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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Identify the singular and non-singular matrix.
(i) $\left[\begin{array}{rrr}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{array}\right]$;
(ii) $\left[\begin{array}{rrr}1 & 2 & 3 \\ 4 & 5 & 6 \\ -2 & -4 & -6\end{array}\right]$
Solution:
If the determinant value of a square matrix is zero it is called a singular matrix. Otherwise it is non-singular.
(i) Now
$
\begin{aligned}
\left|\begin{array}{rrr}
1 & 4 & 9 \\
4 & 9 & 16 \\
9 & 16 & 25
\end{array}\right| & =1\left|\begin{array}{rr}
9 & 16 \\
16 & 25
\end{array}\right|-4\left|\begin{array}{rr}
4 & 16 \\
9 & 25
\end{array}\right|+9\left|\begin{array}{rr}
4 & 9 \\
9 & 16
\end{array}\right| \\
& =1(225-256)-4(100-144)+9(64-81) \\
& =1(-31)-4(-44)+9(-17) \\
& =-31+176-153=-184+176=-8 \neq 0
\end{aligned}
$
$\therefore\left[\begin{array}{rrr}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{array}\right]$ is a non-singular matrix.
(ii)
$
\left|\begin{array}{rrr}
1 & 2 & 3 \\
4 & 5 & 6 \\
-2 & -4 & -6
\end{array}\right|=(-2)\left|\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
1 & 2 & 3
\end{array}\right|=(-2)(0)=0\left[\because \mathrm{R}_1=\mathrm{R}_3\right]
$
$
\Rightarrow\left|\begin{array}{rrr}
1 & 2 & 3 \\
4 & 5 & 6 \\
-2 & -4 & -6
\end{array}\right|
$
is a singular matrix.
Question 2.
Show that
$
\left|\begin{array}{lll}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{array}\right|^2=\left|\begin{array}{ccc}
1-2 a^2 & -a^2 & -a^2 \\
-a^2 & -1 & a^2-2 a \\
-a^2 & a^2-2 a & -1
\end{array}\right|
$
Solution:

$\begin{aligned}
& \text { LHS }=\left|\begin{array}{lll}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{array}\right|^2=\left|\begin{array}{ccc}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{array}\right|\left|\begin{array}{lll}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{array}\right| \\
& =\left|\begin{array}{lll}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{array}\right|(-)(-)\left|\begin{array}{lll}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{array}\right| \\
& =\left|\begin{array}{lll}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{array}\right| \begin{array}{rrr}
1 & a & a \\
-a & -1 & -a \\
-a & -a & -1
\end{array} \mid \\
& =\left|\begin{array}{ccc}
1-2 a^2 & -a^2 & -a^2 \\
-a^2 & -1 & a^2-2 a \\
-a^2 & a^2-2 a & -1
\end{array}\right|=\text { RHS } \\
&
\end{aligned}$