Exercise 3.16 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.16$
Question 1.
In the matrix $A=\left[\begin{array}{cccc}8 & 9 & 4 & 3 \\ -1 & \sqrt{7} & \frac{\sqrt{3}}{2} & 5 \\ 1 & 4 & 3 & 0 \\ 6 & 8 & -11 & 1\end{array}\right]$
(i) The number of elements
(ii) The order of the matrix
(iii) Write the elements $\mathrm{a}_{22}, \mathrm{a}_{23}, \mathrm{a}_{24}, \mathrm{a}_{34}, \mathrm{a}_{43}, \mathrm{a}_{44}$
Solution:
(i) 16
(ii) $4 \times 4$
(iii) $\sqrt{7}, \frac{\sqrt{3}}{2}, 5,0,-11,1$
 

Question $2 .$
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Solution:
$1 \times 18,2 \times 9,3 \times 6,6 \times 3,9 \times 2,18 \times 1$ and $1 \times 6,2 \times 3,3 \times 2,6 \times 1$
 

Question $3 .$
Construct a $3 \times 3$ matrix whose elements are given by then find the transpose of A.
(i) $\mathrm{a}_{\mathrm{ij}}=|\mathrm{i}-2 \mathrm{j}|$
(ii) $\mathrm{a}_{\mathrm{ij}}=\frac{(i+j)^{3}}{3}$
Solution:
(i) $\mathrm{a}_{\mathrm{ij}}=|\mathrm{i}-2 \mathrm{j}|$
$\mathrm{a}_{11}=|1-2 \times 1|=|1-2|=|-1|=1$
$\mathrm{a}_{12}=|1-2 \times 2|=|1-4|=|-3|=3$
$a_{13}=|1-2 \times 3|=|1-6|=|-5|=5$

$\begin{aligned}
&a_{21}=|2-2 \times 1|=|2-2|=0 \\
&a_{22}=|2-2 \times 2|=|2-4|=|-2|=2 \\
&a_{23}=|2-2 \times 3|=|2-6|=|-4|=4 \\
&a_{31}=|3-2 \times 1|=|3-2|=|1|=1 \\
&a_{32}=|3-2 \times 2|=|3-4|=|-1|=1 \\
&a_{33}=|3-2 \times 3|=|3-6|=|-3|=3 \\
&\therefore\left[\begin{array}{lll}
1 & 3 & 5 \\
0 & 2 & 4 \\
1 & 1 & 3
\end{array}\right] \text { is the required } 3 \times 3 \text { matrix }
\end{aligned}$

$\begin{aligned}
&\text { (ii) } a_{i j}=\frac{(i+j)^{3}}{3}\\
&a_{11}=\frac{(1+1)^{3}}{3}=\frac{(2)^{3}}{3}=\frac{8}{3}\\
&a_{12}=\frac{(1+2)^{3}}{3}=\frac{(3)^{3}}{3}=\frac{27}{3}=9\\
&a_{13}=\frac{(1+3)^{3}}{3}=\frac{(4)^{3}}{3}=\frac{64}{3}\\
&a_{21}=\frac{(2+1)^{3}}{3}=9\\
&a_{22}=\frac{(2+2)^{3}}{3}=\frac{(4)^{3}}{3}=\frac{64}{3}\\
&a_{23}=\frac{(2+3)^{3}}{3}=\frac{(5)^{3}}{3}=\frac{125}{3}\\
&a_{31}=\frac{(3+1)^{3}}{3}=\frac{(4)^{3}}{3}=\frac{64}{3}\\
&a_{32}=\frac{(3+2)^{3}}{3}=\frac{125}{3}\\
&a_{33}=\frac{(3+3)^{3}}{3}=\frac{216}{3}=72
\end{aligned}$

$a_{33}=\frac{(3+3)^{3}}{3}=\frac{216}{3}=72$
$\left[\begin{array}{ccc}\frac{8}{3} & 9 & \frac{64}{3} \\ 9 & \frac{64}{3} & \frac{125}{3}\end{array}\right]$ is the required $3 \times 3$ matrix
$\left[\frac{64}{3} \quad \frac{125}{3} \quad 72\right.$
 

Question $4 .$
If $A=\left[\begin{array}{ccc}5 & 4 & 3 \\ 1 & -7 & 9 \\ 3 & 8 & 2\end{array}\right]$ then find the transpose of $A .$
Solution:
If $\mathrm{A}=\left[\begin{array}{ccc}5 & 4 & 3 \\ 1 & -7 & 9 \\ 3 & 8 & 2\end{array}\right]$
Transpose of $\mathrm{A}=\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ccc}5 & 1 & 3 \\ 4 & -7 & 8 \\ 3 & 9 & 2\end{array}\right]$

Question $5 .$

Solution:
$\begin{aligned}
\text { If } \mathrm{A} &=\left[\begin{array}{cc}
\sqrt{7} & -3 \\
-\sqrt{5} & 2 \\
\sqrt{3} & -5
\end{array}\right] \\
-\mathrm{A} &=\left[\begin{array}{cc}
-\sqrt{7} & 3 \\
\sqrt{5} & -2 \\
-\sqrt{3} & 5
\end{array}\right] \\
\text { Transpose of }-\mathrm{A} &=(-\mathrm{A})^{\mathrm{r}}=\left[\begin{array}{ccc}
-\sqrt{7} & +\sqrt{5} & -\sqrt{3} \\
+3 & -2 & +5
\end{array}\right]
\end{aligned}$

Question $6 .$
If $A=\left[\begin{array}{ccc}5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1\end{array}\right]$ then verify $\left(A^{T}\right)^{T}=A$
Solution:
If $\mathrm{A}=\left[\begin{array}{ccc}5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1\end{array}\right], \mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ccc}5 & -\sqrt{17} & 8 \\ 2 & 0.7 & 3 \\ 2 & \frac{5}{2} & 1\end{array}\right]$
$\left(A^{\top}\right)^{\top}=\left[\begin{array}{ccc}5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1\end{array}\right]=$ A. . verified
 

Question 7.
Find the values of $x, y$ and $z$ from the following equations
(i) $\left[\begin{array}{cc}12 & 3 \\ x & \frac{3}{2}\end{array}\right]=\left[\begin{array}{ll}y & z \\ 3 & 5\end{array}\right]$
(ii) $\left[\begin{array}{cc}x+y & 2 \\ 5+z & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$
(iii) $\left[\begin{array}{l}x+y+z \\ x+z \\ y+z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]$

Solution:

$\begin{aligned}
&\text { (i) }\left[\begin{array}{ll}
12 & 3 \\
x & \frac{3}{2}
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
3 & 5
\end{array}\right]\\
&x=3\\
&y=12\\
&z=3\\
&\text { (ii) }\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\\
&\Rightarrow \quad 5+z=5\\
&z=0\\
&x+y=6\\
&x=6-y\\
&x y=8\\
&(6-y) y=8\\
&\Rightarrow 6 y-y^{2}-8=0\\
&y^{2}-6 y+8=0\\
&(y-4)(y-2)=0\\
&y=4,2, x=2,4\\
&x=2, y=4, z=0\\
&\text { (or) }\\
&x=4, y=2, z=0
\end{aligned}$

(iii)
$\left[\begin{array}{c}x+y+z \\ x+z \\ y+z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]$
$\Rightarrow \quad x+y+z=9$
$\Rightarrow \quad x+z=5$
$\Rightarrow \quad y+z=7$
Sub. $y=4$ in (3)
$\begin{array}{r}
4+z=7 \\
z=3
\end{array}$
Sub. $z=3$ in (2)
$\begin{array}{r}
x+3=5 \\
x=2 \\
x=2, y=4, z=3
\end{array}$