Exercise 3.17 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.17$
Question 1.
If $\mathrm{A}=\left[\begin{array}{cc}1 & 9 \\ 3 & 4 \\ 8 & -3\end{array}\right]=\left[\begin{array}{ll}5 & 7 \\ 3 & 3 \\ 1 & 0\end{array}\right]$ then verify that
(i) $\mathrm{A}+\mathrm{B}=\mathrm{B}+\mathrm{A}$
(ii) $\mathrm{A}+(-\mathrm{A})=(-\mathrm{A})+\mathrm{A}=0$
Solution:
$\left.\begin{array}{rl}
\text { L.H.S }=\mathrm{A}+\mathrm{B}= & {\left[\begin{array}{cc}
1 & 9 \\
3 & 4 \\
8 & -3
\end{array}\right]+\left[\begin{array}{ll}
5 & 7 \\
3 & 3 \\
1 & 0
\end{array}\right]} \\
= & {\left[\begin{array}{cc}
6 & 16 \\
6 & 7 \\
9 & -3
\end{array}\right]} \\
\text { R.H.S }=\mathrm{B}+\mathrm{A}= & {\left[\begin{array}{cc}
5 & 7 \\
3 & 3 \\
1 & 0
\end{array}\right]+\left[\begin{array}{cc}
1 & 9 \\
3 & 4 \\
8 & -3
\end{array}\right]} \\
= & {\left[\begin{array}{cc}
6 & 16 \\
6 & 7 \\
9 & -3
\end{array}\right]} \\
= & 4
\end{array}\right]$
$(1)=(2) \Rightarrow$ L.H.S $=$ R.H.S. Hence verified.
(ii) $\mathrm{A}+(-\mathrm{A})=(-\mathrm{A})+\mathrm{A}=0$
L.H.S $=\mathrm{A}+(-\mathrm{A})$

$=\left[\begin{array}{cc}
1 & 9 \\
3 & 4 \\
8 & -3
\end{array}\right]+\left[\begin{array}{cc}
-1 & -9 \\
-3 & -4 \\
-8 & 3
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}\right]$
$\begin{aligned}
&\text { R.H.S }=(-\mathrm{A})+\mathrm{A} \\
&=\left[\begin{array}{cc}
-1 & -9 \\
-3 & -4 \\
-8 & , 3
\end{array}\right]+\left[\begin{array}{cc}
1 & 9 \\
3 & 4 \\
8 & -3
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}$
$(1)=(2) \Rightarrow$ L.H.S. $=$ R.H.S. Hence verified.

Question 2.
If $A=\left[\begin{array}{ccc}4 & 3 & 1 \\ 2 & 3 & -8 \\ 1 & 0 & -4\end{array}\right], B=\left[\begin{array}{ccc}2 & 3 & 4 \\ 1 & 9 & 2 \\ -7 & 1 & -1\end{array}\right]$ and
$\begin{aligned}
&C=\left[\begin{array}{ccc}
8 & 3 & 4 \\
1 & -2 & 3 \\
2 & 4 & -1
\end{array}\right] \text { then verify that } \\
&A+(B+C)=(A+B)+C
\end{aligned}$
Solution:
$(B+C)=\left[\begin{array}{ccc}
2 & 3 & 4 \\
1 & 9 & 2 \\
-7 & 1 & -1
\end{array}\right]+\left[\begin{array}{ccc}
8 & 3 & 4 \\
1 & -2 & 3 \\
2 & 4 & -1
\end{array}\right]$

$\begin{aligned}
&=\left[\begin{array}{ccc}
10 & 6 & 8 \\
2 & 7 & 5 \\
-5 & 5 & -2
\end{array}\right] \\
A+(B+C) &=\left[\begin{array}{ccc}
4 & 3 & 1 \\
2 & 3 & -8 \\
1 & 0 & -4
\end{array}\right]+\left[\begin{array}{ccc}
10 & 6 & 8 \\
2 & 7 & 5 \\
-5 & 5 & -2
\end{array}\right] \\
&=\left[\begin{array}{ccc}
14 & 9 & 9 \\
4 & 10 & -3 \\
-4 & 5 & -6
\end{array}\right]
\end{aligned}$
R.H.S. $(A+B)+C$
$\begin{aligned}
(A+B) &=\left[\begin{array}{ccc}
4 & 3 & 1 \\
2 & 3 & -8 \\
1 & 0 & -4
\end{array}\right]+\left[\begin{array}{ccc}
2 & 3 & 4 \\
1 & 9 & 2 \\
-7 & 1 & -1
\end{array}\right] \\
&=\left[\begin{array}{ccc}
6 & 6 & 5 \\
3 & 12 & -6 \\
-6 & 1 & -5
\end{array}\right]
\end{aligned}$
$(A+B)+C=\left[\begin{array}{ccc}
6 & 6 & 5 \\
3 & 12 & -6 \\
-6 & 1 & -5
\end{array}\right]+\left[\begin{array}{ccc}
8 & 3 & 4 \\
1 & -2 & 3 \\
2 & 4 & -1
\end{array}\right]$

$=\left[\begin{array}{ccc}
14 & 9 & 9 \\
4 & 10 & -3 \\
-4 & 5 & -6
\end{array}\right]$
$(1)=(2) \Rightarrow$ L.H.S. $=$ R.H.S. Hence verified.
 

Question $3 .$
Find $\mathrm{X}$ and $\mathrm{Y}$ if $\mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll}7 & 0 \\ 3 & 5\end{array}\right]$ and $\mathrm{X}-\mathrm{Y}=\left[\begin{array}{ll}3 & 0 \\ 0 & 4\end{array}\right]$ Solution:
$\begin{aligned}
&X+Y=\left[\begin{array}{ll}
7 & 0 \\
3 & 5
\end{array}\right] \\
&X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]
\end{aligned}$
$(1)+(2) \Rightarrow 2 x=\left[\begin{array}{cc}
10 & 0 \\
3 & 9
\end{array}\right] \Rightarrow x=\frac{1}{2}\left[\begin{array}{cc}
10 & 0 \\
3 & 9
\end{array}\right]$

$\begin{aligned}
&=\left[\begin{array}{ll}
5 & 0 \\
\frac{3}{2} & \frac{9}{2}
\end{array}\right]\\
&\text { (1) }-(2) \Rightarrow \mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll}
7 & 0 \\
3 & 5
\end{array}\right]\\
&X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]\\
&\frac{(-)}{2 \mathrm{Y}}=\left[\begin{array}{ll}
4 & 0 \\
3 & 1
\end{array}\right] \Rightarrow \mathrm{Y}=\frac{1}{2}\left[\begin{array}{ll}
4 & 0 \\
3 & 1
\end{array}\right]\\
&\therefore \mathrm{Y}=\left[\begin{array}{ll}
2 & 0 \\
\frac{3}{2} & \frac{1}{2}
\end{array}\right]\\
&\mathrm{X}=\left[\begin{array}{cc}
5 & 0 \\
\frac{3}{2} & \frac{9}{2}
\end{array}\right], y=\left[\begin{array}{cc}
2 & 0 \\
\frac{3}{2} & \frac{1}{2}
\end{array}\right]
\end{aligned}$

Question $4 .$
If $\mathrm{A}=\left[\begin{array}{lll}0 & 4 & 9 \\ 8 & 3 & 7\end{array}\right], \mathrm{B}=\left[\begin{array}{lll}7 & 3 & 8 \\ 1 & 4 & 9\end{array}\right]$ find the value of
(i) $\mathrm{B}-5 \mathrm{~A}$
(ii) $3 \mathrm{~A}-9 \mathrm{~B}$
Solution:
$A=\left[\begin{array}{lll}
0 & 4 & 9 \\
8 & 3 & 7
\end{array}\right], B=\left[\begin{array}{lll}
7 & 3 & 8 \\
1 & 4 & 9
\end{array}\right]$
(i)
$\begin{aligned}
\mathrm{B}-5 \mathrm{~A} &=\left[\begin{array}{lll}
7 & 3 & 8 \\
1 & 4 & 9
\end{array}\right]-5\left[\begin{array}{lll}
0 & 4 & 9 \\
8 & 3 & 7
\end{array}\right] \\
&=\left[\begin{array}{lll}
7 & 3 & 8 \\
1 & 4 & 9
\end{array}\right]-\left[\begin{array}{ccc}
0 & 20 & 45 \\
40 & 15 & 35
\end{array}\right] \\
&=\left[\begin{array}{ccc}
7 & -17 & -37 \\
-39 & -11 & -26
\end{array}\right]
\end{aligned}$
(ii)
$\begin{aligned}
3 \mathrm{~A}-9 \mathrm{~B} &=3\left[\begin{array}{lll}
0 & 4 & 9 \\
8 & 3 & 7
\end{array}\right]-9\left[\begin{array}{lll}
7 & 3 & 8 \\
1 & 4 & 9
\end{array}\right] \\
&=\left[\begin{array}{ccc}
0 & 12 & 27 \\
24 & 9 & 21
\end{array}\right]-\left[\begin{array}{ccc}
63 & 27 & 72 \\
9 & 36 & 81
\end{array}\right] \\
&=\left[\begin{array}{ccc}
-63 & -15 & -45 \\
15 & -27 & -60
\end{array}\right]
\end{aligned}$

Question $5 .$
Find the values of $x, y, z$ if
(i) $\left(\begin{array}{cc}x-3 & 3 x-z \\ x+y+7 & x+y+z\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 1 & 6\end{array}\right)$
(ii) $\left[\begin{array}{lll}x & y-z & z+3\end{array}\right]+\left[\begin{array}{lll}y & 4 & 3\end{array}\right]$
Solution:
(i) $\left(\begin{array}{cc}x-3 & 3 x-z \\ x+y+7 & x+y+z\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 1 & 6\end{array}\right)$
$x-3=1 \Rightarrow x=4$ $3 x-z=0$ $3(4)-z=0$ $-z=-12 \Rightarrow z=12$ $x+y+7=1$ $x+y=-6$ $4+y=-6$ $y=-10$ $x=4, y=-10, z=12$

$\begin{aligned}
&\text { (ii) }\left[\begin{array}{llll}
x & y-z & z+3
\end{array}\right]+\left[\begin{array}{lll}
y & 4 & 3
\end{array}\right]=\left[\begin{array}{lll}
4 & 8 & 16
\end{array}\right]\\
&x+y=4 \ldots \ldots \ldots \ldots \ldots \text { (1) }\\
&\begin{aligned}
&y-z+4=8 \ldots \ldots \ldots \ldots(2) \\
&z+3+3=16
\end{aligned}\\
&\begin{aligned}
&z+3+3=16 \ldots \ldots \ldots \ldots \ldots \\
&\text { From }(3) \text {, we get } z-10
\end{aligned}\\
&\text { From (2), we get } y=14\\
&\text { From (1) we get } x+14=4\\
&\mathrm{x}=-10\\
&x=-10, y=14, z=10\\
&x=4, y=6
\end{aligned}$

Question 7.
Find the non-zero values of $x$ satisfying the matrix equation $x\left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right]+2\left[\begin{array}{cc}8 & 5 x \\ 4 & 4 x\end{array}\right]=2\left[\begin{array}{cc}x^{2}+8 & 24 \\ 10 & 6 x\end{array}\right]$
Solution:
$\begin{aligned}
{\left[\begin{array}{cc}
2 x^{2} & 2 x \\
3 x & x^{2}
\end{array}\right]+\left[\begin{array}{cc}
16 & 10 x \\
8 & 8 x
\end{array}\right] } &=\left[\begin{array}{cc}
2 x^{2}+16 & 48 \\
20 & 12 x
\end{array}\right] \\
{\left[\begin{array}{cc}
2 x^{2}+16 & 12 x \\
3 x+8 & x^{2}+8 x
\end{array}\right]=\left[\begin{array}{cc}
2 x^{2}+16 & 48 \\
20 & 12 x
\end{array}\right] } \\
\Rightarrow 12 x=48 & \Rightarrow x=4
\end{aligned}$

Question $8 .$
Solve for $x, y:\left[\begin{array}{l}x^{2} \\ y^{2}\end{array}\right]+2\left[\begin{array}{c}-2 x \\ -y\end{array}\right]=\left[\begin{array}{l}5 \\ 8\end{array}\right]$
Solution:
$\begin{aligned}
&x^{2}-4 x=5 \\
&y^{2}-2 y=8 \\
&y^{2}-2 y-8=0 \\
&(y-4)(y+2)=0 \\
&y=4,-2 \\
&x^{2}-4 x-5=0 \\
&(x-5)(x+1)=0 \\
&x=5,-1 \\
&x=-1,5, y=4,-2
\end{aligned}$