Exercise 1.2 - Chapter 1 Applications of Matrices and Determinants 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.2$
Question 1.
Find the rank of the following matrices by the minor method:
(i) $\left[\begin{array}{cc}2 & -4 \\ -1 & 2\end{array}\right]$
(ii) $\left[\begin{array}{rr}-1 & 3 \\ 4 & -7 \\ 3 & -4\end{array}\right]$
(iii) $\left[\begin{array}{rrrr}1 & -2 & -1 & 0 \\ 3 & -6 & -3 & 1\end{array}\right]$
(iv) $\left[\begin{array}{ccc}1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1\end{array}\right](v)\left[\begin{array}{llll}0 & 1 & 2 & 1 \\ 0 & 2 & 4 & 3 \\ 8 & 1 & 0 & 2\end{array}\right]$
Solution:
(i) $A=\left[\begin{array}{cc}2 & -4 \\ -1 & 2\end{array}\right] A$ is of order $2 \times 2 . \quad \Rightarrow \rho(A) \leq 2$
Now $|A|=\left|\begin{array}{cc}2 & -4 \\ -1 & 2\end{array}\right|=0 \quad \Rightarrow \rho(A) \neq 2$
So $\quad \rho(A) \leq 1$
Now in A a minor of order 1 is $a_{11}=2 \neq 0$
So
$
\begin{aligned}
\rho(A) & =1 \\
A & =\left[\begin{array}{rr}
-1 & 3 \\
4 & -7 \\
3 & -4
\end{array}\right]
\end{aligned}
$
(ii) Let $A$ is of order $3 \times 2$. So $\rho(A) \leq$ minimum of $3 \times 2$. (i.e) $\rho(A) \leq 2$.

$\begin{aligned}
& \text { Consider a minor of order } 2 \times 2=\left|\begin{array}{cc}
-1 & 3 \\
4 & -7
\end{array}\right|=7-12=-5 \neq 0 \\
& \text { So } \rho(A)=2 \text {. }
\end{aligned}$

(iii) Let $A=\left(\begin{array}{llll}1 & -2 & -1 & 0 \\ 3 & -6 & -3 & 1\end{array}\right)$
$A$ is of order $2 \times 4$. So $\rho(A) \leq \min \{2,4\}$ (i.e) $\rho(A) \leq 2$
(iv) Let $A=\left[\begin{array}{rrr}1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1\end{array}\right]$
$\mathrm{A}$ is of order $3 \times 3$. So $\rho(\mathrm{A}) \leq 3$.
The only minor of order $3 \times 3$ is $\left|\begin{array}{rrr}1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1\end{array}\right|=1(-4+6)+2(-2+30)+3(2-20)$ $=1(2)+2(28)+3(-18)=2+56-54=4 \neq 0$
$\operatorname{So} \rho(A)=3$.
(v) Let
$
A=\left[\begin{array}{llll}
0 & 1 & 2 & 1 \\
0 & 2 & 4 & 3 \\
8 & 1 & 0 & 2
\end{array}\right]
$
$\mathrm{A}$ is of order $3 \times 4$. So $\rho(\mathrm{A}) \leq \min \{3,4\}$ (i.e) $\rho(\mathrm{A}) \leq 3$
Consider a mi So $\rho(A)=3$
$\operatorname{rder} 3 \times 3=\left|\begin{array}{lll}0 & 1 & 1 \\ 0 & 2 & 3 \\ 8 & 1 & 2\end{array}\right|=8(3-2)=8(1)=8 \neq 0$

Question 2.
Find the rank of the folowing matrices by row reduction method:
(i) $\left[\begin{array}{cccc}1 & 1 & 1 & 3 \\ 2 & -1 & 3 & 4 \\ 5 & -1 & 7 & 11\end{array}\right]$
(ii) $\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & -1 & 2 \\ 1 & -2 & 3 \\ 1 & -1 & 1\end{array}\right]$
(iii) $\left[\begin{array}{cccc}3 & -8 & 5 & 2 \\ 2 & -5 & 1 & 4 \\ -1 & 2 & 3 & -2\end{array}\right]$
Solution:

(ii) Let 

(iii) Let, 

The last equivalent matrix is in row-echelon form. It has three non zero rows. $\rho(A)=3$
Question 3.
Find the inverse of each of the following by Gauss-Jordan method:
(i) $\left[\begin{array}{ll}2 & -1 \\ 5 & -2\end{array}\right]$
(ii) $\left[\begin{array}{ccc}1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3\end{array}\right]$
(iii) $\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8\end{array}\right]$
Solution:
(i) Let $A=\left(\begin{array}{ll}2 & -1 \\ 5 & -2\end{array}\right)$
Applying Gauss-Jordan method we get

(ii) Let, 

(iii) Let