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

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On April 24, 2024, 11:35 AM

Miscellaneous Problems
Question 1.
Find the rank of the matrix $A=\left(\begin{array}{rrrr}1 & -3 & 4 & 7 \\ 9 & 1 & 2 & 0\end{array}\right)$
Solution:
$
A=\left(\begin{array}{rrrr}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)
$
Order of $A$ is $2 \times 4$. So $\rho(A) \leq 2$
Consider the second-order minor
$
\left|\begin{array}{cc}
1 & -3 \\
9 & 1
\end{array}\right|=1+27=28 \neq 0
$
There is a minor of order 2 which is not zero.
So $\rho(\mathrm{A})=2$
Question 2.
Find the rank of the matrix $A=\left(\begin{array}{rrrr}-2 & 1 & 3 & 4 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 4 & 7\end{array}\right)$
Solution:
Given $A=\left(\begin{array}{rrrr}-2 & 1 & 3 & 4 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & 4 & 7\end{array}\right)$
Order of $A$ is $3 \times 4$. So $\rho(A) \leq 3$
Consider the third-order minor
$
\begin{aligned}
& \left|\begin{array}{ccc}
-2 & 1 & 3 \\
0 & 1 & 1 \\
1 & 3 & 4
\end{array}\right| \\
& =-2(4-3)-1(0-1)+3(0-1) \\
& =-2+1-3= \\
& =-4 \neq 0
\end{aligned}
$
There exists a minor of order 3 which is not zero. So $\rho(A)=3$
Question 3.
Find the rank of the matrix $A=\left(\begin{array}{cccc}4 & 5 & 2 & 2 \\ 3 & 2 & 1 & 6 \\ 4 & 4 & 8 & 0\end{array}\right)$

Solution:
Given $\mathrm{A}=\left(\begin{array}{llll}4 & 5 & 2 & 2 \\ 3 & 2 & 1 & 6 \\ 4 & 4 & 8 & 0\end{array}\right)$
Order of $\mathrm{A}$ is $3 \times 4$. So $\rho(\mathrm{A}) \leq 3$
Consider the third-order minor,
$
\begin{aligned}
& \left|\begin{array}{ccc}
4 & 5 & 2 \\
3 & 2 & 1 \\
4 & 4 & 8
\end{array}\right| \\
& =4(16-4)-5(24-4)+2(12-8) \\
& =48-100+8 \\
& =-44 \neq 0
\end{aligned}
$
There is a minor of order 3 which is not zero.
$
\rho(\mathrm{A})=3
$
Question 4.
Examine the consistency of the system of equations:
$
x+y+z=7, x+2 y+3 z=18, y+2 z=6
$
Solution:
The given system is $x+y+z=7, x+2 y+3 z=18, y+2 z=6$.
The matrix equation corresponding to the given system
$
\left(\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
0 & 1 & 2
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{c}
7 \\
18 \\
6
\end{array}\right)
$
A $\quad \mathrm{X}=\mathrm{B}$
Augmented matrix [A, B]
$
\begin{aligned}
& \left(\begin{array}{cccc}
1 & 1 & 1 & 7 \\
1 & 2 & 3 & 18 \\
0 & 1 & 2 & 6
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 7 \\
0 & 1 & 2 & 11 \\
0 & 1 & 2 & 6
\end{array}\right) \quad R_2 \rightarrow R_2-R_1 \\
& \sim\left(\begin{array}{cccc}
0 & 1 & 1 & 7 \\
0 & 1 & 2 & 11 \\
0 & 0 & 0 & -5
\end{array}\right) \quad \quad R_3 \rightarrow R_3-R_2 \\
&
\end{aligned}
$

The last equivalent matrix is in the echelon form. [A, B] has 3 non-zero rows and $[\mathrm{A}]$ has 2 non-zero rows.
$\rho([\mathrm{A}, \mathrm{B}])=3, \rho(\mathrm{A})=2$
$\rho(A) \neq \rho([A, B])$
Hence the given system is inconsistent and has no solution.
Question 5.
Find $\mathrm{k}$ if the equations $2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}=5,3 \mathrm{x}-\mathrm{y}+4 \mathrm{z}=2, \mathrm{x}+7 \mathrm{y}-6 \mathrm{z}=\mathrm{k}$ are consistent.
Solution:
The given system is $2 x+3 y-z=5,3 x-y+4 z=2, x+7 y-6 z=k$
It is also given that the system is consistent.
The matrix equation corresponding to the given system is
$
\begin{aligned}
\left(\begin{array}{ccc}
2 & 3 & -1 \\
3 & -1 & 4 \\
1 & 7 & -6
\end{array}\right)\left(\begin{array}{l}
\dot{x} \\
y \\
z
\end{array}\right) & =\left(\begin{array}{l}
5 \\
2 \\
k
\end{array}\right) \\
\mathrm{A} \mathrm{X} & =\mathrm{B}
\end{aligned}
$
Augmented matrix [A, B]
$
\begin{aligned}
& \left(\begin{array}{cccc}
2 & 3 & -1 & 5 \\
3 & -1 & 4 & 2 \\
1 & 7 & -6 & k
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 7 & -6 & k \\
3 & -1 & 4 & 2 \\
2 & 3 & -1 & 5
\end{array}\right) \\
& \mathrm{R}_1 \leftrightarrow \mathrm{R}_3 \\
& \sim\left(\begin{array}{cccc}
1 & 7 & -6 & k \\
0 & -22 & 22 & 2-3 k \\
0 & -11 & 11 & 5-2 k
\end{array}\right) \quad \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_2-3 \mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3-2 \mathrm{R}_1
\end{array} \\
& \sim\left(\begin{array}{cccc}
1 & 7 & -6 & k \\
0 & -22 & 22 & 2-3 k \\
0 & 0 & 0 & k-8
\end{array}\right) \quad \quad \mathrm{R}_3 \rightarrow \mathrm{R}_2-2 \mathrm{R}_3 \\
& \rho(\mathrm{A})=2 ; \rho([\mathrm{A}, \mathrm{B}])=2 \text { or } 3 \text {. } \\
&
\end{aligned}
$

Question 6.
Find $k$ if the equations $x+y+z=1,3 x-y-z=4, x+5 y+5 z=k$ are inconsistent.
Solution:
The given system is $x+y+z=1,3 x-y-z=4, x+5 y+5 z=k$ and it is inconsistent.
The matrix equation corresponding to the given system is
$
\begin{aligned}
& \left(\begin{array}{ccc}
1 & 1 & 1 \\
3 & -1 & -1 \\
1 & 5 & 5
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{l}
1 \\
4 \\
k
\end{array}\right) \\
& \text { A } \quad \mathrm{X}=\mathrm{B} \\
& \text { Augmented matrix [A, B] } \\
& \left(\begin{array}{cccc}
1 & 1 & 1 & 1 \\
3 & -1 & -1 & 4 \\
1 & 5 & 5 & k
\end{array}\right) \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 & -4 & -4 & 1 \\
0 & 4 & 4 & k-1
\end{array}\right) \quad \begin{array}{l}
\mathrm{R}_2 \rightarrow \mathrm{R}_2-3 \mathrm{R}_1 \\
\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1
\end{array} \\
& \sim\left(\begin{array}{cccc}
1 & 1 & 1 & 1 \\
0 & -4 & -4 & 1 \\
0 & 0 & 0 & k
\end{array}\right) \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_2 \\
&
\end{aligned}
$
$\rho(A)=2$ since the equivalent matrix has 2 non-zero rows.
For the equations to be inconsistent
$\rho([\mathrm{A}, \mathrm{B}]) \neq \rho(\mathrm{A})$
$\rho([\mathrm{A}, \mathrm{B}]) \neq 2 \Rightarrow \mathrm{k} \neq 0$
So $\mathrm{k}$ can assume any real value other than 0.

Question 7.
Solve the equations $x+2 y+z=7,2 x-y+2 z=4, x+y-2 z=-1$ by using Cramer's rule.
Solution:
The given system is $x+2 y+z=7,2 x-y+2 z=4, x+y-2 z=-1$
$
\begin{aligned}
& \text { Here } \Delta=\left|\begin{array}{ccc}
1 & 2 & 1 \\
2 & -1 & 2 \\
1 & 1 & -2
\end{array}\right| \\
& =1(2-2)-2(4-2)+1(2+1) \\
& =12+3 \\
& =15 \neq 0
\end{aligned}
$
We can apply Cramer's rule; the system is consistent and has a unique solution.

Question 8.
The cost of $2 \mathrm{~kg}$. of wheat and $1 \mathrm{~kg}$. of sugar is ₹ 100 . The cost of $1 \mathrm{~kg}$. of wheat and $1 \mathrm{~kg}$. of rice is ₹ 80. The cost of $3 \mathrm{~kg}$. of wheat, $2 \mathrm{~kg}$. of sugar and $1 \mathrm{~kg}$ of rice is ₹ 220 . Find the cost of each per kg., using Cramer's rule.
Solution:
Let the cost of wheat per kg be ₹ $\mathrm{x}$, the cost of sugar per kg be ₹ $\mathrm{y}$ and the cost of rice per kg be ₹ $\mathrm{z}$, respectively.
It is given that $2 x+y=100, x+z=80,3 x+2 y+z=220$
Here $\Delta=\left|\begin{array}{ccc}2 & 1 & 0 \\ 1 & 0 & 1 \\ 3 & 2 & 1\end{array}\right|$
$
=2(-2)-1(-2)
$
$
=-2 \neq 0
$
we can apply Cramer's rule and the system is consistent and has a unique solution.

Question 9.
A salesman has the following record of sales for three months for three items A, B, and C, which have different rates of commission.

Find out the rate of commission on items A, B, and C by using Cramer's rule.
Solution:
Let the rate of commission on items $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ be ₹ $\mathrm{x}$, ₹ $\mathrm{y}$ and ₹ $\mathrm{z}$ per unit respectively.
According to the problem we have,
January, $90 \mathrm{x}+100 \mathrm{y}+20 \mathrm{z}=800$
February, $130 \mathrm{x}+50 \mathrm{y}+40 \mathrm{z}=900$
March, $60 \mathrm{x}+100 \mathrm{y}+30 \mathrm{z}=850$

Question 10.
The subscription department of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the
magazine while others do not. From this mailing list, $60 \%$ of those who already subscribe will subscribe again while $25 \%$ of those who do not now subscribe will subscribe. In the last letter, it was found that $40 \%$ of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Solution:
Let $\mathrm{X}$ represent people who subscribe to the magazine and $\mathrm{Y}$ represent people who do not subscribe to the magazine. Then
$(\mathrm{XX}) \Rightarrow$ those who already subscribed will do it again.
$(\mathrm{XY}) \Rightarrow$ those who already subscribed will not do it again.
$(\mathrm{Y} \mathrm{X}) \Rightarrow$ those who have not subscribed will do it now.
$(\mathrm{Y} Y) \Rightarrow$ those who have not subscribed already will not do it now also.
From the question,
$
\begin{aligned}
& (\mathrm{XX})=60 \%=\frac{60}{100}=0.6 \\
& (\mathrm{XY})=1-0.6=0.4 \\
& (\mathrm{YX})=25 \%=\frac{25}{100}=0.25 \\
& (\mathrm{YY})=1-0.25=0.75
\end{aligned}
$
The current position is given by $X=40 \%$ and $Y=60 \%$

We have to predict the value of $\mathrm{X}$ and $\mathrm{Y}$ after the current letter is sent.