-2$
$y=3$ $2x=6$
$x=3$
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
$\begin{bmatrix}x & z \\ y &t \end{bmatrix} = \begin{bmatrix} 3 &9\\ 6 & 6 \end{bmatrix}$
$x=3,y=6,z=9\, \, and\, \, t=6$
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
$3x+y=5$
$3\times 3+y=5$
$9+y=5$
$y=5-9$
$y=-4$
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 then 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 $x=2,y=4,z=1\, \, and\, \, w=3.$
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
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 is not equal to the left-hand side, that is
Question 14(ii): Show that
Answer:
To prove the following multiplication of three by three matrices is 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 out the value of the square of matrix A
$A\times A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}\times \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$
$A^{2} = \begin{bmatrix} 2\times 2+0\times 2+1\times 1 & 2\times 0+0\times 1+1\times -1 & 2\times 1+0\times 3+1\times 0\\ 2\times 2+1\times 2+3\times 1& 2\times 0+1\times 1+3\times -1 &2\times 1+1\times 3+3\times 0 \\ 1\times 2+(-1)\times 2+0\times 1 & 1\times 0+(-1)\times 1+0\times -1 & 1\times 1+(-1)\times 3+0\times 0 \end{bmatrix}$
$A^{2} = \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$
$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$
$\therefore$ $A^2 -5A + 6I$
$= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$ $-5 \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$ $+6 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$
$= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$ $- \begin{bmatrix} 10 & 0 & 5\\ 10 & 5 &15 \\ 5 & -5 & 0 \end{bmatrix}$ $+\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 &0 \\ 0 & 0 & 6 \end{bmatrix}$
$= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0\\ 9-10+0 & -2-5+6 &5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix}$
$= \begin{bmatrix} 1 & -1 & -3\\ -1 & -1 &-10 \\ -5 & 4 & 4 \end{bmatrix}$
Question16: 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.
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. $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$ .
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.
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.
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.
The restriction on n, k and p so that PY + WY will be defined are:
(A) $k = 3, p = n$
Answer:
P and Y are of order $p*k$ and $3*k$ respectively.
$\therefore$ 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 $3*k$ respectively.
$\therefore$ 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 solutions Exercise: 3.3 Page number: 66-68 Total questions: 12 |
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
$A^{T} = \begin{bmatrix} -1 & \sqrt3 & 2\\ 5& 5 &3 \\ 6 &6 &-1 \end{bmatrix}$
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.
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
$(A - B)' = A' - B'$ .
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
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. $(A - B)' = A' - B'$ .
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.
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.
Answer:
$A = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$
$A' = \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$
$A+A' = \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha& \cos\alpha \end{bmatrix}$ $+ \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha& \cos\alpha \end{bmatrix}$ $= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$
$A+A' = \begin{bmatrix} 2\cos\alpha & 0\\ 0 & 2\cos\alpha \end{bmatrix}$ $= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$
$2 cos \alpha=1$
$cos \alpha=\frac{1}{2}$
$\alpha=\frac{\pi}{3}$
Option B is correct.
Class 12 Maths chapter 3 solutions Exercise: 3.4 Page number: 69-69 Total questions: 1 |
Question 1: Matrices A and B will be inverse of each other only if
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, matrices A and B will be inverse of each other only if $AB=BA=I$ .
Option D is correct.
Class 12 Maths chapter 3 solutions Miscellaneous Exercise: Page number: 72-73 Total questions: 11 |
Question 1: 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 2: 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.
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
$2x^{2} = 1$ $6y^{2} = 1$ $3z^{2} = 1$
$x^{2} = \frac{1}{2}$ $y^{2} = \frac{1}{6}$ $z^{2}=\frac{1}{3}$
$x = \pm \frac{1}{\sqrt{2}}$ $y= \pm \frac{1}{\sqrt{6}}$ $z=\pm \frac{1}{\sqrt{3}}$
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 5: 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
$A^2 -5A + 7I= 0$ .
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 7(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
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 7(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
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 $= 53000-36000= Rs. 17000$
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 9: Choose the correct answer in the following questions:
If $A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ is such that $A^2 = I$
(A) $1 + \alpha^2 + \beta \gamma = 0$
(B) $1 - \alpha^2 + \beta \gamma = 0$
(C) $1 - \alpha^2 - \beta \gamma = 0$
(D) $1 + \alpha^2 - \beta \gamma = 0$
Answer:
$A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$
$A^2 = I$
$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ $\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ $= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$
$\begin{bmatrix} \alpha^{2} +\beta \gamma&\alpha \beta-\alpha \beta\\\alpha \gamma-\alpha \gamma&\beta \gamma+\alpha^{2} \end{bmatrix}$ $= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$
$\begin{bmatrix} \alpha^{2} +\beta \gamma&0\\0&\beta \gamma+\alpha^{2} \end{bmatrix}$ $= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$
Thus we obtained that
$\alpha^{2} +\beta \gamma=1$
$\Rightarrow 1-\alpha^{2} -\beta \gamma=0$
Option C is correct.
Question 10: If the matrix A is both symmetric and skew-symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
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 11: If A is square matrix such that $A^{2}=A$ , then $(I + A)^3 - 7 A$ is equal to
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.
Also Read,
Matrices Class 12 NCERT Solutions Exercise 3.1
Matrices Class 12 NCERT Solutions Exercise 3.2
Matrices Class 12 NCERT Solutions Exercise 3.3
Question: If $\left[\begin{array}{cc}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]$ then the value of $\mathrm{x}+\mathrm{y}$ is:
Solution:
We are given that,
$\left[\begin{array}{cc}
2 x+y & 4 x \\
5 x-7 & 4 x
\end{array}\right]=\left[\begin{array}{cc}
7 & 7 y-13 \\
y & x+6
\end{array}\right]$
By equating the two matrices, we get-
$4 x=x+6 \Rightarrow 3 x=6 \Rightarrow x=2$
Also, $2x+y=7$
$\Rightarrow y=7-2 x=7-4=3$
Therefore, the value of ($x+y$) is (2 + 3) = 5.
Hence, the correct answer is 5.
The Class 12 Maths Chapter 3 Solutions (Matrices), cover the following topics in detail.
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.
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.
Matrix Addition:
If A = [aij]m × n and B = [bij]m × n, then A + B = [aij + bij]m × n.
Matrix Subtraction:
If A = [aij]m × n and B = [bij]m × n, then A - B = [aij - bij]m × n.
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.
The transpose of a matrix A, denoted as $A^T$, is obtained by interchanging its rows and columns.
A matrix A is symmetric if A =$A^T$ (i.e., it is equal to its transpose).
A matrix A is skew-symmetric if $A^T$ = -A (i.e., the transpose of A is equal to the negative of A).
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.
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.
Matrices play a significant role in Class 12 mathematics, and here are some key steps on how to approach matrix-related questions effectively:
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We at Careers360 compiled all the NCERT class 12 Maths solutions in one place for easy student reference. Access them by using the following links.
Also Read,
Here are the subject-wise links for the NCERT solutions of class 12:
Given below are the class-wise solutions of class 12 NCERT:
Here are some useful links for the NCERT books and the NCERT syllabus for class 12
Frequently Asked Questions (FAQs)
The adjoint of a matrix is the transpose of its cofactor matrix, and it's used to find the inverse of a matrix by dividing the adjoint by the determinant of the original matrix. The inverse matrix is also found using the following equation:
A-1 =adj(A)/det(A),
where adj(A) refers to the adjoint of a matrix A,det(A) refers to the determinant of a matrix A.
A square matrix A is said to be symmetric if aij = aji for all i and j, where aij is an element present at (i,j)th position (ith row and jth column in matrix A) and aji is an element present at (j,i)th position (jth row and ith column in matrix A) whereas square matrix A is said to be skew-symmetric if aij =−aji for all i and j. In other words, we can say that matrix A is said to be skew-symmetric if the transpose of matrix A is equal to the negative of matrix A i.e (AT =−A)
The rank of a matrix is equal to the number of linearly independent rows or columns in it. It cannot be more than its number of rows and columns. To find the rank of a matrix, we can transform the matrix to its row echelon form and count the number of non-zero rows.
To find the inverse of a matrix A using elementary transformations, we can use elementary row operations on A = IA, in a sequence, until we get I = BA. We can also use elementary column operations on A = AI, in a sequence, till we get I = AB. If the inverse of matrix A exists, we can write A = IA and apply a sequence of row operations till we get an identity matrix on the LHS and use the same elementary operations on the RHS to get I = BA
The topics covered in matrices for class 12 include the following topics:
On Question asked by student community
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Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.
Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
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