Exercise 7.4 - Chapter 7 - Matrices and Determinants - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 7.4 - Chapter 7 - Matrices and Determinants - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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}
$