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

Let P represent the vapour pressure and T represent the vapour temperature.
Given \(\frac { dp }{ dt } \alpha \quad p.\frac { 1 }{ { T }^{ 2 } } \)
[∴ Inversely proportional to the square of the  temperature]
\(\Rightarrow \frac { dp }{ dt } =\frac { kP }{ { T }^{ 2 } } \) where k is a constant.

 sin^-1 [sin5]
We know that \({ sin }^{ -1 }\left( sin5 \right) =0\quad if\quad \theta \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ e } \right] \)
Consiideerning the approximation \(\frac { \pi }{ 2 } =\frac { 11 }{ 7 } \)
\(5=5\times \frac { 11 }{ 7 } \times \frac { 7 }{ 11 } =5\times \frac { \pi }{ 2 } \times \frac { 7 }{ 11 } =\frac { 35\pi }{ 22 } \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
But \(5-2\pi =5-2\times \frac { 22 }{ 7 } =5-\frac { 44 }{ 7 } =\frac { 35z44 }{ 7 } \)
= \(\frac { -11 }{ 7 } \varepsilon \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(5-2\pi \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\({ sin }^{ -1 }\left( sin5 \right) ={ sin }^{ -1 }\left( sin(5-2\pi ) \right) \)
= \(5-2\pi \quad \) 

 \(\overline { 3i } +\frac { 1 }{ 2-i } \)
= - 3i + \(\frac { 1 }{ 2-i } \times \frac { 2+i }{ 2+i } \)
[∴ Conjugate of 3i is -3i]
= - 3i + \(\frac { 2+i }{ 2^{ 2 }-{ i }^{ 2 } } =-3i+\frac { 2+i }{ 4+1 } \)
= - 3i + \(\frac { 2+i }{ 5 } \)
= \(\frac { -15i+2+i }{ 5 } =\frac { -14i+2 }{ 5 } \)
\(=\frac { 2 }{ 5 } -\frac { 14}{ 5 }i \).

Let A = \(\left[ \begin{matrix} 6 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -9 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \). Then A is a matrix of order 4 × 3 and ρ(A) ≤ 3.
The last two rows are zero rows. There are several second order minors.
We find that there is a second order minor, for example, \(\left| \begin{matrix} 6 & 0 \\ 0 & 2 \end{matrix} \right| \) = (6)(2) = 12 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third and fourth rows are zero rows.

Let A = \(\left[ \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3.
The only third order minor is |A| = \(\left| \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right| \) = (-2)(5)(0) = 0. So ρ(A) ≤ 2.
There are several second order minors. We find that there is a second order minor, for example, \(\left| \begin{matrix} -2 & 2 \\ 0 & 5 \end{matrix} \right| \) = (-2)(5) = -10 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third row is a zero row.

We know that the formula for the distance between two parallel ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is \(\delta =\frac { |{ d }_{ 1 }-{ d }_{ 2 }| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \). Rewrite the second equation as x + 2y - 2z + \(\frac { 5 }{ 2 } \) = 0. 
Comparing the given equations with the general equations, we get a = 1, b = 2, c = -2, d₁ = 1, d₂ = \(\frac { 5 }{ 2 } \). Substituting these values in the formula, we get the distance
\(\delta =\frac { |{ d }_{ 1 }-{ d }_{ 2 }| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } =\frac { |1-\frac { 5 }{ 2 } | }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+(-2^{ 2 }) } } =\frac { 1 }{ 2 } \) units.

We compute \(\left| 6-8i \right| =\sqrt { { 6 }^{ 2 }+\left( -8 \right) ^{ 2 } } =10\)
and applying the formula for square root, we get
\(\sqrt { 6-8i } =\pm \left( \sqrt { \frac { 10+6 }{ 2 } } -i\sqrt { \frac { 10-6 }{ 2 } } \right) \) (\(\therefore\) b is negative\( \frac{b}{|b|}=-1 \))
= \(\pm \left( \sqrt { 8 } +i\sqrt { 2 } \right) \)
= \(\pm \left( 2\sqrt { 2 } -i\sqrt { 2 } \right) \)

Let A = \(\left[ \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3
The third order minor |A| = \(\left| \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right| \) = (2)(3)(1) = 6 ≠ 0. So, ρ(A) = 3.
Note that there are three non-zero rows.

Let sec^-1\(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\theta \).
Then, sec\(\theta\) = \(-\frac{2}{\sqrt3}\) where \(\theta\in[0,\pi]\)\{\(\frac{\pi}{2}\)}.
Thus, cos \(\theta =-\frac{\sqrt{3}}{2}\).
Now, \(cos\frac { 5\pi }{ 6 } =cos\left( \pi -\frac { \pi }{ 6 } \right) =-cos\left( \frac { \pi }{ 6 } \right) =-\frac { \sqrt { 3 } }{ 2 } .\) 
Hence, Sec-¹ \(\left( -\frac { 2\sqrt { 3 } }{ 3 } \right) =\frac { 5\pi }{ 6 } \)

 We know that sin^-1: [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus,
 \({ Sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)= \(\frac{\pi}{4}\), Since \(\frac{\pi}{4}\)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)and sin \(\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)

Since the domain of y = sin^-1 is −[11], and 2\(\notin \)[-1, 1], sin⁻¹(2) does not exist.

\(y=2tan\frac { x }{ 2 } -x+c\)

prove

\(\left( \frac { dy }{ dx } \right) ^{ 2 }=4y\left( x\frac { dy }{ dx } -2y \right) \)

\(\frac { 128 }{ 315 } \)

0

\(\frac { 1 }{ 4 } \)

\(L(z)={ 2 }^{ 1/4 }+\frac { 1 }{ 4 } \left( { 2 }^{ -3/4 } \right) \left( z-2 \right) \)

Let x_o = 3 and dx = 0.02
f(x_o) = f(3) = 3 (3²) + 5 (3) + 3
= 27 + 15 + 3 = 45
f'(x) = 6x + 5
f'(x) = f'(3) = 6 (3) + 5 = 23
∴ f(3. 02) = f(x_o) +f'(x_o) dx
= 45 + 23 (.02)
= 45 + 0.46 = 45.46

Let X be the random variable denotes the total C score is two throws of a die.
Sample space S

n (S) = 36
X = {4,5,6,7,8}
From the sample space

Let X be the random variable of getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Expending along R₁,
|A| = \(5\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| +1\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \)
= 5 (25 - 1)-1 (5 - 1)+ 1 (1 - 5)
= 5 (24) - 1(4) + 1(- 4)
= 120 - 4 - 4 = 120 - 8 = 112 ≠ 0
Since A is non singular, A^-1 exit
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 5 & 1 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(25-1)-(5-1)+(1-5) \\ -(5-1)+(25-1)-(5-1) \\ +(1-5)+(5-1)+(25-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -4 \\ -4 & -4 & 24 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -24 \\ -4 & -4 & 24 \end{matrix} \right] \)
Taking 4 common from every entry we get,
adj A = \(4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 112 } .4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
= \(\frac { 1 }{ 28 } \left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \).

Let A = \(\left[ \begin{matrix} 3 & 1 & 4 \\ 2 & 0 & -1 \\ 5 & 2 & 1 \end{matrix} \right] \). Then, |A| = 3(0 - 2) - 1(2 + 5) + 4(4 - 0) = 6 - 7 + 16 = 15 ≠ 0. So, A is non-singular.
Keeping the identity matrix as our goal, we perform the row operations sequentially on A as follows:
\(\left[ \begin{matrix} 3 & 1 & 4 \\ 2 & 0 & -1 \\ 5 & 2 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \frac { 1 }{ 3 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \frac { 1 }{ 3 } & \frac { 4 }{ 3 } \\ 2 & 0 & -1 \\ 5 & 2 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 },{ R }_{ 3 }\longrightarrow { R }_{ 3 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \frac { 1 }{ 3 } & \frac { 4 }{ 3 } \\ 0 & -\frac { 2 }{ 3 } & -\frac { 11 }{ 3 } \\ 0 & \frac { 1 }{ 3 } & -\frac { 17 }{ 3 } \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow \left( \frac { 3 }{ 2 } \right) { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \frac { 1 }{ 3 } & \frac { 4 }{ 3 } \\ 0 & 1 & \frac { 11 }{ 2 } \\ 0 & \frac { 1 }{ 3 } & -\frac { 17 }{ 3 } \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }-\frac { 1 }{ 3 } { R }_{ 2 },{ R }_{ 3 }-\frac { 1 }{ 3 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & -\frac { 1 }{ 2 } \\ 0 & 1 & \frac { 11 }{ 2 } \\ 0 & 0 & -\frac { 15 }{ 2 } \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow \left( -\frac { 2 }{ 15 } \right) { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & -\frac { 1 }{ 2 } \\ 0 & 1 & \frac { 11 }{ 2 } \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }+\frac { 1 }{ 2 } { R }_{ 3 }.{ R }_{ 2 }-\frac { 11 }{ 2 } { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)

\(\left[ \begin{matrix} 3 & -1 & 2 \\ -6 & 2 & 4 \\ -3 & 1 & 2 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \)
Note
\(\left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }/8 }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \).
This is also a row-echelon form of the given matrix.
So, a row-echelon form of a matrix is not necessarily unique.

Given s is a non-empty set and x 0 y = x, x,y \(\in \) s y₀x = y
x₀u ≠ y₀x ⇒ is not commutative
Now, x₀(y₀z) = x₀y = x
and (x₀y) 0 z = x₀z = x
x₀(y₀z) = (y₀z) 0 z
0 is associative.

Given \(a*b=\frac { a+b }{ 2 } \), where a.b ∈Q Let a,b ∈Q
An element e has to found out such that 
a*e = e*a = a
Let a = 5, Then 5*e = 5
\(\Rightarrow \frac { 5+e }{ 2 } =\)5 ⇒ 5 + e = 10
Let a = \(\frac{2}{3}\). Then \(\frac{2}{3}\)*e = \(\frac{2}{3}\)
\(\Rightarrow \frac { \frac { 2 }{ 3 } +e }{ 2 } =\frac { 2 }{ 3 } \)
\(\Rightarrow \frac { 2 }{ 3 } +e=\frac { 4 }{ 3 } \)
\(\Rightarrow e=\frac { 4 }{ 3 } -\frac { 2 }{ 3 } =\frac { 2 }{ 3 } \)
It is seen that for the binary operation * defined on Q, identity element e is not unique. Hence identity element not defined for the binary operation * on Q. 
The identity does not exist. Hence inverse also does not exist for the operation * on Q.

The table for the operation x₁₁ is as follows.

Following the same kind of procedure as explained in the previous example, a brief outline of the process of verification of the properties of ×₁₁ on A is given below.
(i) Since each box has an unique element of A, ×₁₁ is a binary operation on A.
(ii) The entries are symmetrical about the main diagonal. Hence ×₁₁ has commutative property.
(iii) As usual, the associative property can be seen to be true.
(iv) The entries of both the row and column headed by the element 1 are identical. Hence 1 is the identity element.
(v) Since the identity 1 exists in each row and each column, the existence of inverse property is assured for ×₁₁. The inverse of 1 is 1, that of 3 is 4, that of 4 is 3, 5 is 9, and, that of 9 is 5.

i) Though - is not binary on N; it is binary on Z. To check the validity of any more properties satisfied by – on Z, it is better to check them for some particular simple values.
ii) Take m = 4 , n = 5 and (m− n) = (4 − 5) = −1and (n −m) = (5 − 4) = 1.
Hence (m− n) ≠ (n −m). So the operation - is not commutative on Z.
iii) In order to check the associative property, let us put m = 4, n = 5 and p = 7 in both (m- n) - p and m- (n - p).
(m−n)− p = (4−5)−7 = (−1−7) = −8 …(1)
m−(n− p) = 4−(5−7) = (4+2) = 6  …(2)
From (1) and (2), it follows that (m - n) - p m - (n - p).
Hence – is not associative on Z.
iv) Identity does not exist (why?).
v) Inverse does not exist (why?).

6x⁴- 35x³+ 62x²- 35x + 6 = 0
This equation is type I even degree reciprocal equation.
Hence, it can be rewritten as
\(6\left( { x }^{ 2 }+\frac { 1 }{ x } \right) -35(x+\frac { 1 }{ x } )+62=0 ...(1)\)
putting \(x+\frac { 1 }{ x } =y\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2={ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ y }^{ 2 }-2\)
∴ (1) becomes as,
\(\Rightarrow 6({ y }^{ 2 }-2)-35y+62=0\)
\(\Rightarrow { 6y }^{ 2 }-12-35y+62=0\)
\(\Rightarrow { 6y }^{ 2 }-35y+50=0\)
\(\Rightarrow (3y-10)(2y-5)=0\)
\(\Rightarrow y=\frac { 10 }{ 3 } ,\frac { 5 }{ 2 } \)
Case (i) when \(y=\frac { 10 }{ 3 } ,x+\frac { 1 }{ x } =\frac { 10 }{ 3 } \)


\(\Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { 10 }{ 3 } \)
\(\Rightarrow { 3x }^{ 2 }-10x+3=10x\)
\(\Rightarrow { 3x }^{ 2 }-10x+3=0\)
\(\Rightarrow (x-3)(3x-1)=0\)
\(\Rightarrow x=3,\frac { 1 }{ 3 } \)
Case (ii) when \(y=\frac { 5 }{ 2 } ,x+\frac { 1 }{ x } =\frac { 5 }{ 2 } \Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { 5 }{ 2 } \)
\(\Rightarrow { 2x }^{ 2 }+2=5x\Rightarrow { 2x }^{ 2 }-5x+2=0\)
\(\Rightarrow (x-2)(2x-1)=0\)
Hence the roots are \(2,\frac { 1 }{ 2 } ,3,\frac { 1 }{ 3 } \)

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₂] =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -\frac { 5 }{ 2 } & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 0 & 1 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -5 & 2 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ We get A^-1=\(\left[ \begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)

Applying Gauss-Jordan method, we get
[A | I₃] =\(\left[ \begin{matrix} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \frac { 1 }{ 2 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 0 & \left( 1/2 \right) & -\left( 1/2 \right) \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ -\left( 3/2 \right) & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ -3 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 2 & -1 & 0 \\ -3 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 1 }\longrightarrow { R }_{ 1 }-{ R }_{ 3 } \\ { R }_{ 2 }\longrightarrow { R }_{ 2 }+{ R }_{ 3 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \).
So, A^-1 = \(\left[ \begin{matrix} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \).