Exercise 7.4 - Chapter 7 Matrices and Determinants 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 7.4
Question 1.
Find the area of the triangle whose vertices are $(0,0),(1,2)$ and $(4,3)$
Solution:
Area of triangle with vertices
$0\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is $\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$
$\therefore$ Area of $\mathrm{A}$ with vertices $(0,0),(1,2)$ and $(4,3)$ is
$
\begin{aligned}
\frac{1}{2}\left|\begin{array}{lll}
0 & 0 & 1 \\
1 & 2 & 1 \\
4 & 3 & 1
\end{array}\right| & =\frac{1}{2}\{1(3-8)\} \\
& =\frac{1}{2}(-5)=5 / 2 \text { sq. unit }=2.5 \text { sq. unit }
\end{aligned}
$
(as the area cannot be negative).
Question 2.
If $(\mathrm{k}, 2),(2,4)$ and $(3,2)$ are vertices of the triangle of area 4 square units then determine the value of $\mathrm{k}$.

Solution:
Area of $\Delta$ with vertices $(k, 2)(2,4)$ and $(3,2)=\frac{1}{2}\left|\begin{array}{lll}k & 2 & 1 \\ 2 & 4 & 1 \\ 3 & 2 & 1\end{array}\right|=4$ (given)
$
\begin{aligned}
& \Rightarrow \quad\left|\begin{array}{lll}
k & 2 & 1 \\
2 & 4 & 1 \\
3 & 2 & 1
\end{array}\right|=2(4)=8 \\
& \text { (i.e.,) } k(4-2)-2(2-3)+1(4-12)= \pm 8 \\
& \text { (i.e.,) } 2 k-2(-1)+1(-8)= \pm 8 \\
& \text { (i.e.,) } 2 k+2-8=8 \\
& \text { (i.e.,) } 2 k=8+8-2=14 \\
& k=14 / 2=7 \\
& \therefore k=7 \text {. } \\
& \begin{array}{l}
2 k+2-8=-8 \\
\Rightarrow 2 k=-8+8-2 \\
2 k=-2 \\
k=-1 \\
\text { So } k=7 \text { (or) } k=-1
\end{array} \\
&
\end{aligned}
$
Question 3
Identify the singular and non-singular matrices:
(i) $\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{array}\right|$
$\left|\begin{array}{rrr}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$
(iii) $\left|\begin{array}{ccc}0 & a-b & k \\ b-a & 0 & 5 \\ -k & -5 & 0\end{array}\right|$
Solution:
(i) For a given square matrix A,
1. If $|\mathrm{A}|=0$ then it is a singular matrix.
2. If $|\mathrm{A}| \neq 0$ then it is a non singular matrix.
(i)
$
\begin{aligned}
A & =\left(\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right) \\
|\mathrm{A}|=\left|\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right| & =1(45-48)-2(36-42)+3(32-35) \\
& =1(-3)-2(-6)+3(-3) \\
& =-3+12-9=0
\end{aligned}
$
$\Rightarrow A$ is a singular matrix.

(ii) Let
$
\begin{aligned}
& A=\left(\begin{array}{ccc}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right) \\
& |A|=\left|\begin{array}{ccc}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right| \text { (expanding along } R_2 \text { ) } \\
& =-6(21-25)-4(10+3) \\
& =-6(-4)-4(13) \\
& =24-52=-28 \neq 0 \quad \therefore \mathrm{A} \text { is a non singular matrix } \\
&
\end{aligned}
$

(iii) Let $\mathrm{A}\left[\begin{array}{ccc}0 & a-b & k \\ b-a & 0 & 5 \\ -k & -5 & 0\end{array}\right]$
Which is a skew symmetric matrix
$\therefore|\mathrm{A}|=0 \Rightarrow \mathrm{A}$ is a singular matrix.
Question 4.
Determine the value of $a$ and $b$ so that the following matrices are singular:
(i) $A=\left[\begin{array}{rr}7 & 3 \\ -2 & a\end{array}\right]$
(ii) $\mathrm{B}=\left[\begin{array}{ccc}b-1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4\end{array}\right]$
Solution:

$(i)$ $A=\left(\begin{array}{cc}7 & 3 \\ -2 & a\end{array}\right)$
Given $A$ is a singular matrix
$
\begin{array}{rlrl}
\Rightarrow & |\mathrm{A}| & =0 \Rightarrow\left|\begin{array}{cc}
7 & 3 \\
-2 & a
\end{array}\right|=0 \\
\text { (i.e.) } 7 a+6=0 & 7 a= & -6 \Rightarrow a=-6 / 7 . \\
& \mathrm{B} & =\left(\begin{array}{ccc}
b-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right)
\end{array}
$
Given $B$ is a singular matrix $\Rightarrow|B|=0$
$
\Rightarrow \text { Now } \quad|\mathrm{B}|=\left[\begin{array}{ccc}
b-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]
$
expanding along $\mathrm{R}_1$
$
\begin{aligned}
& \mathrm{b}(4+4)+7(-6-1)=0 \text { (given) } \\
& 8 \mathrm{~b}+7(-7)=0 \\
& \text { (i.e.) } 8 \mathrm{~b}-49=0 \Rightarrow 8 \mathrm{~b}=49 \Rightarrow \mathrm{b}=49 / 8
\end{aligned}
$
Question 5.
If $\cos 2 \theta=0$, determine $\left[\begin{array}{ccc}\theta & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right]^2$

Solution:
Given $\cos 2 \theta=0$
$
\Rightarrow \begin{aligned}
2 \theta & =\pi / 2 \Rightarrow \theta=\pi / 4 \\
\therefore \quad \cos \theta & =\cos \pi / 4=1 / \sqrt{2}
\end{aligned}
$
and
$
\sin \theta=\sin \pi / 4=1 / \sqrt{2}
$
$
\text { Let } \begin{aligned}
\Delta=\left|\begin{array}{ccc}
0 & \cos \theta & \sin \theta \\
\cos \theta & \sin \theta & 0 \\
\sin \theta & 0 & \cos \theta
\end{array}\right| & =\left|\begin{array}{ccc}
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}
\end{array}\right| \\
& =0()-\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}-0\right)+\frac{1}{\sqrt{2}}\left(0-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\right) \\
, \quad & -\frac{1}{\sqrt{2}} \times \frac{1}{2}-\frac{1}{\sqrt{2}} \frac{1}{2}=\frac{-1}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}=\frac{-2}{2 \sqrt{2}} \\
& =-\frac{1}{\sqrt{2}} \\
\therefore \quad \Delta^2 & =\left(\frac{-1}{\sqrt{2}}\right)^2=\frac{1}{2}
\end{aligned}
$
Question 6.
Find the value of the product; $\left[\begin{array}{ll}\log _3 64 & \log _4 3 \\ \log _3 8 & \log _4 9\end{array}\right] \times\left[\begin{array}{ll}\log _2 3 & \log _8 3 \\ \log _3 4 & \log _3 4\end{array}\right]$

Sol:

$
\begin{aligned}
& \left|\begin{array}{rr}
\log _3 64 & \log _4 3 \\
\log _3 8 & \log _4 9
\end{array}\right|\left|\begin{array}{ll}
\log _2 3 & \log _8 3 \\
\log _3 4 & \log _3 4
\end{array}\right| \\
& =\left|\begin{array}{cc}
\log _3 64 \times \log _2 3+\log _4 3 \times \log _3 4 & \log _3 64 \times \log _8 3+\log _3 64+\log _3 4 \\
\log _3 8 \times \log _2 3+\log _4 9 \times \log _3 4 & \log _3 8 \times \log _8 3+\log _4 9 \times \log _3 4
\end{array}\right| \\
& =\left|\begin{array}{cc}
\log _2 64+1 & \log _8 64+1 \\
\log _2 8+\log _3 9 & 1+\log _3 9
\end{array}\right| \\
& {\left[\begin{array}{l}
\log _2 64=\log _2 2^6=6 \log _2 2=6 \\
\log _8 64=\log _8 8^2=2 \log _8 8=2 \\
\log _2 8=\log _2 2^3=3 \log _2 2=3 \\
\log _3 9=\log _3 3^2=2 \log _3 3=2
\end{array}\right]} \\
& =\left|\begin{array}{ll}
6+1 & 2+1 \\
3+2 & 1+2
\end{array}\right|=\left|\begin{array}{ll}
7 & 3 \\
5 & 3
\end{array}\right|=21-15=6 \\
&
\end{aligned}
$