Quadratic Equations: Word Problems

₹ 7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received ₹ 100 more. Find the original number of children.

Let the original number of children $=x$

Total amount $=$ ₹ 7500

$\therefore$ Amount received by each child $=$ ₹ $\dfrac{7500}{x}$

New number of children $=x-20$

$\therefore$ New amount received by each child $=$ ₹ $\dfrac{7500}{x-20}$

$\therefore$ As per question,

$\dfrac{7500}{x-20} = \dfrac{7500}{x}+100$

i.e. $\dfrac{75}{x-20} = \dfrac{75}{x}+1$

i.e. $75x=75\left(x-20\right)+x\left(x-20\right)$

i.e. $75x=75x-1500+x^2-20x$

i.e. $x^2-20x-1500=0$

i.e. $x^2 -50x+30x-1500=0$

i.e. $x\left(x-50\right)+30\left(x-50\right)=0$

i.e. $\left(x-50\right) \left(x+30\right)=0$

i.e. $x=50$ or $x=-30$

Since the number of children cannot be negative, therefore, $x=-30$ is not an acceptable solution.

$\therefore$ Original number of children $=50$

Quadratic Equations: Word Problems

The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers.

Let the two consecutive natural numbers, which are multiples of 3, be $x$ and $x+3$

Then, as per question

$x \left(x+3\right) = 810$

i.e. $x^2 + 3x - 810 = 0$

i.e. $x^2 -27x + 30x -810 = 0$

i.e. $x \left(x-27\right)+30\left(x-27\right)=0$

i.e. $\left(x-27\right)\left(x+30\right)=0$

$\implies$ $x=27$ or $x=-30$

Since the required numbers are natural numbers, therefore, $x=27$ is the acceptable solution.

$\therefore$ The required numbers are 27, 30

Mensuration

On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.

Radius of each circular design $=r=7 \ cm$

$\therefore$ Diameter of each circular design $= 14 \ cm$

$\therefore$ Length of side of the square handkerchief $=s= 14 \times 3 = 42 \ cm$


$\therefore$ Area of the square handkerchief $=s^2 = 42^2 = 1764 \ cm^2$ $\cdots$ (1)

Area of each circular design $=\pi r^2 = \dfrac{22}{7} \times 7^2 = 154 \ cm^2$

$\therefore$ Area of circular designs $= 9 \times 154 = 1386 \ cm^2$ $\cdots$ (2)

$\therefore$ Required area [from equations (1) and (2)] $=1764-1386=378 \ cm^2$

Mensuration

A solid metallic cone of slant height 13 cm and radius 5 cm is melted and recast into solid spheres each of radius 1 cm. Find the number of spheres recast.

Slant height of cone $= \ell = 13 \ cm$

Radius of cone $=r=5 \ cm$

Let height of cone $= h$

Now, $\ell^2 = h^2 + r^2$

$\implies$ $h = \sqrt{\ell^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 \ cm$

$\therefore$ Volume of cone $= \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \times \pi \times 5^2 \times 12 = 100 \pi \ cm^3$

Radius of sphere $=R=1 \ cm$

$\therefore$ Volume of sphere $= \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 1^3 = \dfrac{4}{3} \pi \ cm^3$

Let number of spheres $=N$

Since the cone is recast into a number of small spheres,

volume of cone $= N \times$ volume of 1 sphere

i.e. $100 \pi = N \times \dfrac{4}{3} \pi$

$\implies$ $N = \dfrac{100 \times 3}{4} = 75$

$\therefore$ Number of spheres recast = 75

Mensuration

The total surface area of a hollow metal cylinder, open at both ends, of external radius 8 cm and height 10 cm, is $338 \pi \ cm^2$. Taking r to be the internal radius, write down an equation in r and use it to determine the thickness of the metal in the cylinder.

For the hollow cylinder,

outer radius $=R=8 \ cm$

inner radius $=r$

height $=h=10 \ cm$

total surface area (TSA) $=338 \pi \ cm^2$

$\begin{aligned} \text{TSA} & = \text{Lateral surface area} + \text{Base area} \\ & = 2 \pi h \left(R+r\right) + 2 \pi \left(R^2 - r^2\right) \end{aligned}$

$\begin{aligned} \text{i.e. } & 2 \times \pi \times 10 \times \left(8+r\right) + 2 \times \pi \left(8^2 - r^2\right) = 338 \pi \\ \text{i.e. } & 20 \left(8+r\right) + 2 \left(8+r\right) \left(8-r\right) = 338 \\ \text{i.e. } & \left(8+r\right) \left(10+8-r\right) = 169 \\ \text{i.e. } & \left(8+r\right) \left(18-r\right) = 169 \\ \text{i.e. } & 144-8r+18r-r^2=169 \\ \text{i.e. } & r^2 - 10r+25 = 0 \\ \text{i.e. } & \left(r-5\right)^2 = 0 \\ \implies & r = 5 \end{aligned}$

$\therefore$ Inner radius $= 5 \ cm$

$\therefore$ Thickness of metal in the cylinder $= R-r=8-5=3 \ cm$

Mensuration

An association offered to contribute 50% of the cost for making 100 tents. The lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m. If the canvas to be used costs ₹ 100 per square meter, find the amount the association pays.

For the conical portion of the tent:

Diameter $= 4.2 \ m$

$\therefore$ Radius $= r = 2.1 \ m$

Height $=h=2.8 \ m$

$\begin{aligned} \therefore \text{Slant height} = \ell & = \sqrt{h^2+r^2} \\\\ & = \sqrt{2.8^2 + 2.1^2} \\\\ & = \sqrt{12.25} = 3.5 \ m \end{aligned}$

$\therefore$ Curved surface area of the conical portion

$= \pi r \ell = \dfrac{22}{7} \times 2.1 \times 3.5 = 23.1 \ m^2$ $\cdots$ (1)

For the cylindrical portion of the tent:

Diameter $= 4.2 \ m$

$\therefore$ Radius $= R = 2.1 \ m$

Height $=H = 4 \ m$

$\therefore$ Curved surface area of the cylindrical portion $=2\pi R H = 2 \times \dfrac{22}{7} \times 2.1 \times 4 = 52.8 \ m^2$ $\cdots$ (2)

For the entire tent:

$\therefore$ From equations (1) and (2),

curved surface area of the entire tent $=23.1+52.8 = 75.9 \ m^2$

Cost of canvas per square meter $=$ ₹ 100

$\therefore$ Cost of canvas for 1 tent $= 75.9 \times 100 =$ ₹ 7,590

$\therefore$ Cost of canvas for 100 tents $=7590 \times 100 = $ ₹ 7,59,000

$\therefore$ Amount paid by the association $= 50\%$ of ₹ 7,59,000 $= \dfrac{759000}{2}=$ ₹ 3,79,500

Mensuration

PQRS is a square lawn with side $PQ=42 \ m$. Two circular flower beds are there on the sides PS and QR with center at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded portion).

From the figure, $PQ=QR=RS=SP=42 \ m$

$\therefore$ Area of the square lawn $= \left(\text{side}\right)^2 = 42^2 = 1764 \ m^2$ $\cdots$ (1)

$\begin{aligned} \text{Length of diagonal of the square lawn}& =SQ=PR & = & \sqrt{PQ^2 + QR^2} \\ & & & \\ & & = & \sqrt{42^2 + 42^2} = 42 \sqrt{2} \ m \end{aligned}$

$\therefore$ Diameter of the circular flower bed $=SQ=PR=42\sqrt{2} \ m$

$\therefore$ Radius of the circular flower bed $=r=21 \sqrt{2} \ m$

$\therefore$ Area of the circular flower bed $= \pi r^2 = \dfrac{22}{7} \times \left(21\sqrt{2}\right)^2 = 2772 \ m^2$ $\cdots$ (2)

$\therefore$ Area of the circular flower bed NOT covered by the square lawn

$= 2772-1764 = 1008 \ m^2$ [from equations (1) and (2)]

$\therefore$ Area of the shaded portion $= \dfrac{1008}{2} = 504 \ m^2$

Mensuration

The cost of fencing a circular field at the rate of ₹ 24 per meter is ₹ 5,280. The field is to be ploughed at the rate of ₹ 0.50 per square meter. Find the cost of ploughing the field.

Let radius of the circular field $=r$

Cost of fencing the circular field at the rate of ₹ 24 per meter = ₹ 5,280

$\implies$ Circumference of the circular field $= \dfrac{5280}{24} = 220 \ m$

i.e. $2 \pi r = 220$

$\implies$ $r = \dfrac{220}{2\pi} = \dfrac{110 \times 7}{22} = 35 \ m$

$\therefore$ Area of the circular field $= \pi r^2 = \dfrac{22}{7} \times 35^2 = 3850 \ m^2$

Cost of ploughing the field per square meter = ₹ 0.50

$\therefore$ Cost of ploughing the entire field $= 3850 \times 0.50 =$ ₹ 1,925

Trigonometric Functions

Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$

$\cot \left(\alpha - \beta\right) = \dfrac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}$

$\implies$ $\cot \beta - \cot \alpha = \dfrac{\cot \alpha \cot \beta + 1}{\cot \left(\alpha - \beta\right)}$

$\begin{aligned} \text{LHS} & = \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x \\ & \\ & = \cot 2x \left(\cot x - \cot 3x\right) - \cot 3x \cot x \\ & \\ & = \cot 2x \left[\dfrac{\cot x \cot 3x + 1}{\cot \left(3x-x\right)}\right] - \cot 3x \cot x \\ & \\ & = \dfrac{\cot 2x \left(\cot x \cot 3x + 1\right)}{\cot 2x} - \cot 3x \cot x \\ & \\ & = \cot 3x \cot x + 1 - \cot 3x \cot x \\ & \\ & = 1 = \text{RHS} \end{aligned}$

Trigonometric Functions

Prove that
$\dfrac{\sin 5x - 2 \sin 3x + \sin x}{\cos 5x - \cos x} = \tan x$

$\sin \alpha + \sin \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha -\beta}{2}\right)$

$\cos \alpha - \cos \beta = -2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha - \beta}{2}\right)$

$\begin{aligned} \text{LHS} & = \dfrac{\sin 5x - 2 \sin 3x + \sin x}{\cos 5x - \cos x} \\ & \\ & = \dfrac{2 \sin \left(\dfrac{5x+x}{2}\right) \cos \left(\dfrac{5x-x}{2}\right)-2\sin 3x}{-2\sin \left(\dfrac{5x+x}{2}\right) \sin \left(\dfrac{5x-x}{2}\right)} \\ & \\ & = \dfrac{2 \sin 3x \cos 2x - 2 \sin 3x}{-2 \sin 3x \cdot \sin 2x} \\ & \\ & = \dfrac{2 \sin 3x \left(\cos 2x - 1\right)}{-2\sin 3x \cdot \sin 2x} \\ & \\ & = \dfrac{\cos 2x - 1}{-\sin 2x} \\ & \\ & = \dfrac{-2 \sin^2 x}{-2\sin x \cos x} \qquad \left[\text{Note: }\cos 2\theta - 1 = -2 \sin^2 \theta \; ; \quad \sin 2x = 2 \sin x \cos x\right] \\ & \\ & = \dfrac{\sin x}{\cos x} \\ & \\ & = \tan x = \text{RHS} \end{aligned}$

Heights and Distances

Two poles of equal heights are standing opposite to each other on either side of the road which is 80m wide. From a point P between them on the road, the angle of elevation of the top of one pole is $60^{\circ}$ and the angle of depression from the top of the other pole at point P is $30^{\circ}$. Find the heights of the poles and the distance of point P from the poles.

AB, CD: Poles of height 'h'

AC: Road 80m wide

P: Point of observation

In the figure, $AP=x$ meter, $PC=80-x$ meter

In triangle PAB,

$\dfrac{BA}{AP}=\tan 60^{\circ} = \sqrt{3}$

i.e. $\dfrac{h}{x} = \sqrt{3}$ $\implies$ $x = \dfrac{h}{\sqrt{3}}$ $\cdots$ (1)

In triangle PCD,

$\dfrac{CD}{PC} = \tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$

i.e. $\dfrac{h}{80-x} = \dfrac{1}{\sqrt{3}}$ $\implies$ $80-x = h\sqrt{3}$ $\cdots$ (2)

Therefore, from equations (1) and (2) we have,

$80-\dfrac{h}{\sqrt{3}} = h\sqrt{3}$

i.e. $80=\left(\dfrac{1}{\sqrt{3}}+\sqrt{3}\right)h$

i.e. $\dfrac{4h}{\sqrt{3}} = 80$ $\implies$ $h = \dfrac{80\sqrt{3}}{4} = 20\sqrt{3} = 34.64$

$\therefore$ Height of poles = 34.64 m

$\therefore$ From equation (1), $x = \dfrac{20\sqrt{3}}{\sqrt{3}} = 20$

$\therefore$ Distance of P from AB $=20$ m

and distance of P from CD $= 80-20=60$ m

Heights and Distances

The angle of elevation of a jet fighter from a point O on the ground is $60^{\circ}$. After a flight of 10 seconds, the angle of elevation changes to $30^{\circ}$. If the jet is flying at a speed of 648 kmph, find the constant height at which the jet is flying.

$J_1$: Initial position of the jet

$J_2$: Position of jet after 10 seconds $\left[\text{i.e. } \dfrac{10}{3600}=\dfrac{1}{360} \text{hr}\right]$

O: Point of observation

Speed of jet = 648 km/hr

$\therefore$ Distance covered by the jet in 10s $= \text{speed}\times \text{time}=648\times\dfrac{1}{360}=1.8$ km

$\therefore$ From the figure, $J_1J_2 = 1.8$ $\cdots$ (1)

Also from the figure, $J_1J_2 = QP$ $\cdots$ (2) and $QO=QP+PO$ $\cdots$ (3)

From triangle $J_1PO$,

$\dfrac{J_1P}{PO}=\tan 60^{\circ} = \sqrt{3}$

$\implies$ $PO=\dfrac{J_1P}{\sqrt{3}} = \dfrac{h}{\sqrt{3}}$ $\cdots$ (4)

From triangle $J_2QO$,

$\dfrac{J_2Q}{QO} = \tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$

$\therefore$ $QO = J_2Q \times \sqrt{3} = h\sqrt{3}$

i.e. $QP + PO = h\sqrt{3}$ $\cdots$ (5) [from equation (3)]

In view of equations (1), (2) and (4), equation (5) becomes

$1.8+\dfrac{h}{\sqrt{3}} = h \sqrt{3}$

i.e. $\left(\sqrt{3}-\dfrac{1}{\sqrt{3}}\right)h= 1.8$

i.e. $\dfrac{2}{\sqrt{3}}h = 1.8$

$\implies$ $h=\dfrac{1.8\times \sqrt{3}}{2}=1.558$

$\therefore$ Height at which the jet is flying = 1.558 km

Heights and Distances

From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20m high building are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower.

OB: Building of height 20m

BT: Transmission tower of height 'h'

C: Point of observation

From the figure, in triangle BOC,

$\dfrac{BO}{OC} = \tan 45^{\circ} = 1$ $\implies$ $BO = OC = 20$

In triangle TOC,

$\dfrac{TO}{OC} = \tan 60^{\circ} = \sqrt{3}$

$\implies$ $TO = OC\sqrt{3} = 20\sqrt{3}$

i.e. $h+20 = 20\sqrt{3}$

i.e. $h=14.64$

Therefore, height of the tower is 14.64m

Heights and Distances

The angle of elevation of the top of a building from the foot of the tower is $30^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $60^{\circ}$. If the tower is 60m high, find the height of the building.

AB: Building of height 'h'

OT: Tower of height 60m

From the figure, in triangle ATO,

$\dfrac{TO}{AO} = \tan 60^{\circ} = \sqrt{3}$

$\therefore$ $AO = \dfrac{TO}{\sqrt{3}} = \dfrac{60}{\sqrt{3}} = 20\sqrt{3}$

In triangle BAO,

$\dfrac{BA}{OA} = \tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$

$\therefore$ $BA = h = \dfrac{OA}{\sqrt{3}} = \dfrac{20\sqrt{3}}{\sqrt{3}} = 20$

$\therefore$ The height of the building is 20m.

Heights and Distances

The angle of elevation of the top of a vertical tower from a point on the ground is $60^{\circ}$. From another point 10m vertically above the first, its angle of elevation is $45^{\circ}$. Find the height of tower.

OT: Tower of height 'h'

B: Point 10m vertically above the first point A

From the figure, $OT=h$, $PT=h-10$, $BA=10$, $OA=PB$

From triangle TOA, $\dfrac{OT}{OA} = \tan 60^{\circ} = \sqrt{3}$

$\implies$ $OA=\dfrac{h}{\sqrt{3}}$ $\cdots$ (1)

From triangle TPB, $\dfrac{PT}{PB} = \tan 45^{\circ} = 1$

$\implies$ $PT=PB=OA = \dfrac{h}{\sqrt{3}}$ $\cdots$ (2) [From equation (1)]

From the figure $PT=h-10$

$\therefore$ From equation (2) we have

$h-10 = \dfrac{h}{\sqrt{3}}$

i.e. $h-\dfrac{h}{\sqrt{3}} = 10$

i.e. $\dfrac{\left(\sqrt{3}-1\right)h}{\sqrt{3}}=10$

i.e. $0.732h = 17.32$

$\implies$ $h = \dfrac{17.32}{0.732} = 23.66$

$\therefore$ Height of the tower is 23.66m

Heights and Distances

A 2m tall boy is standing at some distance from a 29m tall building. The angle of elevation, from his eyes to the top of the building increases from $30^{\circ}$ to $60^{\circ}$ as he walks towards the building. Find the distance he walks towards the building.

AB: Building of height 29m

CD: Initial position of boy of height 2m at a distance 'x' from the building

EF: Final position of boy after walking a distance 'y' towards the building

From the figure,

$AB=29m$; $CD=EF=PA=2m$; $AC=PD=x$

$AE=PF=x-y$; $BP=BA-PA=29-2=27m$

In triangle BPD, $\tan 30^{\circ} = \dfrac{BP}{PD}$

$\implies$ $PD=x=\dfrac{BP}{\tan 30^{\circ}} = \dfrac{27}{1/\sqrt{3}} = 27\sqrt{3}m$ $\cdots$ (1)

In triangle BPF, $\tan 60^{\circ} = \dfrac{BP}{PF}$

$\implies$ $PF = x-y = \dfrac{BP}{\tan 60^{\circ}} = \dfrac{27}{\sqrt{3}}= 9\sqrt{3}m$ $\cdots$ (2)

Substituting the value of x from equation (1) in equation (2) gives

$27\sqrt{3} - y = 9\sqrt{3}$

$\implies$ $y = 27\sqrt{3} - 9\sqrt{3} = 18\sqrt{3} = 31.18m$

Therefore, the boy walks 31.18m towards the building.

Matrices

If $A=$ $\begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$, then verify that $A^t A = I$

$A^t =$ $\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

$\begin{aligned} \therefore A^t A & = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \\ & \\ & = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & -\cos \alpha \sin \alpha + \sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \\ & \\ & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ & \\ & = I \end{aligned}$

Matrices

If $A=$ $\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}$ and $I =$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, find k so that $A^2 = 5A + kI$

$A^2 = A \times A$
$=\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}$ $\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} =$ $\begin{pmatrix} 3 \times 3 + 1 \times \left(-1\right) & 3 \times 1 + 1 \times 2 \\ -1 \times 3 + 2 \times \left(-1\right) & -1 \times 1 + 2 \times 2 \end{pmatrix}$

$=\begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix}$

$5A = 5$ $\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} = $ $\begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix}$

$kI = k$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} =$ $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

$\therefore$ $A^2 = 5A + kI \implies$ $\begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} =$ $\begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} +$ $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

i.e. $\begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} =$ $\begin{pmatrix} 15+k & 5 \\ -5 & 10+k \end{pmatrix}$

$\implies 8 = 15+k \implies k = -7$

Matrices

Given: $\begin{pmatrix} 4 & 2 \\ -1 & 1 \end{pmatrix} M = 6I$, where M is a matrix and I is an unit matrix of order $2 \times 2$.
State the order of matrix M.
Find the matrix M.

Order of matrix $M = 2 \times 2$

Let $M=$ $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $I=$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} =$ unit matrix

$\therefore$ $\begin{pmatrix} 4 & 2 \\ -1 & 1 \end{pmatrix} M = 6 I \implies$ $\begin{pmatrix} 4 & 2 \\ -1 & 1 \end{pmatrix}$ $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 6$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

i.e. $\begin{pmatrix} 4a + 2c & 4b+2d \\ -a+c & -b+d \end{pmatrix} =$ $\begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}$

i.e. $4a + 2c = 6$ $\implies$ $2a + c = 3 \cdots (1)$

$-a+c = 0$ $\implies$ $a=c$ $\cdots$ (2)

$4b+2d=0$ $\implies$ $d = -2b$ $\cdots$ (3)

$-b+d=6$ $\cdots$ (4)

From equations (1) and (2) we have

$2c + c = 3$ $\implies$ $c=1$

Therefore, from equation (2), $a = 1$

From equations (3) and (4) we have

$-b-2b = 6$ $\implies$ $b=-2$

Therefore, from equation (3), $d=4$

$\therefore$ $M =$ $\begin{pmatrix} 1 & -2 \\ 1 & 4 \end{pmatrix}$

Matrices

Find the inverse of the matrix $\begin{bmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{bmatrix}$ using elementary transformations.

Let $A=$ $\begin{bmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{bmatrix}$, $I=$ $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = $ Identity matrix of order 3

Let $A^{-1}$ be the inverse of A.

Then, $A \times A^{-1} = I$

i.e. $\begin{bmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{bmatrix} A^{-1} =$ $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Using row transformations:

$R_1 \rightarrow R_1+R_3 \implies$ $\begin{bmatrix} -1 & 1 & 6 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{bmatrix} A^{-1} =$ $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$R_2 \rightarrow R_2-R_3 \implies$ $\begin{bmatrix} -1 & 1 & 6 \\ -2 & 1 & -2 \\ -3 & 2 & 3 \end{bmatrix} A^{-1}=$ $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}$

$R_1 \rightarrow R_1-R_2 \implies$ $\begin{bmatrix} 1 & 0 & 8 \\ -2 & 1 & -2 \\ -3 & 2 & 3 \end{bmatrix} A^{-1}=$ $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}$ $R_2 \rightarrow R_2+2R_1 \implies$ $\begin{bmatrix} 1 & 0 & 8 \\ 0 & 1 & 14 \\ -3 & 2 & 3 \end{bmatrix} A^{-1} =$ $\begin{bmatrix} 1 & -1 & 2 \\ 2 & -1 & 3 \\ 0 & 0 & 1 \end{bmatrix}$

$R_3 \rightarrow R_3 + 3R_1 \implies$ $\begin{bmatrix} 1 & 0 & 8 \\ 0 & 1 & 14 \\ 0 & 2 & 27 \end{bmatrix} A^{-1} =$ $\begin{bmatrix} 1 & -1 & 2 \\ 2 & -1 & 3 \\ 3 & -3 & 7 \end{bmatrix}$

$R_3 \rightarrow R_3 - 2R_2 \implies$ $\begin{bmatrix} 1 & 0 & 8 \\ 0 & 1 & 14 \\ 0 & 0 & -1 \end{bmatrix} = A^{-1}$ $\begin{bmatrix} 1 & -1 & 2 \\ 2 & -1 & 3 \\ -1 & -1 & 1 \end{bmatrix}$

$R_1 \rightarrow R_1+8R_3 \, , R_2 \rightarrow R_2+14R_3 \implies$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} A^{-1} =$ $\begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ -1 & -1 & 1 \end{bmatrix}$

$R_3 \rightarrow \left(-1\right)R_3 \implies$ $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} =$ $\begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix}$

$\therefore A^{-1} =$ $\begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix}$

Matrices

Simplify: $\sin A$ $\begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A$ $\begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix}$

$\begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} = \sin^2 A + \cos^2 A = 1$

$\begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix} = \cos^2 A + \sin^2 A = 1$

Therefore, the given expression becomes:

$\sin A \times 1 + \cos A \times 1 = \sin A + \cos A$

Matrices

If $A = \begin{bmatrix} 3 & 1 & 2 \\ 3 & 2 & -3 \\ 2 & 0 & -1 \end{bmatrix}$, find $A^{-1}$.
Hence solve the system of equations
$3x+3y+2z=1$, $x+2y=4$, $2x-3y-z=5$

$\left|A\right| =$ $ \begin{vmatrix} 3 & 1 & 2 \\ 3 & 2 & -3 \\ 2 & 0 & -1 \end{vmatrix} = 3 \times \left(-2\right) -1 \times \left(-3+6\right) + 2 \times \left(-4\right) = -17 \neq 0 $ Therefore, $A^{-1}$ exists.

Cofactors of A are:
$A_{11} = \left(-1\right)^{1+1}$ $\begin{vmatrix} 2 & -3 \\ 0 & -1 \end{vmatrix}=-2$

$A_{12} = \left(-1\right)^{1+2}$ $\begin{vmatrix} 3 & -3 \\ 2 & -1 \end{vmatrix}=- \left(-3+6\right)=-3$

$A_{13} = \left(-1\right)^{1+3}$ $\begin{vmatrix} 3 & 2 \\ 2 & 0 \end{vmatrix}=-4$

$A_{21} = \left(-1\right)^{2+1}$ $\begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix}=-\left(-1\right)=1$

$A_{22} = \left(-1\right)^{2+2}$ $\begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix}=-3-4=-7$

$A_{23} = \left(-1\right)^{2+3}$ $\begin{vmatrix} 3 & 1 \\ 2 & 0 \end{vmatrix}=- \left(-2\right)=2$

$A_{31} = \left(-1\right)^{3+1}$ $\begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix}=-3-4=-7$

$A_{32} = \left(-1\right)^{3+2}$ $\begin{vmatrix} 3 & 2 \\ 3 & -3 \end{vmatrix}=-\left(-9-6\right)=15$

$A_{33} = \left(-1\right)^{3+3}$ $\begin{vmatrix} 3 & 1 \\ 3 & 2 \end{vmatrix}=6-3=3$

$\text{adj A}=$ $\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}^t = $ $\begin{bmatrix} -2 & -3 & -4 \\ 1 & -7 & 2 \\ -7 & 15 & 3 \end{bmatrix}^t=$ $ \begin{bmatrix} -2 & 1 & -7 \\ -3 & -7 & 15 \\ -4 & 2 & 3 \end{bmatrix} $

$A^{-1} = \dfrac{\text{adj A}}{\left|A\right|} = \dfrac{-1}{17}$ $ \begin{bmatrix} -2 & 1 & -7 \\ -3 & -7 & 15 \\ -4 & 2 & 3 \end{bmatrix} $

Let $A_1 =$ $\begin{bmatrix} 3 & 3 & 2 \\ 1 & 2 & 0 \\ 2 & -3 & -1 \end{bmatrix}$, $B=$ $\begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix}$ and $X=$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$

Then $A_1 X = B \implies X = A_1^{-1}B$

But $A_1 = A^t$ (Comparing A and $A_1$)

Therefore, $X = \left(A^t\right)^{-1}B = \left(A^{-1}\right)^t B$ $\left[\text{Note: } \left(A^t\right)^{-1} = \left(A^{-1}\right)^t\right]$

$X= \dfrac{-1}{17}$ $\begin{bmatrix} -2 & 1 & -7 \\ -3 & -7 & 15 \\ -4 & 2 & 3 \end{bmatrix}^t$ $\begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix} = \dfrac{-1}{17}$ $\begin{bmatrix} -2 & -3 & -4 \\ 1 & -7 & 2 \\ -7 & 15 & 3 \end{bmatrix}$ $\begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix}$

$= \dfrac{-1}{17}$ $\begin{bmatrix} -2-12-20 \\ 1-28+10 \\ -7+60+15 \end{bmatrix}$

$= \dfrac{-1}{17}$ $\begin{bmatrix} -34 \\ -17 \\ 68 \end{bmatrix}$

i.e. $\begin{bmatrix} x \\ y \\ z \end{bmatrix} = $ $\begin{bmatrix} 2 \\ 1 \\ -4 \end{bmatrix}$

$\implies x=2 \,, y =1 \,, z=-4$

Determinants

By using properties of determinants prove that the determinant $ \begin{vmatrix} a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a \end{vmatrix} $ is independent of x.

Let $\Delta = $ $ \begin{vmatrix} a & \sin x & \cos x \\ -\sin x & -a & 1 \\ \cos x & 1 & a \end{vmatrix} $

$C_1 \rightarrow C_1 + C_2 \implies \Delta =$ $ \begin{vmatrix} a+\sin x & \sin x & \cos x \\ -a-\sin x & -a & 1 \\ 1+\cos x & 1 & a \end{vmatrix} $

$R_1 \rightarrow R_1 + R_2, R_2 \rightarrow R_2+aR_3 \implies$

$\Delta =$ $ \begin{vmatrix} 0 & -a+\sin x & 1+\cos x \\ a\cos x-\sin x & 0 & 1+a^2 \\ 1+\cos x & 1 & a \end{vmatrix} $

Expanding along $R_1$ gives $\begin{aligned} \Delta = & -\left(-a+\sin x\right) \left[a\left(a\cos x -\sin x\right)-\left(1+a^2\right)\left(1+\cos x\right)\right] + \\ & \left(1+\cos x\right)\left(a\cos x - \sin x\right) \\ \text{i.e. } \Delta = & -\left(-a+\sin x\right)\left(a^2\cos x -a\sin x -1-\cos x -a^2 -a^2 \cos x\right) + \\ & a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & \left(-a+\sin x\right) \left(a\sin x + \cos x +1 +a^2\right) \\ & a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & -a^2 \sin x - a\cos x -a - a^3 +a \sin^2 x +\sin x \cos x +\sin x + \\ & a^2 \sin x + a\cos x - \sin x + a \cos^2 x - \sin x \cos x \\ \text{i.e. } \Delta = & -a^3 \text{ which is independent of x.} \end{aligned}$

Determinants

If $A+B+C=\pi$, then find the value of $\begin{vmatrix} \sin \left(A+B+C\right) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos \left(A+B\right) & -\tan A & 0 \end{vmatrix}$

Since $A+B+C = \pi$, $\sin \left(A+B+C\right) = \sin \pi = 0$ and $\cos \left(A+B\right) = \cos \left(\pi-C\right) = -\cos C$

$ \begin{aligned} \text{Let } \Delta & = \begin{vmatrix} \sin \left(A+B+C\right) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos \left(A+B\right) & -\tan A & 0 \end{vmatrix} \\ & \\ & = \begin{vmatrix} 0 & \sin B & \cos C \\ -\sin B & 0 & \tan A \\ -\cos C & -\tan A & 0 \end{vmatrix} \\ & \\ & = -\sin B \times \left[0- \tan A \times \left(-\cos C\right)\right] + \cos C \sin B \tan A \\ & \\ & = -\tan A \sin B \cos C + \tan A \sin B \cos C = 0 \end{aligned} $

Determinants

Using properties of determinants, show that $p \alpha^2 + 2q \alpha + r = 0$ given that p, q and r are not in G.P and $ \begin{vmatrix} 1 & \dfrac{q}{p} & \alpha + \dfrac{q}{p} \\ 1 & \dfrac{r}{q} & \alpha + \dfrac{r}{q} \\ p \alpha + q & q \alpha + r & 0 \end{vmatrix} = 0 $

$R_1 \rightarrow R_1-R_2 \implies$ $ \begin{vmatrix} 0 & \dfrac{q}{p}-\dfrac{r}{q} & \dfrac{q}{p}-\dfrac{r}{q} \\ 1 & \dfrac{r}{q} & \alpha+\dfrac{r}{q} \\ p\alpha + q & q \alpha + r & 0 \end{vmatrix} = 0 $

i.e. $\left(\dfrac{q}{p}-\dfrac{r}{q}\right)$ $ \begin{vmatrix} 0 & 1 & 1 \\ 1 & \dfrac{r}{q} & \alpha+\dfrac{r}{q} \\ p\alpha+q & q\alpha+r & 0 \end{vmatrix} = 0 $

$C_2 \rightarrow C_2 - C_3 \implies \left(\dfrac{q^2-pr}{pq}\right)$ $ \begin{vmatrix} 0 & 0 & 1 \\ 1 & -\alpha & \alpha+\dfrac{r}{q} \\ p\alpha+q & q\alpha+r & 0 \end{vmatrix} = 0 $

Expanding along $R_1$ gives

$\dfrac{\left(q^2-pr\right) \left[q\alpha+r +\alpha \left(p\alpha+q\right)\right]}{pq}= 0$

i.e. $\left(q^2-pr\right)\left(p\alpha^2 + 2q\alpha + r\right) = 0$ $\cdots$ (1)

Since p, q and r are not in G.P $\implies$ $q^2 \neq pr$

i.e. $q^2 - pr \neq 0$ $\cdots$ (2)

Therefore, from equations (1) and (2) we have

$p\alpha^2 + 2q \alpha + r = 0$

Determinants

If $\Delta = \begin{vmatrix} 1 & a & a^2 \\ a & a^2 & 1 \\ a^2 & 1 & a \end{vmatrix} = -4$, then find the value of $\begin{vmatrix} a^3 - 1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix}$

$\left[\text{Note: If } C_{ij} \text{ are the cofactors of } a_{ij} \text{ in } \left|A\right| = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \\ \text{ then } \begin{vmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_23 \\ C_{31} & C_{32} & C_{33} \end{vmatrix} = \left|A\right|^2\right] $
Cofactors of determinant $\Delta$ are
$ C_{11} = \left(-1\right)^{1+1} \begin{vmatrix} a^2 & 1 \\ 1 & a \end{vmatrix} = a^3 - 1 $

$ C_{12} = \left(-1\right)^{1+2} \begin{vmatrix} a & 1 \\ a^2 & a \end{vmatrix} = a^2-a^2=0$

$C_{13} = \left(-1\right)^{1+3} \begin{vmatrix} a & a^2 \\ a^2 & 1 \end{vmatrix} = a - a^4$

$C_{21} = \left(-1\right)^{2+1} \begin{vmatrix} a & a^2 \\ 1 & a \end{vmatrix} = a^2 - a^2 = 0$

$C_{22} = \left(-1\right)^{2+2} \begin{vmatrix} 1 & a^2 \\ a^2 & a \end{vmatrix} = a - a^4$

$C_{23} = \left(-1\right)^{2+3} \begin{vmatrix} 1 & a \\ a^2 & 1 \end{vmatrix} = -\left(1-a^3\right) = a^3 - 1$

$C_{31} = \left(-1\right)^{3+1} \begin{vmatrix} a & a^2 \\ a^2 & 1 \end{vmatrix} = a - a^4$

$C_{32} = \left(-1\right)^{3+2} \begin{vmatrix} 1 & a^2 \\ a & 1 \end{vmatrix} = -\left(1-a^3\right) = a^3 - 1$

$C_{33} = \left(-1\right)^{3+3} \begin{vmatrix} 1 & a \\ a & a^2 \end{vmatrix} = a^2 - a^2 = 0$

Now, $\begin{vmatrix} a^3 - 1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix} = \begin{vmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{vmatrix}$

$\therefore$ $\begin{vmatrix} a^3 - 1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix} = \left|\Delta\right|^2 = \left(-4\right)^2 = 16$

Determinants

Using properties of determinants prove that $\begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} = a^3$

Let $\Delta =$ $ \begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} $ $R_2 \rightarrow R_2 - 2R_1, R_3 \rightarrow R_3 - 3R_1 \implies \Delta =$ $ \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 3a & 7a+3b \end{vmatrix} $ $R_3 \rightarrow R_3 - 3R_2 \implies \Delta =$ $ \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 0 & a \end{vmatrix} $
Expanding along $R_1$ gives
$\Delta = a \times a \times a = a^3$