Exercise 8 A
Question 1
Ans:
(i) The order of matrix is $2 \times 2$ and the number of elements is 4
(ii) The order of matrix is $1 \times 2$ and the number of elements is 2
(iii) The order of matrix is $1 \times 4$ and the number of elements is 4 .
(iv)The order of matrix is $3 \times 1$ and the number of elements is 3
(v) The order of matrix is $3 \times 4$ and the number of elements is 12 .
(vi)The order of matrix is $3 \times 5$ and the number of elements
(vii) The order of matrix is $4 \times 2$ and the number of elements is 8 .
Question 2
Ans: (i) No, these matrices are not equal equal because their corresponding element are not same.(ii) $\left[\begin{array}{cc}4 & 7 \\ 3 & 2\end{array}\right]$ and $\left[\begin{array}{cc}3+1 & -\sqrt{45} \\ 5-2 & \frac{6}{3}\end{array}\right]$$\left[\begin{array}{ll}4 & 7 \\ 3 & 2\end{array}\right]$ and $\left[\begin{array}{ll}4 & 7 \\ 3 & 2\end{array}\right]$These matrices are equal because their corresponding elements are same
Question 3
Ans: Let the matrix $2 \times 2$ be $A\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$.
According to the question,
$\begin{aligned} a_{i j} &=(2 i-j)^{2} \\ a_{11} &=(2 \times 1-1)^{2} \\ &=(2-1)^{2} \\ &=(1)^{2} \\ &=1 . \end{aligned}$
$\begin{aligned} a_{12} &=(2 \times 1-2)^{2} \\ &=(2-2)^{2} \\ &=0 \\ a_{21} &=(2 \times 2-1)^{2} . \\ &=(4-1)^{2} \\ &=(3)^{2} \\ &=9 \end{aligned}$
$\begin{aligned}{9}_{22} &=(2 \times 2-2)^{2} \\ &=(4-2)^{2} \\ &=(2)^{2} \\ &=4 . \\ \therefore \text { Matrix } A &=\left[\begin{array}{ll}1 & 0 \\ 9 & 4\end{array}\right] \end{aligned}$
Question 4
Ans: Given,$\text { Matrix } B=\left[\begin{array}{cc}3 & 11 \\-5 & 6 \\8 & 0\end{array}\right]$(i) The order of matrix B is $3 \times 2$(ii) The elements $a_{12}=11, a_{31}=8, a_{22}=6 .$(iii) False, matrix B is not a square matrix because no of rows ≠ no. of columns
Question 5
Ans: (a) A matrix which has 6 elements can be of the following possible orders$6 \times 1,2 \times 3,3 \times 2,1 \times 6$(b) A matrix which has 3 rows and 4 columns will have $3 \times 4=12$ elchemts.
Question 6
Ans:$(i)\left[\begin{array}{cc}x & y \\ -1 & 5\end{array}\right]=\left[\begin{array}{cc}-2 & 0 \\ -1 & 5\end{array}\right]$On comparing their corresponding elements x=-2 and y =0 (ii) $\left[\begin{array}{ll}x & y\end{array}\right]=\left[\begin{array}{ll}-2 & y\end{array}\right]$On comparing their corresponding elements x=-2 and y=4
(iii)
$\left[\begin{array}{cc}2 x & 3 \\0 & y-1\end{array}\right]=\left[\begin{array}{cc}x-4 & 3 \\0 & 5\end{array}\right]$On comparing their corresponding elements,
$\begin{array}{cl}2 x=x-4 & \text { and } y-1=5 \\2 x-x=-4 & \text { and } y=5+1 . \\x=-4 & \text { and } y=6\end{array}$
Question 7
Ans: $\left[\begin{array}{ll}p+4 & 2 q-7 \\ s-3 & r+2 s\end{array}\right]=\left[\begin{array}{cc}6 & -3 \\ 2 & 14\end{array}\right]$On comparing their corresponding elements,$\begin{array}{ll}P+4=6, & 2 q-7=-3, \quad 5-3=2 \text { and } 8+2 \rho=14 \\p=6-4, & 2 q=-3+2, \quad 5=2+3 \text { and } 8+2 \times 5=14 \\p=2 ., & q=\frac{4}{2}=2 ., 5=5 \text { and } r=14-10=4 \text {. }\end{array}$Hence p, q , r and s are 2, 2, 4 and 5 respectively
On comparing their corresponding elements,
$\begin{array}{ll}P+4=6, & 2 q-7=-3, \quad 5-3=2 \text { and } 8+2 \rho=14 \\p=6-4, & 2 q=-3+2, \quad 5=2+3 \text { and } 8+2 \times 5=14 \\p=2 ., & q=\frac{4}{2}=2 ., 5=5 \text { and } r=14-10=4 \text {. }\end{array}$
Hence p, q , r and s are 2, 2, 4 and 5 respectively
Question 8
Ans: (i) False, because It is Lot compulsary.(ii) True(iii) False.(iv) False.(v) False.(vi) False(ví) True.
Question 9
Ans: Given The length width and height of one box are 6,5,10And that of second box are 5, 2, 8 and of third box are 7, 3, 3$\therefore\left[\begin{array}{ccc}6 & 5 & 10 \\ 5 & 2 & 8 \\ 7 & 3 & 3\end{array}\right]$
Question 10
Ans: (i) Given Three pupils Akhil, Rajnish and Sudha got marks in three test algebra. Their scores are Akhil= 79, 87 ,92 ; Rajnish = 95, 98, 91 and Sudha = 76, 88, 77$\therefore\left[\begin{array}{ccc}79 & 87 & 92 \\ 95 & 98 & 91 \\ 76 & 88 & 77\end{array}\right]$
(ii) The price of a record is $R_{9} .30$, of a blade packet if Rs. $1.50$ and a soap cake is RS. $1.70$.
$\therefore\left[\begin{array}{c}30 \\1.50 \\1.70\end{array}\right]$
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