Case study (4 Marks)

Question 14 Marks

Show that the matrix, $A=\left[\begin{array}{rrr}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]$ satisfies the equation, $A^3-A^2-3 A-I_3=O$. Hence, find $A^{-1}$.

Answer

Here, we have:
$A=\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]$
$\begin{array}{l}A^3=A^2 A \\ A^2=\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]\end{array}$
$\begin{array}{l}=\left[\begin{array}{ccc}1+0-6 & 0+0-8 & -2+0-2 \\ -2+2+6 & 0+1+8 & 4-2+2 \\ 3-8+3 & 0-4+4 & -6+8+1\end{array}\right]=\left[\begin{array}{ccc}-5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3\end{array}\right] \\ A ^2 A=\left[\begin{array}{ccc}-5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right] \\ =\left[\begin{array}{ccc}-5^2+16-12 & 0-8+16 & 10-16-4 \\ 6-18+12 & 0-9+16 & -12+18+4 \\ -2-0+9 & 0-0-12 & 4+0+3\end{array}\right] \\ =\left[\begin{array}{ccc}-1 & -8 & -10 \\ 0 & 7 & 10 \\ 7 & 12 & 7\end{array}\right]\end{array}$
Now, $A ^3- A ^2-3 A- I$
$\begin{array}{l}=\left[\begin{array}{ccc}-1 & -8 & -10 \\ 0 & 7 & 10 \\ 7 & 12 & 7\end{array}\right]-\left[\begin{array}{ccc}-5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3\end{array}\right]-3\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]-\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \\ =\left[\begin{array}{ccc}-1+5 & -8+8 & -10+4 \\ 0-6 & 7-9 & 10-4 \\ 7+2 & 12-0 \\ 7-3\end{array}\right]+\left[\begin{array}{ccc}-3-1 & -0-0 & 6-0 \\ 6-0 & +3-1 & -6-0 \\ -9-0 & -12+0 & -3-1\end{array}\right] \\ =\left[\begin{array}{ccc}4 & 0 & -6 \\ -6 & -2 & 6 \\ 9 & 12 & 4\end{array}\right]+\left[\begin{array}{ccc}-4 & 0 & 6 \\ 6 & 2 & -6 \\ -9 & -12 & -4\end{array}\right] \\ =\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\end{array}$
Thus, $A^3-A^2-3 A-I=0$
Multiply both sides by $A^{-1}$, we get
$\begin{array}{l}A^{-1} A^3-A^1 A^2-3 A^{-1} A-I A^{-1}=0 \\ A^2-A-3 I=A^{-1} \ldots\left(\text { since } A^{-1} A=I\right) \\ \Rightarrow A^{-1}=\left(A^2-A-3 I\right)\end{array}$
$\begin{array}{l}=\left[\begin{array}{rrr}-5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3\end{array}\right]-\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]-3\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \\ =\left[\begin{array}{ccc}-5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3\end{array}\right]-\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]-\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right] \\ =\left[\begin{array}{ccc}-5-1-3 & -8-0-0 & -4+2-0 \\ 6+2-0 & 7+1-3 & 4-2-0 \\ -2-3-0 & 0-4-0 & 3-1-3\end{array}\right] \\ =\left[\begin{array}{ccc}-9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1\end{array}\right] \\ \text { Hence, } A ^{-1}=\left[\begin{array}{ccc}-9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1\end{array}\right]\end{array}$

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Question 24 Marks

An amount of ₹ 5000 is put into three investments at the rate of interest of 6%, 7% and 8% per annum respectively. The total annual income is ₹ 358. If the combined income from the first two investments is ₹ 70 more than the income from the third, find the amount of each investment by matrix method.

Answer

Let x, y and z ₹ be the investments at the rates of interest of 6%, 7% and 8% per annum respectively. Then,
Total investment = ₹ 5000
$\Rightarrow$ x + y + z = 5000
Now, Income from first investment of ₹ x $= ₹ \frac{6 x}{100}$
Income from second investment of $ ₹ y = ₹ \frac{7 y}{100}$
Income from third investment of $ ₹ z = ₹ \frac{8 z}{100}$
$\therefore$ Total annual income $= ₹\left(\frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}\right)$
$\Rightarrow \frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}=358[\because$ Total annual income $= ₹ 358]$
It is given that the combined income from the first two investments is ₹ 70 more than the income from the third
$\therefore \frac{6 x}{100}+\frac{7 y}{100}=70+\frac{8 z}{100} \Rightarrow 6 x+7 y-8 z=7000$
Thus, we obtain the following system of simultaneous linear equations:
x + y + z = 5000
6x + 7y + 8z = 35800
6x + 7y - 8z = 7000
This system of equations can be written in matrix form as follows:
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}5000 \\ 35800 \\ 7000\end{array}\right]$
or, $AX = B$, where $A =\left[\begin{array}{rrr}1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{r}5000 \\ 35800 \\ 7000\end{array}\right]$
Now, $| A |=\left|\begin{array}{rrr}1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8\end{array}\right|=1(-56-56)-(-48-48)+(42-42)=-16 \neq 0$
So, $A ^{-1}$exists and the solution of the given system of equations is given by X = $A ^{-1} B$
Let $C _{ ij }$ be the cofactor of $a _{ ij }$ in $A =\left[ a _{ ij }\right]$. Then,
$\begin{array}{l}C_{11}=-112, C_{12}=96, C_{13}=0, C_{21}=15, C_{22}=-14, \\ C_{23}=-1, C_{31}=1, C_{32}=-2 \text { and } C_{33}=1\end{array}$
$\therefore$ adj $A=\left[\begin{array}{rrr}-112 & 96 & 0 \\ 15 & -14 & -1 \\ 1 & -2 & 1\end{array}\right]=\left[\begin{array}{rrr}-112 & 15 & 1 \\ 96 & -14 & -2 \\ 0 & -1 & 1\end{array}\right]$
So, $A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=-\frac{1}{16}\left[\begin{array}{rrr}-112 & 15 & 1 \\ 96 & -14 & -2 \\ 0 & -1 & 1\end{array}\right]$
Hence, the solution is given by
$X = A ^{-1} B=-\frac{1}{16}\left[\begin{array}{rrr}-112 & 15 & 1 \\ 96 & -14 & -2 \\ 0 & -1 & 1\end{array}\right]\left[\begin{array}{r}5000 \\ 35800 \\ 7000\end{array}\right]=-\frac{1}{16}\left[\begin{array}{rrr}-560000 & +537000 & +7000 \\ 480000 & -501200 & -14000 \\ 0 & -35800 & +7000\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1000 \\ 2200 \\ 1800\end{array}\right]$
$\Rightarrow$x = 1000, y = 2200 and z = 1800
Hence, three investments are of ₹ 1000, ₹ 2200 and ₹ 1800 respectively.

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Question 34 Marks

How much was paid by Mr. Aggarwal to repay the entire amount of home loan?
$\left[\right.$ Use $\left.(1.00625)^{240}=4.4608 ;(1.00625)^{91}=1.7629\right]$

Answer

Total amount paid = n * EMI
= 240 * 2416.81
= ₹ 5800276.8

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Question 44 Marks

In the year 2010, Mr. Aggarwal took a home loan of ₹ 30,00,000 from State Bank of India at 7.5% p.a. compounded monthly for 20 years.
Based on the above information, answer the following questions:
(a) Determine the EMI.
(b) Find the principal paid by Mr. Aggarwal in the $150^{\text {th }}$ instalment.
(c) Find the total interest paid by Mr. Aggarwal.

Answer

In the year 2010, Mr. Aggarwal took a home loan of ₹ 30,00,000 from State Bank of India at 7.5% p.a. compounded monthly for 20 years.
Based on the above information, answer the following questions:
(i) Given $P = ₹ 30,00,000, i =\frac{7.5}{1200}=0.00625$ and $n =12 \times 20=240$ months
$\begin{array}{l}\text { EMI }==\frac{P i}{1-(1+i)^{-n}} \\ =\frac{30,00,000 \times 0.00625}{1-(1.00625)^{-240}-1} \\ =\frac{30,00,000 \times 0.00625 \times 4.4608}{3.4608} \\ ₹ 24167.82\end{array}$
(ii) Given $P = ₹ 30,00,000, i =\frac{7.5}{1200}=0.00625$ and $n =12 \times 20=240$ months
Interest paid on $150^{\text {th }}$ instalment
$\begin{array}{l}=\frac{E M I \times\left[(1+i)^{240-150+1}-1\right]}{(1+i)^{240-150+1}} \\ =\frac{24167 \times[1.7629-1]}{1.7629} \\ = ₹ 10458.70 \\ \Rightarrow \text { Principal paid in } 150^{\text {th }} \text { instalment }=\text { EMI }- \text { interest } \\ = ₹ (24167.82-10458.70)\end{array}$
(iii) Total Interest paid = n x EMI - P
= ₹ (240 x 24167.82 - 30,00,000)
= ₹ 28,00,276.80

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Question 54 Marks

The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semicircular ends. Find the value of x for which the whole area is maximum.

Answer

Let Z be the area of the whole floor, then
$Z = xy +2 \cdot \frac{1}{2} \pi\left(\frac{y}{2}\right)^2= xy +\frac{\pi}{4} \cdot y^2$
$\begin{array}{l}=x \cdot \frac{200-2 x}{\pi}+\frac{\pi}{4} \cdot\left(\frac{200-2 x}{\pi}\right)^2 \\ =\frac{2}{\pi}\left(100 x - x ^2\right)+\frac{(100-x)^2}{\pi}\end{array}$
$\begin{array}{l}\Rightarrow \frac{d Z}{d x}=\frac{2}{\pi}(100-2 x )+\frac{2}{\pi}(100- x )(-1)=\frac{2 x}{\pi} \text { and } \frac{d^2 Z}{d x^2}=-\frac{2}{\pi} \\ \frac{d Z}{d x}=0 \Rightarrow-\frac{2 x}{\pi}=0 \Rightarrow x =0\end{array}$
When $x =0, \frac{d^2 Z}{d x^2}=-\frac{2}{\pi}<0 \Rightarrow Z$ is maximum when $x =0$.

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Question 64 Marks

An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as shown below:

(a)If x and y represents the length and breadth of the rectangular region, then find the relation between the variables.
(b) Find the area of the rectangular region A expressed as a function of x.
(c) Find the maximum value of area A.

Answer

An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as shown below:

(i) According to given information,
perimeter of floor of the building = $2 x +2 .\left(\pi \cdot \frac{y}{2}\right)=200$
$\Rightarrow 2 x +\pi y=200$
(ii) Area of rectangular region of the floor = A = xy
$= x \left(\frac{200-2 x}{\pi}\right)=\frac{2}{\pi}\left(100 x - x ^2\right)$
(iii) $_{ A }=\frac{2}{\pi}\left(100 x - x ^2\right)\left(\right.$ from $\left( x \left(\frac{200-2 x}{\pi}\right)=\frac{2}{\pi}\left(100 x - x ^2\right)\right)$
$\begin{array}{l}\Rightarrow \frac{d A}{d x}=\frac{2}{\pi}(100 x -2 x ) \text { and } \frac{d^2 A}{d x^2}=\frac{2}{\pi}(0-2)=-\frac{4}{\pi} . \\ \text { Now, } \frac{d A}{d x}=0 \Rightarrow \frac{2}{\pi}(100-2 x )=0 \Rightarrow x =50 . \\ \text { When } x =50, \frac{d^2 A}{d x^2}=\frac{4}{\pi}<0\end{array}$
$\Rightarrow$ A is maximum when x = 50.
Maximum value of A $=\frac{2}{\pi}\left(100 \times 50-50^2\right)=\frac{5000}{\pi} m^2$.

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Applied Maths Case study (4 Marks)

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