Case study (4 Marks)

Question 14 Marks

Express the following matrices as the sum of symmetric and skew-symmetric matrices: $\left[\begin{array}{lll}3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7\end{array}\right]$

Answer

Let $\mathrm{A}=\left[\begin{array}{lll}3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7\end{array}\right] \Rightarrow A^{\prime}=\left[\begin{array}{lll}3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7\end{array}\right]$
$\therefore A+\mathrm{A}^{\prime}=\left[\begin{array}{lll}3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7\end{array}\right]+\left[\begin{array}{lll}3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7\end{array}\right]=\left[\begin{array}{rrr}6 & 6 & 5 \\ 6 & 2 & 9 \\ 5 & 9 & 14\end{array}\right]$
$\Rightarrow \frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)=\left[\begin{array}{ccc}3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7\end{array}\right]$, which is a symmetric matrix.
$\mathrm{A}-\mathrm{A}^{\prime}=\left[\begin{array}{lll}3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7\end{array}\right]-\left[\begin{array}{lll}3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7\end{array}\right]=\left[\begin{array}{rrr}0 & -2 & 5 \\ 2 & 0 & -3 \\ -5 & 3 & 0\end{array}\right]$
$\Rightarrow \frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right)=\left[\begin{array}{rrr}0 & -1 & \frac{5}{2} \\ 1 & 0 & -\frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 0\end{array}\right]$, which is a skew-symmetric.
Since $A=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)$,
$\therefore\left[\begin{array}{lll}3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7\end{array}\right]=\left[\begin{array}{lll}3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7\end{array}\right]+\left[\begin{array}{rrr}0 & -1 & \frac{5}{2} \\ 1 & 0 & -\frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 0\end{array}\right]$

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Question 24 Marks

An automobile company uses three types of steel $\mathrm{S}_{1}, \mathrm{~S}_{2}$ and $\mathrm{S}_{3}$ for producing three types of cars $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$.
Steel requirements (in tons) for each type of cars are given below:

	Cars
steel	$\mathrm{C}_{1}$	$\mathrm{C}_{2}$	$\mathrm{C}_{3}$
$\mathrm{~S}_{1}$	2	3	4
$\mathrm{~S}_{2}$	1	1	2
$\mathrm{~S}_{3}$	3	2	1

Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tonnes of steel of three types respectively.

Answer

Let $\mathrm{x}, \mathrm{y}$ and z be the number of cars produced by steel type $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ respectively.
Now, we can arrange this model in linear equation system
Thus, we have
$2 x+3 y+4 z=29$
$x+y+2 z=13$
$3 x+2 y+z=16$
It can be written as $\left[\begin{array}{lll}2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}29 \\ 13 \\ 16\end{array}\right]$ or $\mathrm{AX}=\mathrm{B}$
where $\mathrm{A}=\left[\begin{array}{lll}2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right], \mathrm{B}=\left[\begin{array}{l}29 \\ 13 \\ 16\end{array}\right]$
Here
$\Rightarrow \mathrm{D}=\left|\begin{array}{lll}2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1\end{array}\right|$
Expand by $\mathrm{R}_{1}$
$\Rightarrow \mathrm{D}=2(1-4)-3(1-6)+4(2-3)$
$=-6+15-4$
$\Rightarrow \mathrm{D}=5$
Again, Solve $D_{1}$ formed by replacing $1^{\text {st }}$ column by B matrices
$\Rightarrow \mathrm{D}_{1}=\left|\begin{array}{ccc}29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1\end{array}\right|$
$=29(1-4)-3(13-32)+4(26-16)$
$=-87+57+40$
$\Rightarrow \mathrm{D}_{1}=10$
Again, Solve $\mathrm{D}_{2}$ formed by replacing $2^{\text {nd }}$ column by B matrices
$\Rightarrow \mathrm{D}_{2}=\left|\begin{array}{lll}2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1\end{array}\right|$
$=2(13-32)-29(1-6)+4(16-39)=-38+145-92$
$\Rightarrow \mathrm{D}_{2}=15$
And, Solve $\mathrm{D}_{3}$ formed by replacing $3^{\text {rd }}$ column by $B$ matrices
$\Rightarrow \mathrm{D}_{3}=\left|\begin{array}{lll}2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16\end{array}\right|$
$=2(16-26)-3(16-39)+29(2-3)$
$=-20+69-29$
$\Rightarrow \mathrm{D}_{3}=20$
Thus by Cramer's Rule, we have
$\Rightarrow \mathrm{x}=\frac{\mathrm{D}_{1}}{\mathrm{D}}, \mathrm{y}=\frac{\mathrm{D}_{2}}{\mathrm{D}}$ and $\mathrm{z}=\frac{\mathrm{D}_{3}}{\mathrm{D}}$
$\Rightarrow \mathrm{x}=\frac{10}{5}, y=\frac{15}{5}$ and $\mathrm{z}=\frac{20}{5}$
$\Rightarrow \mathrm{x}=2, \mathrm{y}=3$ and $\mathrm{z}=4$
Thus the Number of cars produced by type $C_{1}, C_{2}$ and $C_{3}$ are 2,3 and 4 respectively.

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Question 34 Marks

Read the text carefully and answer the questions:
A sinking fund contains money set aside or saved to pay off a debt or bond. A company that issues debt will need to pay that debt off in the future, and the sinking fund helps to soften the hardship of a large outlay of revenue. A sinking fund allows companies that have floated debt in the form of bonds gradually save money and avoid a large lump-sum payment at maturity.
Example
- Cost of Machine: ₹2,00,000/-
- Effective Life: 7 Years
- Scrap Value: ₹30,000/-
- Sinking Fund Earning Rate: 5%
- The Expected Cost of New Machine: ₹3,00,000/-
(a) What is the money required for a new machine after 7 years?
(b) What is the value of $\mathrm{A}, \mathrm{i}$ and n here?
(c) What formula will you use to get the requisite amount?
or
What amount should the company put into a sinking fund earning 5% per annum to replace the machine after its useful life?

Answer

Read the text carefully and answer the questions:
What Is a Sinking Fund?
A sinking fund contains money set aside or saved to pay off a debt or bond. A company that issues debt will need to pay that debt off in the future, and the sinking fund helps to soften the hardship of a large outlay of revenue. A sinking fund allows companies that have floated debt in the form of bonds gradually save money and avoid a large lump-sum payment at maturity.
Example:
- Cost of Machine: ₹2,00,000/-
- Effective Life: 7 Years
- Scrap Value: ₹30,000/-
- Sinking Fund Earning Rate: 5%
- The Expected Cost of New Machine: ₹ $ 3,00,000 /-$
(i) Cost of new machine ₹$= 300000$
Scrap value of old machine = ₹ $30000$
Hence, the money required for new machine after 7 years
$
\text ₹ 300000 \text { - ₹ } 30000 \text { = }$ ₹ 270000

(ii) $\mathrm{A}$= ₹ 270000, $\mathrm{i}=\frac{5}{100}=0.05, \mathrm{n}=7$
(iii) $\mathrm{A}=R\left[\frac{(1+i)^{n}-1}{i}\right]$
or
Cost of new machine = ₹ $300000$
Scrap value of old machine = ₹ $30000$
Hence, the money required for new machine after 7 years
= ₹ $300000-$ ₹ $30000=$ ₹ $270000$
So, we have $\mathrm{A}=$ ₹ 270000, $\mathrm{i}=\frac{5}{100}=0.05, \mathrm{n}=7$
Using formula, $\mathrm{A}=R\left[\frac{(1+i)^{n}-1}{i}\right]$, we get
$270000=\mathrm{R}\left[\frac{(1.05)^{7}-1}{0.05}\right]$
$\left[\right.$ Let $\mathrm{x}=(1.05)^{7}$
$\Rightarrow \log \mathrm{x}=7 \log 1.05=7 \times 0.0212=0.1484$
$\Rightarrow \mathrm{x}=$ antilog 0.1484
$\Rightarrow \mathrm{x}=1.407$
$\Rightarrow R=\frac{270000 \times 0.05}{(1.05)^{7}-1}$
$\Rightarrow R=\frac{13500}{1.407-1}=\frac{13500}{0.407}$
$\Rightarrow \mathrm{R}=33169.53$
Hence, the company should deposit ₹ 33169.53 at the end of each year for 7 years.

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Question 44 Marks

Read the text carefully and answer the questions:
The sum of the length of hypotenuse and a side of a right-angled triangle is given by AC + BC = 10

(a) Base $\mathrm{BC}=$ ?
(b) If $\mathbf{S}$ be the area of the triangle, then find the value of $\frac{d S}{d c}$ ?
(c) What is the values of c when $\frac{d s}{d c}=0$ ?
or
Find the value of $\frac{d^{2} S}{d c^{2}}$ at $\mathrm{C}=\frac{10 \sqrt{3}}{3}$?

Answer

Read the text carefully and answer the questions:
The sum of the length of hypotenuse and a side of a right-angled triangle is given by AC $+\mathrm{BC}=10$

(i) $\frac{100-c^{2}}{20}$
(ii) $\frac{100-3 c^{2}}{40}$
(iii) $\frac{10 \sqrt{3}}{3}$
or
$
\frac{-\sqrt{3}}{2}
$

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Applied Maths Case study (4 Marks)

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