\displaystyle \text{Question 1: If a Matix has 8 elements, what are the possible orders it can have?} \\ \text{What if it has 5 elements?} 

\displaystyle \text{Answer:}

\displaystyle \text{If a matrix is of order } m \times n \text{ elements, it has } mn \text{ elements. }

\displaystyle \text{So, if the matrix has 8 elements, we will find the ordered pairs } m \text{ and } n.

\displaystyle mn = 8

\displaystyle \text{Then, ordered pairs } m \text{ and } n \text{ can be }  m \times n \text{ be } (8 \times 1),(1 \times 8),(4 \times 2),(2 \times 4).

\displaystyle \text{Now, if it has 5 elements. }

\displaystyle \text{Possible orders are } (5 \times 1), (1 \times 5).

\\

\displaystyle \text{Question 2: If }  A = [a_{ij}] = \begin{bmatrix}   2 & 3 & -5 \\ 1 & 4 & 9 \\ 0 & 7 & -2  \end{bmatrix}  \text{ and } B = [b_{ij}] = \begin{bmatrix}   2 & -1 \\ -3 & 4 \\ 1 & 2   \end{bmatrix} \text{ then find }

\displaystyle \text{(i) } a_{22}+ b_{21}

\displaystyle \text{(ii) } a_{11} \ b_{11} + a_{22} \ b_{22}

\displaystyle \text{Answer:}

\displaystyle \text{Given } A = [a_{ij}] = \begin{bmatrix}   2 & 3 & -5 \\ 1 & 4 & 9 \\ 0 & 7 & -2  \end{bmatrix}  \text{ and } B = [b_{ij}] = \begin{bmatrix}   2 & -1 \\ -3 & 4 \\ 1 & 2   \end{bmatrix}

\displaystyle \text{(i)}

\displaystyle a_{22}+ b_{21} = 4+(-3) = 1

\displaystyle \text{(ii)}

\displaystyle a_{11} \ b_{11} + a_{22} \ b_{22} = 2 \times 2 + 4 \times 4 = 20

\\

\displaystyle \text{Question 3: Let } A \text{ be a matrix of order }  3 \times 4. \text{ If } R_1  \text{ denotes the first row of } A \text{ and } \\ C_2 \text{ denotes its second column, then determine the order of matrices } R_1 \text{ and } C_2.

\displaystyle \text{Answer:}

\displaystyle A \text{ is a matrix of order } 3 \times 4

\displaystyle \Rightarrow A = \begin{bmatrix}   a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}  \end{bmatrix}

\displaystyle \text{Now, } R_1  = \begin{bmatrix}   a_{11} & a_{12} & a_{13} & a_{14}   \end{bmatrix}. \ \text{It has 4 columns and 1 row. }

\displaystyle \text{Therefore order of } R_1 \text{ is } 1 \times 4.

\displaystyle \text{Now, } C_2 = \begin{bmatrix}    a_{12}  \\  a_{22} \\  a_{32}  \end{bmatrix}. \ \text{ It has 1 column and 3 rows. }

\displaystyle \text{Therefore order of } R_1 \text{ is } 3 \times 1.

\\

\displaystyle \text{Question 4: Construct a } 2 \times 3 \text{ matrix } A = [a_{ij}] \text{ whose elements } a_{ij} \text{ are given by }

\displaystyle \text{(i) } a_{ij} = i \times j           \displaystyle \text{(ii) } a_{ij} = 2i - j           \displaystyle \text{(iii) } a_{ij} = i + j           \displaystyle \text{(iv) } a_{ij}= \frac{(i+j)^2}{2}

\displaystyle \text{Answer:}

\displaystyle \text{So, the elements in a } 2 \times 3 \text{ matrix are } a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

\displaystyle \text{Therefore } A =  \begin{bmatrix}   a_{11} & a_{12} & a_{13}  \\ a_{21} & a_{22} & a_{23}   \end{bmatrix}

\displaystyle \text{(i)}

\displaystyle \text{When } a_{ij} = i \times j \text{ we get }

\displaystyle a_{11}= 1 \times 1 = 1      \displaystyle a_{12} = 1 \times 2 = 2      \displaystyle a_{13} = 1 \times 3 = 3

\displaystyle a_{21} = 2 \times 1 = 2      \displaystyle a_{22} = 2 \times 2 = 4      \displaystyle a_{23} = 2 \times 3 = 6

\displaystyle \text{Therefore }A =  \begin{bmatrix}   1 & 2 & 3  \\   2 & 4 & 6  \end{bmatrix}

\displaystyle \text{(ii)}

\displaystyle \text{When } a_{ij} = 2i - j \text{ we get }

\displaystyle a_{11}= 2 \times 1 - 1 = 1      \displaystyle a_{12} = 2 \times 1 - 2 = 0      \displaystyle a_{13} = 2 \times 1 - 3 = -1

\displaystyle a_{21} = 2 \times 2 - 1 = 3      \displaystyle a_{22} = 2 \times 2 - 2 = 2      \displaystyle a_{23} = 2 \times 2 - 3 = 1

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

\displaystyle \text{(iii)}

\displaystyle \text{When } a_{ij} = i + j \text{ we get }

\displaystyle a_{11}= 1 + 1 = 2      \displaystyle a_{12} = 1 + 2 = 3      \displaystyle a_{13} = 1 + 3 = 4

\displaystyle a_{21} = 2 + 1 = 3      \displaystyle a_{22} = 2 + 2 = 4      \displaystyle a_{23} = 2 + 3 = 5

\displaystyle \text{Therefore }A =  \begin{bmatrix}   2 & 3 & 4  \\   3 & 4 & 5  \end{bmatrix}

\displaystyle \text{(iv)}

\displaystyle \text{When } a_{ij} = \frac{(i+j)^2}{2} \text{ we get }

\displaystyle a_{11}= \frac{(1+1)^2}{2} = 2      \displaystyle a_{12} = \frac{(1+2)^2}{2} = 4.5      \displaystyle a_{13} = \frac{(1+3)^2}{2} = 8

\displaystyle a_{21} = \frac{(i+j)^2}{2} = 4.5      \displaystyle a_{22} = \frac{(2+2)^2}{2} = 8      \displaystyle a_{23} = \frac{(i+j)^2}{2} = 12.5

\displaystyle \text{Therefore }A =  \begin{bmatrix}   2 & 4.5 & 8  \\   4.5 & 8 & 12.5  \end{bmatrix}

\\

\displaystyle \text{Question 5: Construct a } 2 \times 2 \text{ matrix } A = [a_{ij}] \text{ whose elements } a_{ij} \text{ are given by }

\displaystyle \text{(i) } a_{ij} = \frac{(i+j)^2}{2}       \displaystyle \text{(ii) } a_{ij} = \frac{(i-j)^2}{2}       \displaystyle \text{(iii) } a_{ij} = \frac{(i-2j)^2}{2} 

\displaystyle \text{(iv) } a_{ij} = \frac{(2i+j)^2}{2}{\hspace{8.5cm} [\text{CBSE 2002}] } 

\displaystyle \text{(v) } a_{ij} = \frac{|2i-3j|}{2}       \displaystyle \text{(vi) } a_{ij} = \frac{|-3i+j|}{2}       \displaystyle \text{(vii) } a_{ij} = e^{2ix} \sin xj

\displaystyle \text{Answer:}

\displaystyle \text{So, the elements in a } 2 \times 2 \text{ matrix are } a_{11}, a_{12},  a_{21}, a_{22}

\displaystyle \text{Therefore } A =  \begin{bmatrix}   a_{11} & a_{12}   \\ a_{21} & a_{22}   \end{bmatrix}

\displaystyle \text{(i)}

\displaystyle \text{When } a_{ij} = \frac{(i+j)^2}{2} \text{ we get }

\displaystyle a_{11}= \frac{(1+1)^2}{2} = 2      \displaystyle a_{12} = \frac{(1+2)^2}{2} = 4.5

\displaystyle a_{21} = \frac{(2+1)^2}{2} = 4.5      \displaystyle a_{22} = \frac{(2+2)^2}{2} = 8

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

\displaystyle \text{(ii)}

\displaystyle \text{When } a_{ij} = \frac{(i-j)^2}{2} \text{ we get }

\displaystyle a_{11}= \frac{(1-1)^2}{2} = 0      \displaystyle a_{12} = \frac{(1-2)^2}{2} = 0.5

\displaystyle a_{21} = \frac{(2-1)^2}{2} = 0.5      \displaystyle a_{22} = \frac{(2-2)^2}{2} = 0

\displaystyle \text{Therefore }A =  \begin{bmatrix}   0 & 0.5   \\   0.5 & 0  \end{bmatrix}

\displaystyle \text{(iii)}

\displaystyle \text{When } a_{ij} = \frac{(i-2j)^2}{2} \text{ we get }

\displaystyle a_{11}= \frac{(1-2 \times 1)^2}{2} = 0.5      \displaystyle a_{12} = \frac{(1- 2 \times 2)^2}{2} = 4.5  

\displaystyle a_{21} = \frac{(2- 2 \times 1)^2}{2} = 0      \displaystyle a_{22} = \frac{(2- 2 \times 2)^2}{2} = 2   

\displaystyle \text{Therefore }A =  \begin{bmatrix}   0.5 & 4.5   \\   0 & 2  \end{bmatrix}

\displaystyle \text{(iv)}

\displaystyle \text{When } a_{ij} = \frac{(2i+j)^2}{2} \text{ we get }

\displaystyle a_{11}= \frac{(2 \times 1+1)^2}{2} = 4.5      \displaystyle a_{12} = \frac{(2 \times 1+2)^2}{2} = 8  

\displaystyle a_{21} = \frac{(2 \times 2+1)^2}{2} = 12.5      \displaystyle a_{22} = \frac{(2 \times 2+2)^2}{2} = 18   

\displaystyle \text{Therefore }A =  \begin{bmatrix}   4.5 & 8   \\   12.5 & 18  \end{bmatrix}

\displaystyle \text{(v)}

\displaystyle \text{When } a_{ij} = \frac{|2i-3j|}{2} \text{ we get }

\displaystyle a_{11}= \frac{|2 \times 1-3 \times 1|}{2} = 0.5      \displaystyle a_{12} = \frac{|2 \times 1-3 \times 2|}{2} = 2  

\displaystyle a_{21} = \frac{|2 \times 2-3 \times 1|}{2} = 0.5      \displaystyle a_{22} = \frac{|2 \times 2-3 \times 2|}{2} = 1   

\displaystyle \text{Therefore }A =  \begin{bmatrix}   0.5 & 2   \\   0.5 & 1  \end{bmatrix}

\displaystyle \text{(vi)}

\displaystyle \text{When } a_{ij} = \frac{|-3i+j|}{2} \text{ we get }

\displaystyle a_{11}= \frac{|-3 \times 1+1 |}{2} = 1      \displaystyle a_{12} = \frac{|-3 \times 1+2|}{2} = 0.5  

\displaystyle a_{21} = \frac{|-3 \times 2+1|}{2} = 2.5      \displaystyle a_{22} = \frac{|-3 \times 2+2|}{2} = 2   

\displaystyle \text{Therefore }A =  \begin{bmatrix}   1 & 0.5   \\   2.5 & 2  \end{bmatrix}

\displaystyle \text{(vii)}

\displaystyle \text{When } a_{ij} = e^{2ix} \sin xj \text{ we get }

\displaystyle a_{11}= e^{2 \times 1 x} \sin (x \times 1) = e^{2x} \sin x       \displaystyle a_{12} = e^{2 \times 1 x} \sin (x \times 2) = e^{2x} \sin 2x

\displaystyle a_{21} = e^{2 \times 2 x} \sin (x \times 1) = e^{4x} \sin x      \displaystyle a_{22} = e^{2 \times 2 x} \sin (x \times 2) = e^{2x} \sin 2x

\displaystyle \text{Therefore }A =  \begin{bmatrix}  e^{2x} \sin x & e^{2x} \sin 2x  \\  e^{4x} \sin x & e^{4x} \sin 2x  \end{bmatrix}

\\

\displaystyle \text{Question 6: Construct a } 3 \times 4 \text{ matrix } A = [a_{ij}] \text{ whose elements } a_{ij} \text{ are given by }

\displaystyle \text{(i) } a_{ij} = i + j      \displaystyle \text{(ii) } a_{ij} = i - j      \displaystyle \text{(iii) } a_{ij} = 2i      \displaystyle \text{(iv) } a_{ij} = j      \displaystyle \text{(v) } a_{ij} = \frac{1}{2} |-3i+j|

\displaystyle \text{Answer:}

\displaystyle \text{So, the elements in a } 3 \times 4 \text{ matrix are } \\ a_{11}, a_{12}, a_{13}, a_{14}, a_{21}, a_{22}, a_{23}, a_{24}, a_{31}, a_{32}, a_{33}, a_{34}, a_{41}, a_{42}, a_{43}, a_{44}

\displaystyle \text{Therefore } A =  \begin{bmatrix}  a_{11} &  a_{12} & a_{13} & a_{14} \\  a_{21} & a_{22} & a_{23} & a_{24} \\  a_{31} & a_{32} & a_{33} & a_{34} \\  a_{41} & a_{42} & a_{43} & a_{44}   \end{bmatrix}

\displaystyle \text{(i)}

\displaystyle \text{When } a_{ij} = i + j \text{ we get }

\displaystyle a_{11}= 1 + 1 = 2      \displaystyle a_{12} = 1 + 2 = 3      \displaystyle a_{13} = 1 + 3 = 4     \displaystyle a_{14} = 1 + 4 = 5

\displaystyle a_{21}= 2 + 1 = 3      \displaystyle a_{22} = 2 + 2 = 4      \displaystyle a_{23} = 2 + 3 = 5     \displaystyle a_{24} = 2 + 4 = 6

\displaystyle a_{31}= 3 + 1 = 4      \displaystyle a_{32} = 3 + 2 = 5      \displaystyle a_{33} = 3 + 3 = 6     \displaystyle a_{34} = 3 + 4 = 7

\displaystyle \text{Therefore }A =  \begin{bmatrix}   2 & 3 & 4 & 5 \\   3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7   \end{bmatrix}

\displaystyle \text{(ii)}

\displaystyle \text{When } a_{ij} = i - j \text{ we get }

\displaystyle a_{11}= 1 - 1 = 0      \displaystyle a_{12} = 1 - 2 = -1      \displaystyle a_{13} = 1 - 3 = -2     \displaystyle a_{14} = 1 - 4 = -3

\displaystyle a_{21}= 2 - 1 = 1      \displaystyle a_{22} = 2 - 2 = 0      \displaystyle a_{23} = 2 - 3 = -1     \displaystyle a_{24} = 2 - 4 = -2

\displaystyle a_{31}= 3 - 1 = 2      \displaystyle a_{32} = 3 - 2 = 1      \displaystyle a_{33} = 3 - 3 = 0     \displaystyle a_{34} = 3 - 4 = -1

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

\displaystyle \text{(iii)}

\displaystyle \text{When } a_{ij} = 2i \text{ we get }

\displaystyle a_{11}= 2 \times 1 = 2      \displaystyle a_{12} = 2 \times 1 = 2      \displaystyle a_{13} = 2 \times 1 = 2     \displaystyle a_{14} = 2 \times 1 = 2

\displaystyle a_{21}= 2 \times 2 = 4      \displaystyle a_{22} = 2 \times 2 = 4      \displaystyle a_{23} = 2 \times 2 = 4     \displaystyle a_{24} = 2 \times 2 = 4

\displaystyle a_{31}= 2 \times 3 = 6      \displaystyle a_{32} = 2 \times 3 = 6      \displaystyle a_{33} = 2 \times 3 = 6     \displaystyle a_{34} = 2 \times 3 = 6

\displaystyle \text{Therefore }A =  \begin{bmatrix}   2 & 2 & 2 & 2 \\   4 & 4 & 4 & 4 \\ 6 & 6 & 6 & 6   \end{bmatrix}

\displaystyle \text{(iv)}

\displaystyle \text{When } a_{ij} = i \text{ we get }

\displaystyle a_{11}= 1      \displaystyle a_{12} = 1      \displaystyle a_{13} = 1     \displaystyle a_{14} = 1

\displaystyle a_{21}= 2      \displaystyle a_{22} = 2      \displaystyle a_{23} = 2     \displaystyle a_{24} = 2

\displaystyle a_{31}= 3      \displaystyle a_{32} = 3      \displaystyle a_{33} = 3     \displaystyle a_{34} = 3

\displaystyle \text{Therefore }A =  \begin{bmatrix}   1 & 1 & 1 & 1 \\   2 & 2 & 2 & 2 \\ 3 & 3 & 3 & 3   \end{bmatrix}

\displaystyle \text{(v)}

\displaystyle \text{When } a_{ij} = \frac{|-3i+j|}{2} \text{ we get }

\displaystyle a_{11}= \frac{|-3 \times 1+1 |}{2} = 1      \displaystyle a_{12} = \frac{|-3 \times 1+2|}{2} = 0.5   \displaystyle a_{13}= \frac{|-3 \times 1+3 |}{2} = 0      \displaystyle a_{14} = \frac{|-3 \times 1+4|}{2} = 0.5  

\displaystyle a_{21} = \frac{|-3 \times 2+1|}{2} = 2.5      \displaystyle a_{22} = \frac{|-3 \times 2+2|}{2} = 2   \displaystyle a_{23} = \frac{|-3 \times 2+3|}{2} = 1.5      \displaystyle a_{24} = \frac{|-3 \times 2+4|}{2} = 1   

\displaystyle a_{31} = \frac{|-3 \times 3+1|}{2} = 4      \displaystyle a_{32} = \frac{|-3 \times 3+2|}{2} = 3.5   \displaystyle a_{33} = \frac{|-3 \times 3+3|}{2} = 3      \displaystyle a_{34} = \frac{|-3 \times 3+4|}{2} = 2.5   

\displaystyle \text{Therefore }A =  \begin{bmatrix}   1 & 0.5 & 0 & 0.5   \\   2.5 & 2  & 1.5 & 1 \\ 4 & 3.5 & 3 & 2.5 \end{bmatrix}

\\

\displaystyle \text{Question 7: Construct a } 4 \times 3 \text{ matrix } A = [a_{ij}] \text{ whose elements } a_{ij} \text{ are given by }

\displaystyle \text{(i) } a_{ij} =  2i + \frac{i}{j}                   \displaystyle \text{(ii) } a_{ij} = \frac{i-j}{i+j}                 \displaystyle \text{(iii) } a_{ij} = i

\displaystyle \text{Answer:}

\displaystyle \text{So, the elements in a } 4 \times 3 \text{ matrix are } \\ a_{11}, a_{12}, a_{13},  a_{21}, a_{22}, a_{23},  a_{31}, a_{32}, a_{33},  a_{41}, a_{42}, a_{43}

\displaystyle \text{Therefore } A =  \begin{bmatrix}  a_{11} &  a_{12} & a_{13}  \\  a_{21} & a_{22} & a_{23}  \\  a_{31} & a_{32} & a_{33}  \\  a_{41} & a_{42} & a_{43}    \end{bmatrix}

\displaystyle \text{(i)}

\displaystyle \text{When } a_{ij} =  2i + \frac{i}{j}  \text{ we get }

\displaystyle a_{11}= 2 \times 1 + \frac{1}{1} = 3      \displaystyle a_{12} = 2 \times 1 + \frac{1}{2} = \frac{5}{2}     \displaystyle a_{13} = 2 \times 1 + \frac{1}{3} = \frac{7}{3}

\displaystyle a_{21}= 2 \times 2 + \frac{2}{1} = 6      \displaystyle a_{22} = 2 \times 2 + \frac{2}{2} = 5      \displaystyle a_{23} = 2 \times 2 + \frac{2}{3} = \frac{14}{3}

\displaystyle a_{31}= 2 \times 3+ \frac{3}{1} = 9      \displaystyle a_{32} = 2 \times 3 + \frac{3}{2} = \frac{15}{2}      \displaystyle a_{33} = 2 \times 3 + \frac{3}{3} = 7

\displaystyle a_{41}= 2 \times 4 + \frac{4}{1} = 12      \displaystyle a_{42} = 2 \times 4 + \frac{4}{2} = 10      \displaystyle a_{33} = 2 \times 4 + \frac{4}{3} = \frac{28}{3}

\displaystyle \text{Therefore } A =  \begin{bmatrix}  3 &  2 & \frac{7}{3}  \\  6 & 5 & \frac{14}{3}  \\  9 & \frac{15}{2} & 7  \\  12 & 10 & \frac{28}{3}    \end{bmatrix}

\displaystyle \text{(ii)}

\displaystyle \text{When } a_{ij} =  \frac{i-j}{i+j} \text{ we get }

\displaystyle a_{11}= \frac{1-1}{1+1} = 0      \displaystyle a_{12} = \frac{1-2}{1+2} = \frac{-1}{3}      \displaystyle a_{13} = \frac{1-3}{1+3} = \frac{-1}{2}

\displaystyle a_{21}= \frac{2-1}{2+1} = \frac{1}{3}      \displaystyle a_{22} = \frac{2-2}{2+2} = 0        \displaystyle a_{23} = \frac{2-3}{2+3} = \frac{-1}{5}

\displaystyle a_{31}= \frac{3-1}{3+1} =  \frac{1}{2}      \displaystyle a_{32} = \frac{3-2}{3+2} = \frac{1}{5}        \displaystyle a_{33} = \frac{3-3}{3+2} = 0

\displaystyle a_{41}= \frac{4-1}{4+1} = \frac{3}{5}      \displaystyle a_{42} = \frac{4-2}{4+2} = \frac{1}{3}        \displaystyle a_{43} = \frac{4-3}{4+3} = \frac{1}{7}

\displaystyle \text{Therefore } A =  \begin{bmatrix}  0 &  \frac{-1}{3} & \frac{-1}{2} \\ & &   \\  \frac{1}{3} & 0 & \frac{-1}{5} \\ & &   \\  \frac{1}{2} & \frac{1}{5} & 0  \\ & &  \\  \frac{3}{5} & \frac{1}{3} & \frac{1}{7}   \end{bmatrix}

\displaystyle \text{(iii)}

\displaystyle \text{When } a_{ij} =  2i \text{ we get }

\displaystyle a_{11}= 2 \times 1 = 2      \displaystyle a_{12} = 2 \times 1 = 2      \displaystyle a_{13} = 2 \times 1 = 2

\displaystyle a_{21}= 2 \times 2 = 4      \displaystyle a_{22} = 2 \times 2 = 4        \displaystyle a_{23} = 2 \times 2 = 4

\displaystyle a_{31}= 2 \times 3 = 6      \displaystyle a_{32} = 2 \times 3 = 6        \displaystyle a_{33} = 2 \times 3 = 6

\displaystyle a_{41}= 2 \times 4 = 8      \displaystyle a_{42} = 2 \times 4 = 8        \displaystyle a_{43} = 2 \times 4 = 8

\displaystyle \text{Therefore } A =  \begin{bmatrix} 2 & 2 & 2 \\ 4 & 4 & 4 \\ 6 & 6 & 6 \\ 8 & 8 & 8 \end{bmatrix}

\\

\displaystyle \text{Question 8: Find } x, y, a \text{ and }b  \text{ if }  \begin{bmatrix}   3x+4y & 2 & x-2y \\ a+b & 2a-b & -1   \end{bmatrix}  =  \begin{bmatrix}   2 & 2 & 4 \\ 5 & -5 & -1   \end{bmatrix}

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle 3x + 4y = 2 \ldots \ldots  (1)

\displaystyle x - 2y = 4 \ldots \ldots  (2)

\displaystyle a + b = 5 \ldots \ldots  (3)

\displaystyle 2a - b = - 5 \ldots \ldots  (4)

\displaystyle \text{Multiplying equation (2) by 2 and adding to equation (1) }

\displaystyle 3x + 4y + 2x - 4y = 2 + 8

\displaystyle \Rightarrow  5x = 10

\displaystyle \Rightarrow  x = 2

\displaystyle \text{Now, Putting the value of } x \text{ in equation (1) }

\displaystyle 3 \times 2 + 4y = 2

\displaystyle \Rightarrow  6 + 4y = 2

\displaystyle \Rightarrow  4y = 2 - 6

\displaystyle \Rightarrow  4y = - 4

\displaystyle \Rightarrow  y = - 1

\displaystyle \text{Adding equation (3) and (4), }

\displaystyle a + b + 2a - b = 5 + (-5)

\displaystyle \Rightarrow  3a = 5 - 5 = 0

\displaystyle \Rightarrow  a = 0

\displaystyle \text{Now, Putting the value of a in equation (3) }

\displaystyle 0 + b = 5

\displaystyle \Rightarrow  b = 5

\displaystyle \therefore  a = 0, b = 5,  x = 2 \text{ and } y = - 1

\\

\displaystyle \text{Question 9: Find } x, y, a \text{ and }b  \text{ if }  \begin{bmatrix}   2x-3y & a-b & 3 \\ 1 & x+4 & 3a+4b   \end{bmatrix}  =  \begin{bmatrix}   1 & -2 & 3 \\ 1 & 6 & 29   \end{bmatrix}

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle 2x-3y = 1 \ldots \ldots  (1)

\displaystyle x+ 4y = 6 \ldots \ldots  (2)

\displaystyle \text{Solving (1) and (2) we get}

\displaystyle \Rightarrow  x = 6 - 4y

\displaystyle \Rightarrow  2 (6-4y) - 3y = 1

\displaystyle \Rightarrow  12 - 8 y - 3y = 1

\displaystyle \Rightarrow  11y = 11

\displaystyle \Rightarrow  y = 1

\displaystyle \Rightarrow  x = 6 - 4 ( 1) = 2

\displaystyle a-b = - 2 \ldots \ldots  (3)

\displaystyle 3a+4b = 29 \ldots \ldots  (4)

\displaystyle \text{Solving (3) and (4) we get}

\displaystyle \Rightarrow a = b - 2

\displaystyle \Rightarrow  2 ( b - 2) + 4b = 29

\displaystyle \Rightarrow  - 6 + 3b + 4b = 29

\displaystyle \Rightarrow  7b = 35

\displaystyle \Rightarrow  b = 35

\displaystyle \Rightarrow  b = 5

\displaystyle \Rightarrow  a = 5-2 = 3

\displaystyle \text{Therefore } a = 3, b = 5,  x = 2 \text{ and } y = 1

\\

\displaystyle \text{Question 10: Find the values of } , b, c \text{ and } d \text{ from the following equations:}

\displaystyle \begin{bmatrix}   2a+b & a-2b \\ 5c-d & 4c+3d  \end{bmatrix}  =  \begin{bmatrix}   4 & -3 \\ 11 & 24   \end{bmatrix}

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle 2a + b = 4 \ldots \ldots  (1)

\displaystyle a - 2b = - 3 \ldots \ldots  (2)

\displaystyle 5c - d = 11 \ldots \ldots  (3)

\displaystyle 4c + 3d = 24 \ldots \ldots  (4)

\displaystyle \text{Multiplying equation (1) by 2 and adding to equation (2)}

\displaystyle 4a + 2b + a - 2b = 8 - 3

\displaystyle \Rightarrow  5a = 5

\displaystyle \Rightarrow  a = 1

\displaystyle \text{Now, Putting the value of a in equation (1) }

\displaystyle 2 \times 1 + b = 4

\displaystyle \Rightarrow  2 + b = 4

\displaystyle \Rightarrow  b = 4 - 2

\displaystyle \Rightarrow  b = 2

\displaystyle \text{Multiplying equation (3) by 3 and adding to equation (4),}

\displaystyle 15c - 3d + 4c + 3d = 33 + 24

\displaystyle \Rightarrow  19c = 57

\displaystyle \Rightarrow  c = 3

\displaystyle \text{Now, Putting the value of c in equation (4)}

\displaystyle 4 \times 3 + 3d = 24

\displaystyle \Rightarrow  12 + 3d = 24

\displaystyle \Rightarrow  3d = 24 - 12

\displaystyle \Rightarrow  3d = 12

\displaystyle \Rightarrow  d = 4

\displaystyle \text{Therefore } a = 1, b = 2,  c = 3 \text{ and } d = 4

\\

\displaystyle \text{Question 11: Find } x, y \text{ and }  z \text{ so that } A=B, \text{ where } 

\displaystyle A=\begin{bmatrix}   x-2 & 3 & 2z \\ 18z & x+2 & 6z  \end{bmatrix}, B =  \begin{bmatrix}   y & z & 6\\ 6y & x & 2y   \end{bmatrix}

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle x - 2 = y \ldots \ldots  (1)

\displaystyle z = 3

\displaystyle y + 2 = z \ldots \ldots  (2)

\displaystyle 2y = 6z

\displaystyle \Rightarrow  y = 3z \ldots \ldots  (3)

\displaystyle \text{Putting the value of } z \text{ in equation (3),}

\displaystyle \text{Therefore, }  y = 3z  = 3 \times 3 = 9

\displaystyle \text{Putting the value of y in equation (1),}

\displaystyle x - 2 = 9

\displaystyle \Rightarrow  x - 2 = 9

\displaystyle \Rightarrow  x = 9 + 2

\displaystyle \Rightarrow  x = 11

\displaystyle \text{Therefore } x = 11,  y = 9, z = 3.

\\

\displaystyle \text{Question 12: If }  \begin{bmatrix}   x & 3y  \\ 2x+z & 3y- \omega   \end{bmatrix}  =  \begin{bmatrix}   3 & 2  \\ 4 & 7    \end{bmatrix}  \text{ find } x, y, z, \omega. 

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle x = 3 \ldots \ldots  (1)

\displaystyle 3x - y = 2 \ldots \ldots  (2)

\displaystyle 2x + z = 4 \ldots \ldots  (3)

\displaystyle 3y - \omega = 7 \ldots \ldots  (4)

\displaystyle \text{Putting the value of } x  \text{ in equation (2),}

\displaystyle 3 \times 3 - y = 2

\displaystyle \Rightarrow  9 - y = 2

\displaystyle \Rightarrow  y = 9 - 2

\displaystyle \Rightarrow  y = 7

\displaystyle \text{Now, putting the value of } y  \text{ in equation (4),}

\displaystyle 3 \times 7 - \omega = 7

\displaystyle \Rightarrow  21 - \omega = 7

\displaystyle \Rightarrow  \omega = 21 - 7

\displaystyle \Rightarrow  \omega = 14

\displaystyle \text{Again, Putting the value of } x \text{ in equation (3),}

\displaystyle 2 \times 3 + z = 4

\displaystyle \Rightarrow  6 + z = 4

\displaystyle \Rightarrow  z = 4 - 6

\displaystyle \Rightarrow  z = - 2

\displaystyle \text{Therefore } x = 3, y = 7,  z = - 2 \text{ and } \omega = 14

\\

\displaystyle \text{Question 13: If }  \begin{bmatrix}   x-y & z  \\ 2x-y & \omega   \end{bmatrix}  =  \begin{bmatrix}   -1 & 4  \\ 0 & 5    \end{bmatrix}  \text{ find } x, y, z, \omega.{\hspace{3.5cm} [\text{CBSE 2014}] } 

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle \Rightarrow \omega = 5 \text{ and } z = 4

\displaystyle x-y = -1 \ldots \ldots  (1)

\displaystyle 2x-y = 0 \ldots \ldots  (2)

\displaystyle \Rightarrow 2(y-1) - y = 0

\displaystyle \Rightarrow 2y - 1 - y = 0

\displaystyle \Rightarrow y - 1 = 0

\displaystyle \Rightarrow y = 1

\displaystyle \Rightarrow x = 1-1=0

\displaystyle \text{Therefore } x = 0, y = 1,  z = 4 \text{ and } \omega = 5

\\

\displaystyle \text{Question 14: If } \begin{bmatrix}   x+3 & z+4 & 2y-7 \\ 4x+6 & a-1 & 0 \\ b-3 & 3b & z+2c  \end{bmatrix} = \begin{bmatrix}   0 & 6 & 3y-2 \\ 2x & -3 & 2c+2 \\ 2b+4 & -21 & 0  \end{bmatrix}  \\ \\ \text{obtain the values of } a,b, c, x, y \text{ and } z. 

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle x + 3 = 0

\displaystyle \Rightarrow  x = 0 - 3 = - 3 \ldots \ldots  (1)

\displaystyle z + 4 = 6

\displaystyle \Rightarrow  z = 6 - 4 = 2 \ldots \ldots  (2)

\displaystyle 2y - 7 = 3y - 2

\displaystyle \Rightarrow  2y - 3y = - 2 + 7

\displaystyle \Rightarrow  - y = 5

\displaystyle \Rightarrow  y = - 5 \ldots \ldots  (3)

\displaystyle 4x + 6 = 2x \ldots \ldots  (4)

\displaystyle a - 1 = - 3

\displaystyle \Rightarrow  a = - 3 + 1 = - 2 \ldots \ldots  (5)

\displaystyle 2c + 2 = 0

\displaystyle \Rightarrow  2c = - 2

\displaystyle \Rightarrow  c = - 1 \ldots \ldots  (6)

\displaystyle b - 3 = 2b + 4

\displaystyle \Rightarrow  b - 2b = 4 + 3

\displaystyle \Rightarrow  - b = 7

\displaystyle \Rightarrow  b = - 7 \ldots \ldots  (7)

\displaystyle x = - 3,  y = - 5,  z = 2  \text{ and } a = - 2,  b = - 7,  c = - 1

\\

\displaystyle \text{Question 15: If } \begin{bmatrix}   2x+1 & 5x \\ 0 & y^2+1   \end{bmatrix} = \begin{bmatrix}   x+3 & 10 \\ 0 & 26 \end{bmatrix}  \text{find the values of } (x+y).{\hspace{0.5cm} [\text{CBSE 2012}] }

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle 2x + 1 = x + 3 \ldots \ldots  (1) 

\displaystyle \Rightarrow  2x - x = 3 - 1  

\displaystyle \Rightarrow  x = 2 

\displaystyle 5x = 10 \ldots \ldots  (2) 

\displaystyle y^2 + 1 = 2 \ldots \ldots  (3) 

\displaystyle \Rightarrow  y^2 = 26 - 1  

\displaystyle \Rightarrow  y^2 = 25  

\displaystyle \Rightarrow  y = 5 \text{ or } - 5 

\displaystyle \therefore x = 2, y = 5 \text{ or } - 5  

\displaystyle \therefore x + y = 2 + 5 = 7  

\displaystyle \text{or } x + y = 2 - 5 = - 3 

\\

\displaystyle \text{Question 16: If } \begin{bmatrix}   xy & 4 \\ z+6 & x+y   \end{bmatrix} = \begin{bmatrix}   8 & \omega \\ 0 & 6 \end{bmatrix}  \text{ then find the values of } x, y, z \text{ and } \omega. 

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle w = 4 \text{ and } z = - 6

\displaystyle x+y = 6 \ldots \ldots  (1)

\displaystyle xy = 8 \ldots \ldots  (2)

\displaystyle \Rightarrow  x(6-x) = 8

\displaystyle \Rightarrow  6x - x^2 = 8

\displaystyle \Rightarrow  x^2 - 6x + 8 = 0

\displaystyle \Rightarrow  (x-4)(x-2) = 0

\displaystyle \Rightarrow  x = 4  \Rightarrow   y = 2

\displaystyle \Rightarrow  x = 2  \Rightarrow   y = 4

\displaystyle x = 4,  y = 2,  z = -6  \text{ and } \omega = 4

\displaystyle \text{or}

\displaystyle x = 2,  y = 4,  z = -6  \text{ and } \omega = 4

\\

\displaystyle \text{Question 17: Give an example of}

\displaystyle \text{(i) a row matrix which is also a column matrix } 

\displaystyle \text{(ii) a diagonal matrix which is not scalar } 

\displaystyle \text{(iii) a triangular matrix } 

\displaystyle \text{Answer:}

\displaystyle \text{(i)}

\displaystyle \text{As we know that order of a row matrix } = 1 \times n

\displaystyle \text{And order of a column matrix } = m \times 1

\displaystyle \text{So, order of a row as well as column matrix } = 1 \times 1

\displaystyle \text{Therefore, required matrix } A = [a_{ij}] 1 \times 1

\displaystyle \text{(ii)}

\displaystyle \text{We know that a diagonal matrix has only } a_{11}, a_{22} \text{ and } a_{33}  \text{ for }  a 3 \times 3 \text{ matrix} \\ \text{such that these elements are equal or different and all other entries 0 while scalar} \\ \text{matrix has } a_{11},=a_{22}=a_{33}  = k \text{(say).}

\displaystyle \text{So, a diagonal matrix which is not scalar must have } a_{11} \neq a_{22} \neq a_{33} \\ \text{for } i \neq j

\displaystyle \text{Therefore, and example of such matrix is, }

\displaystyle A = [a_{ij}] = \begin{bmatrix}   1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3  \end{bmatrix}

\displaystyle \text{(iii)}

\displaystyle \text{A triangular matrix is a square matrix, }

\displaystyle A = [a_{ij}] \text{ such that } a_{ij} = 0 \text{ for all } i>j

\displaystyle \text{Therefore, and example of such matrix is, }

\displaystyle A = [a_{ij}] = \begin{bmatrix}   1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6  \end{bmatrix}

\\

Question 18: The sales figure of two car dealers during January 2013 showed that dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 premium and 4 standard cars. Total sales over the 2 month period of January – February revealed that dealer A sold 8 deluxe 7 premium and 6 standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premium and 7 standard cars. Write 2 x 3 matrices summarizing sales data for January and 2 month period for each dealer.

\displaystyle \text{Answer:}

Given Dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 premium and 4 standard cars,

\begin{bmatrix}  & \text{Deluxe} & \text{Premium} & \text{Standrd}     \\  \text{Delear A } & 5 & 3 & 4   \\  \text{Delear B } & 7 & 2 & 3   \\  \end{bmatrix}

Total sales over the 2 month period of January – February revealed that dealer A sold 8 deluxe 7 premium and 6 standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premium and 7 standard cars.

\begin{bmatrix}  & \text{Deluxe} & \text{Premium} & \text{Standrd}     \\  \text{Delear A } & 8 & 7 & 6   \\  \text{Delear B } & 10 & 5 & 7    \\  \end{bmatrix}

\\

\displaystyle \text{Question 19: For what values of } x \text{ and } y \text{ are the following matrices equal?} 

\displaystyle A = \begin{bmatrix}   2x+1 & 2y \\ 0 & y^2-5y   \end{bmatrix}, B = \begin{bmatrix}   x+3 & y^2+2 \\ 0 & -6 \end{bmatrix}  

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle 2x + 1 = x + 3

\displaystyle \Rightarrow  2x - x = 3 - 1

\displaystyle \Rightarrow  x = 2 \ldots \ldots  (1)

\displaystyle 2y = y^2 + 2

\displaystyle \Rightarrow  y^2 - 2y + 2 = 0

\displaystyle \Rightarrow  y = \frac{-2 \pm \sqrt{4-8}}{2}

\displaystyle \Rightarrow  y = \frac{-2 \pm 2i}{2}

\displaystyle \Rightarrow  y = -1 \pm i

\displaystyle \text{No real solutions } \ldots \ldots  (2)

\displaystyle y^2 - 5y = - 6

\displaystyle \Rightarrow  y^2 - 5y + 6 = 0

\displaystyle \Rightarrow  y^2 - 3y - 2y + 6 = 0

\displaystyle \Rightarrow  y(y - 3) - 2(y - 3) = 0

\displaystyle \Rightarrow  (y - 3)(y - 2) = 0

\displaystyle \Rightarrow  y = 3  \text{ or } 2 \ldots \ldots  (3)

\displaystyle \text{From the above equations we can say that A and B can't be equal for any value of } y.

\\

\displaystyle \text{Question 20: Find the values of } x \text{ and } y \text{ if } \begin{bmatrix}   x+10 & y^2+2y \\ 0 & -4   \end{bmatrix} = \begin{bmatrix}   3x+4 & 3 \\ 0 & y^2-5y \end{bmatrix}

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle x + 10 = 3x + 4

\displaystyle \Rightarrow  x - 3x = 4 - 10

\displaystyle \Rightarrow  - 2x = - 6

\displaystyle \Rightarrow  x = 3 \ldots \ldots  (1)

\displaystyle y^2 + 2y = 3

\displaystyle \Rightarrow  y^2 + 2y - 3 = 0

\displaystyle \Rightarrow  y^2 + 3y - y - 3 = 0

\displaystyle \Rightarrow  y(y + 3) - 1(y + 3) = 0

\displaystyle \Rightarrow  (y + 3)(y - 1) = 0

\displaystyle \Rightarrow  y = - 3 \text{ or } 1 \ldots \ldots  (2)

\displaystyle y^2 - 5y = - 4

\displaystyle \Rightarrow  y^2 - 5y + 4 = 0

\displaystyle \Rightarrow  y^2 - 4y - y + 4 = 0

\displaystyle \Rightarrow  y(y - 4) - 1(y - 4) = 0

\displaystyle \Rightarrow  (y - 4)(y - 1) = 0

\displaystyle \Rightarrow  y = 4 \text{ or } 1 \ldots \ldots  (3)

\displaystyle \text{The common value is } x = 3 \text{ and }  y = 1

\\

\displaystyle \text{Question 21: Find the values of } a \text{ and } b \text{ if } A = B, \text{ where}

\displaystyle A = \begin{bmatrix}   a+4 & 3b \\ 8 & -6   \end{bmatrix}, B = \begin{bmatrix}   2a+2 & b^2+2 \\ 8 & b^2-10 \end{bmatrix}  

\displaystyle \text{Answer:}

\displaystyle \text{We know that if two matrices are equal then the corresponding elements of each matrices} \\ \text{are also equal. }

\displaystyle a + 4 = 2a + 2

\displaystyle \Rightarrow  a - 2a = 2 - 4

\displaystyle \Rightarrow  - a = - 2

\displaystyle \Rightarrow  a = 2 \ldots \ldots  (1)

\displaystyle 3b = b^2 + 2

\displaystyle \Rightarrow  b^2 - 3b + 2 = 0

\displaystyle \Rightarrow  b^2 - 2b - b + 2 = 0

\displaystyle \Rightarrow  b(b - 2) - 1(b - 2) = 0

\displaystyle \Rightarrow  (b - 2)(b - 1) = 0

\displaystyle \Rightarrow  b = 2 \text{ or } 1 \ldots \ldots  (2)

\displaystyle - 6 = b^2 - 10

\displaystyle \Rightarrow  b^2 = - 10 + 6

\displaystyle \Rightarrow  b^2 = - 4

\displaystyle \Rightarrow  b = \pm 2i

\displaystyle \text{No real solution} \ldots \ldots  (3)

\displaystyle a = 2, b = 2 \text{ or } 1

\\

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