12th Standard Maths English Medium Reduced Syllabus Model Question paper - 2021 Part - 1

Download Tamil Nadu 12th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Multiple Choice Question

20 x 1 = 20

The sum of the mean and variance of a binomial distribution for 6 total is 2.16. Then the probability of success p =__________

(a)

0.4

(b)

0.6

(c)

0.8

(d)

0.2

If a random variable X has the p.d.f.\(f(x)=\frac { k }{ { x }^{ 2 }+1 }\) ,0 then k is _____________

(a)

\(\pi \)

(b)

\(\frac { 1 }{ \pi } \)

(c)

(d)

\(\frac { 2 }{ \pi } \)

The I.F of y log y \(\frac{dx}{dy}+x-log\ y=0\) is __________

(a)

log(log y)

(b)

log y

(c)

\(\frac{1}{log\ y}\)

(d)

\(\frac{1}{log(log\ y)}\)

Which of the following is a statement?

(a)

7+2<10

(b)

Wish you all success

(c)

All the best

(d)

How old are you?

Define * on Z by a * b = a + b + 1 ∀ a,b \(\in \) Z. Then the identity element of z is ________

(a)

(b)

(c)

(d)

-1

The identity element of \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \right\} \) |x \(\in \) R, x ≠ 0} under matrix multiplication is __________

(a)

\(\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \)

(b)

\(\left( \begin{matrix} \frac { 1 }{ 4x } & \frac { 1 }{ 4x } \\ \frac { 1 }{ 4x } & \frac { 1 }{ 4x } \end{matrix} \right) \)

(c)

\(\left( \begin{matrix} \frac { 1 }{ 2 } & \frac { 1 }{ 2 } \\ \frac { 1 }{ 2 } & \frac { 1 }{ 2 } \end{matrix} \right) \)

(d)

\(\left( \begin{matrix} \frac { 1 }{ 2x } & \frac { 1 }{ 2x } \\ \frac { 1 }{ 2x } & \frac { 1 }{ 2x } \end{matrix} \right) \)

The value of \(\int _{ -\pi }^{ \pi }{ { sin }^{ 3 }x \ { cos }^{ 3 }x \ } dx\) is __________

(a)

(b)

\(\pi \)

(c)

2\(\pi \)

(d)

4\(\pi \)

If the radius of the sphere is measured as 9 cm with an error of 0.03 cm, the approximate error in calculating its volume is _____________

(a)

9.72 cm³

(b)

0.972 cm³

(c)

0.972π cm³

(d)

9.72π cm³

The equation of the tangent to the curve y = x²-4x+2 at (4, 2) is __________

(a)

x + 4y + 12 = 0

(b)

4x + y + 12 = 0

(c)

4x - y - 14 = 0

(d)

x + 4y - 12 = 0

10.

The tangent at any point P on the ellipse \(\frac { { x }^{ 2 } }{ 6 } +\frac { { y }^{ 2 } }{ 3 } \) = 1 whose centre C meets the major axis at T and PN is the perpendicular to the major axis; The CN CT = ______________

(a)

\(\sqrt6\)

(b)

(c)

\(\sqrt3\)

(d)

11.

The length of major and minor axes of 4x² + 3y² = 12 are ____________

(a)

4, 2\(\sqrt3\)

(b)

2, \(\sqrt3\)

(c)

2\(\sqrt3\), 4

(d)

\(\sqrt3\), 2

12.

If the vectors \(a\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \), \(\overset { \wedge }{ i } +b\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and \(\overset { \wedge }{ i } +\overset { \wedge }{ j } +c\overset { \wedge }{ k } \) (a ≠ b ≠ c ≠ 1) are coplaner, then \(\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } =\) _____________

(a)

(b)

(c)

(d)

\(\frac { abc }{ (1-a)(1-b)(1-c) } \)

13.

If \(\lambda \overset { \wedge }{ i } +2\lambda \overset { \wedge }{ j } +2\lambda \overset { \wedge }{ k } \) is a unit vector, then the value of λ is _____________

(a)

土 \(\frac { 1 }{ 3 } \)

(b)

土 \(\frac { 1 }{ 4 } \)

(c)

土 \(\frac { 1 }{ 9 } \)

(d)

\(\frac { 1 }{ 2 } \)

14.

If the foci of the ellipse \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } \) = 1 and the hyperbola \(\frac { { x }^{ 2 } }{ 144 } -\frac { { y }^{ 2 } }{ 81 } =\frac { 1 }{ 25 } \) coincide then b² is __________

(a)

(b)

(c)

(d)

15.

The distance between the foci of a hyperbola is 16 and e = \(\sqrt { 2 } \). Its equation is ____________

(a)

x² - y² = 32

(b)

y²- x² = 32

(c)

x² - y² = 16

(d)

y² - x²= 16

16.

The value of tan \(\left( { cos }^{ -1 }\frac { 3 }{ 5 } +{ tan }^{ -1 }\frac { 1 }{ 4 } \right) \) is ______

(a)

\(\frac { 19 }{ 8 } \)

(b)

\(\frac { 8 }{ 19 } \)

(c)

\(\frac { 19 }{ 12 } \)

(d)

\(\frac { 3 }{ 4 } \)

17.

Equation of tangent at (-4, -4) on x² = -4y is _____________

(a)

2x - y + 4 = 0

(b)

2x + y - 4 = 0

(c)

2x - y - 12 = 0

(d)

2x + y + 4 = 0

18.

If \(4{ cos }^{ -1 }x+{ sin }^{ -1 }x=\pi \) then x is _____________

(a)

\(\frac { 3 }{ 2 } \)

(b)

\(\frac { 1 }{ \sqrt { 2 } } \)

(c)

\(\frac { \sqrt { 3 } }{ 2 } \)

(d)

\(\frac { 2 }{ \sqrt { 3 } } \)

19.

If A^TA⁻¹ is symmetric, then A² =

(a)

A^-1

(b)

(A^T)²

(c)

A^T

(d)

(A^-1)²

20.

If P = \(\left[ \begin{matrix} 1 & x & 0 \\ 1 & 3 & 0 \\ 2 & 4 & -2 \end{matrix} \right] \) is the adjoint of 3 × 3 matrix A and |A| = 4, then x is

(a)

(b)

(c)

(d)

Generic 2

6 x 2 = 12

21.

If w=log(x²+y²),x=cosθ,y=sinθ, find \(\frac { dw }{ d\theta } \)

22.

Write the converse, inverse, and contrapositive of each of the following implication.
If x and y are numbers such that x = y, then x² = y²

23.

Find df for f(x) = x²+ 3x and evaluate it for
x = 3 and dx = 0.02

24.

Evaluate \(\underset{x\rightarrow 1}{lim}(\frac{x^{2}-3x+2}{x^{2}-4x+3})\).

25.

For each of the following differential equations, determine its order, degree (if exists)
\({ \left( \frac { d^2y }{ dx^2 } \right) }^{ 3 }=\sqrt { 1+\left( \frac { dy }{ dx } \right) } \)

26.

If \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ b } =-\overset { \wedge }{ i } +2\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =3\overset { \wedge }{ i } +\overset { \wedge }{ j } \) find \(\frac { \lambda }{ c } \) such that \(\overset { \rightarrow }{ a } +\lambda \overset { \rightarrow }{ b } \) is perpendicular to \(\overset { \rightarrow }{ c } \)

Generic 5

6 x 5 = 30

27.

Find the local maximum and local minimum values for f(x)=12x²-2x²-x⁴.

28.

Find the intervals for which the function f(x)=2x²-9x²-12x+1 is increasing or decfreasing and find the local extermems.

29.

Gas is escaping from a spherical balloon at the rate of 900 cm³/sec. How fast is the surface area and radius of the balloon shrinking when the radius of the balloon is 30 cm?

30.

Prove that p➝(¬q V r) ≡ ¬pV(¬qVr) using truth table.

31.

The girder of a railway bridge is a parabola with its vertex at the highest point 15 m above the ends. If the span is 120 m, find the height of the bridge at 24 m from the middle point.

32.

Solve: (2x² - 3x + 1) (2x² + 5x + 1) = 9x².

Generic 3

7 x 3 = 21

33.

Solve : ydx+(x-y²)dy=0

34.

A ball is thrown vertically upwards, moves according to the law s = 13.8 t - 4.9 t² where s
is in metres and t is in seconds.
(i) Find the acceleration at t = 1
(ii) Find velocity at t = 1
(iii) Find the maximum height reached by the ball?

35.

Let X be a random variable denoting the life time of an electrical equipment having probability density function
\(f(x)=\begin{cases} \begin{matrix} { ke }^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & forx\le 0 \end{matrix} \end{cases}\)
Find
(i) the value of k
(ii) Distribution function
(iii) P(X < 2)
(iv) calculate the probability that X is at least for four unit of time
(v) P(X = 3)

36.

Evaluate: \(\int _{ 0 }^{ 3 }{ (3{ x }^{ 2 }-4x+5) } dx\)

37.

Prove that \({ tan }^{ -1 }\left( \frac { m }{ n } \right) -{ tan }^{ -1 }\left( \frac { m-n }{ m+n } \right) =\frac { \pi }{ 4 } \)

38.

Show that the lines \(\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-1 }{ 5 } \) and \(\frac { x+2 }{ 4 } =\frac { y-1 }{ 3 } =\frac { z+1 }{ -2 } \) do not intersect

39.

Find the rank of the matrix math \(\left[ \begin{matrix} 4 \\ -2 \\ 1 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 0 \\ -1 \\ 8 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \).

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Answers Key:

Multiple Choice Question

20 x 1 = 20

(d)
0.2

(b)
\(\frac { 1 }{ \pi } \)

(b)
log y

(a)
7+2<10

(d)
-1

(c)
\(\left( \begin{matrix} \frac { 1 }{ 2 } & \frac { 1 }{ 2 } \\ \frac { 1 }{ 2 } & \frac { 1 }{ 2 } \end{matrix} \right) \)

(a)
0

(a)
9.72 cm³

10.

(d)
6

11.

12.

(b)
1

13.

(a)
土 \(\frac { 1 }{ 3 } \)

14.

15.

(a)
x² - y² = 32

16.

(b)
\(\frac { 8 }{ 19 } \)

17.

(a)
2x - y + 4 = 0

18.

19.

(b)
(A^T)²

20.

(d)
11

Generic 2

6 x 2 = 12

21.

\(\frac { { \vartheta }^{ 2 }u }{ \vartheta x\vartheta y } \)

22.

If x and y are numbers such that x = y, then x²= y²
Converse statement :
If x and y are numbers such that x²= y² then x = y
Inverse statement :
If x and y are numbers such that x ≠ y then x²≠ y²
Contrapositive statement :
If x and y are numbers such that x²≠ y² then x ≠ y

23.

x = 3 and dx = 0.02
When x = 3 and dx = 0.02,
df = (6 + 3) (0.02)
= 9(0.02) = 0.18

24.

If we put directly x = 1 we observe that the given function is in an indeterminate form \(\frac{0}{0}\). As the numerator and the denominator functions are polynomials of degree 2 they both are differentiable.
Hence, by an application of the l’Hôpital Rule, we get
\(\underset{x\rightarrow 1}{lim}(\frac{x^{2}-3x+2}{x^{2}-4x+3})=\underset{x\rightarrow 1}{lime}(\frac{2x-3}{2x-4})\)
= \(\frac{1}{2}\)
Note that this limit may also be evaluated through the factorization of the numerator and denominator as \(\frac{x^{2}-3x+2}{x^{2}-4x+3}=\frac{(x-1)(x-2)}{(x-1)(x-3)}\)

25.

The given differential equation is
\({ \left( \frac { d^2y }{ dx^2 } \right) }^{ 3\times2 }= { 1+\left( \frac { dy }{ dx } \right) } \)
squaring both sides, we get
\({ \left( \frac { dy }{ dx } \right) }^{ 6 }=1+\left( \frac { dy }{ dx } \right) \)
In this equation, the highest order derivative is 2 and its power is 6.
∴ Order 2, degree 6.

26.

\(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ b } =-\overset { \wedge }{ i } +2\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =3\overset { \wedge }{ i } +\overset { \wedge }{ j } \)
Since \(\overset { \rightarrow }{ a } +\lambda \overset { \rightarrow }{ b } \) ⏊^r to \(\overset { \rightarrow }{ c } \)
\(\Rightarrow \left[ \overset { \wedge }{ i } +\overset { \wedge }{ 2j } +\overset { \wedge }{ 3k } +\lambda \left( \overset { \wedge }{ -i } +\overset { \wedge }{ 2j } +\overset { \wedge }{ k } \right) \right] .\left( \overset { \wedge }{ 3i } +\overset { \wedge }{ j } \right) =0\)
\(\Rightarrow \left[ \left( 1-\lambda \right) \overset { \wedge }{ i } +\left( 2+2\lambda \right) \overset { \wedge }{ j } +\left( 3+\lambda \right) \overset { \wedge }{ k } \right] .\left( \overset { \wedge }{ 3i } +\overset { \wedge }{ j } \right) =0\)
⇒ 3(1 - λ) + (2 + 2λ) = 0 ⇒ 3 - 3λ + 2+ 2λ = 0
⇒ 5 - λ = 0 ⇒ λ = 5

Generic 5

6 x 5 = 30

27.

Local min value =−4
Local max.value = 0

28.

(−∞,−2)decreasing
(−2,−1)increasing
(−1,∞)decreasing
Local maximum value = 6
Local minimum value =5

29.

\(\frac { ds }{ dt } ={ 60cm }^{ 2 }/sec\frac { dt }{ dr } =\frac { 1 }{ 4\pi } cm/sec\)

30.

p	q	r	~ q	~q V r	p➝(¬qVr)	~p	~pV(~qVr)
T	T	T	F	T	T	F	T
T	T	F	F	F	F	F	F
T	F	T	T	T	T	F	T
T	F	F	T	T	T	F	T
F	T	T	F	T	T	T	T
F	T	F	F	F	T	T	T
F	F	T	T	T	T	T	T
F	F	F	T	T	T	T	T

From the table, it is clear that the column of p➝(¬q V ~r) and ~pV(~q V r) are identical
∴ p➝(¬q V ~r) ≡ ~pV(~q V r)
Hence proved.

31.

Let us take the axis AX as the and the tangent AY at A as y-axis.
Equation of the parabola is y² = 4ax
CA 15, FG = 120
CF = CG= 60
F is (15, 60)
Since F lies on (1), 602 = Aa(15) ∴ ⇒ a = 60
y² = 240x
When y = 24, 24²= 240(x)
⇒ x = \(\frac { 24\times 24 }{ 240 } \)
x = \(\frac{24}{10}\) = \(\frac{12}{5}\) = 2.4
From the diagram
BD = BE - ED = 15-2.4 = 12.6m.
Hence the required height is 12.6 m.

32.

Given equation is
(2x² - 3x + 1) (2x² + 5x + 1) = 9x²....(1)
Clearly x = 0 does not satisfy (1),
∴ (1) can be rewritten as
\(\left( 2x+\frac { 1 }{ x } -3 \right) \left( 2x+\frac { 1 }{ x } +5 \right) =9...(2)\)
put \(2x+\frac { 1 }{ x } =y\)
∴ (2)⇒ (y - 3) (y + 5) = 9
⇒ y²+ 2y-15 = 9
or y²+ 2y - 24 = 0
⇒ (y + 6) (y - 4) = 0
⇒ y = -6, -4
Case (i)
When \(y=-6,2x+\frac { 1 }{ x } =-6\)
2x²+6x+1 = 0
\(\Rightarrow =\frac { -6\pm \sqrt { 36-8 } }{ 4 } \)
\(x=-3\pm \frac { \sqrt { 7 } }{ 2 } \)
Case(ii)
When \(y=4,2x+\frac { 1 }{ x } =4\)
⇒ 2x²- 4x+1 = 0
\(x=\frac{4 \pm \sqrt{16-8}}{4}\)
\(x=\frac { 2\pm \sqrt { 2 } }{ 2 } \)
This, the roots are
\(\\ \frac { -3+\sqrt { 7 } }{ 2 } ,\frac { -3-\sqrt { 7 } }{ 2 } ,\frac { 2+\sqrt { 2 } }{ 2 } ,\frac { 2-\sqrt { 2 } }{ 2 } \)

Generic 3

7 x 3 = 21

33.

\(xy=\frac { { y }^{ 4 } }{ 4 } +c\)

34.

s = 13.8 t - 4.9 t²
v = \(\frac { ds }{ dt } \) =13.8 - 4.9 (2t)
=13.8-9.8t
When t = 1, v = 13.8 - 9.8(1)
4 m/sec.
Acceleration = \(\frac { d^{ 2 }x }{ { dt }^{ 2 } } \) = -9.8 m/sec²
At maximum height, v = 0
∴ 13.8 - 9.8 t = 0
⇒ 13.8 = 9.8 t
⇒ t = \(\frac { 13.8 }{ 9.8 } \) = 1.40 sec
At t = 1.4 sec,
distance (s) = 13.8 (1.40) - 4.9 (1.40)²
= 19.32 - 9.604 = 9.716 m

35.

(i) Since f (x) is a probability density function, f (x) ≥ 0 and \(\int _{ - }^{ \infty }{ f(x) } dx=1\)
That is \(\int _{ -\infty }^{ 0 }{ 0dx } +\int _{ 0 }^{ \infty }{ k{ e }^{ -2x }dx } =1\)
\(0+k\left( \frac { { e }^{ -2x } }{ -2 } \right) =1\Rightarrow k\left( \frac { { e }^{ -\infty }-{ e }^{ 0 } }{ -2 } \right) =1\Rightarrow k=2\)
Therefore the probability density function is
\(f\left( x \right) =\begin{cases} \begin{matrix} 2{ e }^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & forx \end{matrix}\le 0 \end{cases}\)
(ii) Distribution function
By definition the distribution function \(F(x)=P\left( x\le x \right) =\int _{ -\infty }^{ x }{ f(u) } du\)
When x≤0 \(F(x)=\int _{ -\infty }^{ x }{ F(u) } du=\int _{ -\infty }^{ x }{ odu=0 } \)
When x > 0 \(F(x)=\int _{ -\infty }^{ x }{ f(u) } du\int _{ -\infty }^{ x }{ 0du } +\int _{ 0 }^{ x }{ { 2e }^{ -2x }du\left( \frac { { e }^{ -2x } }{ -2 } \right) } =1-{ e }^{ 2x }\)
This gives \(F(x)=\begin{cases} \begin{matrix} 0 & forx\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ 2x } & forx>0 \end{matrix} \end{cases}\)
(iii) P(X < ) = P(X ≤2 ) = F(2 ) = 1-e^2\(\times\)2 (since F(x) is continuous)
(iv) The probability that X is at least equal to four unit of time is
P(X ≥ 4 ) = 1 - P(X < 4 ) = 1- F( 4) = 1 - ( 1-e^{-2^\(\times\)4}) = e⁸
(v) In the continuous case, f (x) at x = a is not the probability that X takes the value a, that is f (x) at x = a is not equal to P( X ) a. If X is continuous type, P(X = a) = 0 for a ∈ R. Therefore P(x = 3) = 0.

36.

\(\int _{ 0 }^{ 3 }{ (3{ x }^{ 2 }-4x+5) } dx=\int _{ 0 }^{ 3 }{ 3{ x }^{ 2 } } dx-\int _{ 0 }^{ 3 }{ 4x } dx+\int _{ 0 }^{ 3 }{ 5 } dx\)
\(=3\int _{ 0 }^{ 3 }{ { x }^{ 2 } } dx-4\int _{ 0 }^{ 3 }{ c } dx+5\int _{ 0 }^{ 3 }{ dx } \)
\(=3{ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }-4{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 }+5{ \left[ x \right] }_{ 0 }^{ 3 }\)
= (27 − 0) − 2(9 − 0) + 5(3− 0)
= 27 −18 +15 = 24.

37.

LHS = \({ tan }^{ -1 }\left( \frac { m }{ n } \right) -{ tan }^{ -1 }\left( \frac { m-n }{ m+n } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { m }{ n } -\frac { m-n }{ m+n } }{ 1+\frac { m }{ n } \left( \frac { m-n }{ m+n } \right) } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { m(m+n)-n(m-n) }{ m(m+n) } }{ \frac { n(m+n)+m(m-n) }{ n(m+n) } } \right) \)

= \({ tan }^{ -1 }\left( \frac { { m }^{ 2 }+{ n }^{ 2 } }{ { m }^{ 2 }+{ n }^{ 2 } } \right) ={ tan }^{ -1 }(1)\)
= \(\frac { \pi }{ 4 } =RHS\)
Hence proved.

38.

From the line \(\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-1 }{ 5 } \)
(x₁, y₁, z₁) is (1, -1, 1)
(l₁, m₁, n₁) is (3, 2, 5)
From the line \(\frac { x+2 }{ 4 } =\frac { y-1 }{ 3 } =\frac { z+1 }{ -2 } \)
we get, (x₂, y₂, z₂) is (-2, 1, -1)
(l₂, m₂, n₂) is 4, 3, -2
The Condition for intersecting lines is
\(\left| \begin{matrix} { x }_{ 2 }-{ x }_{ 1 } \\ { l }_{ 1 } \\ { l }_{ 2 } \end{matrix}\begin{matrix} { y }_{ 2 }-{ y }_{ 1 } \\ { m }_{ 1 } \\ { m }_{ 2 } \end{matrix}\begin{matrix} { z }_{ 2 }-{ z }_{ 1 } \\ { n }_{ 1 } \\ { n }_{ 2 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} -2-1 \\ 3 \\ 4 \end{matrix}\begin{matrix} 1+1 \\ 2 \\ 3 \end{matrix}\begin{matrix} -1-1 \\ 5 \\ -2 \end{matrix} \right| \)
\(\Rightarrow \left| \begin{matrix} -3 \\ 3 \\ 4 \end{matrix}\begin{matrix} 2 \\ 2 \\ 3 \end{matrix}\begin{matrix} -2 \\ 5 \\ -2 \end{matrix} \right| \)
= -3 (-4 -15) -2 (-6 -20) -2 (9 - 8)
= -3(-19) - 2(-26) -2 (1)
= 57 + 52 - 2 = 57 + 50
= 107 ≠ 0
Hence the given lines do not intersect

39.

Let A =\(\left[ \begin{matrix} 4 \\ -2 \\ 1 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 0 \\ -1 \\ 8 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
A\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ -2 \\ 4 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 8 \\ -1 \\ 0 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 4 \\ 11 \\ -12 \end{matrix}\begin{matrix} 8 \\ 15 \\ -32 \end{matrix}\begin{matrix} 7 \\ 19 \\ -25 \end{matrix} \right] \)
A is in row - echelon form and it has 3 non-zero rows.
∴ \(\rho\) (A) = 3