MATHEMATICS-I (CIT0100304)¶

Paper Details

Course: BCA (SEM-I)
Full Marks: 60
Time: 2½ hours
Instructions: The figures in the margin indicate full marks for the questions.

(1) Answer the following¶

Marks: \(1 \times 8 = 8\)

a. Define Set.
b. When is a matrix said to be singular?
c. Give an example of a bijective function.
d. What is the Cardinality of the Set \(A=\{0, 1, -1, 8\}\)?
e. What is Sample Space?
f. If the probability of an event A is \(P(A)=\frac{2}{3}\), then find \(P(\overline{A})\).
g. State De Morgan's Law for two sets A and B.
h. Find the values of \(x\) and \(y\) if:

\[ \begin{bmatrix}x & 5 \\ 3 & -2 \end{bmatrix} = \begin{bmatrix} -8 & 5 \\ y & -2 \end{bmatrix} \]

(2) Answer any six from the following¶

Marks: \(2 \times 6 = 12\)

a. Define injective and surjective functions with one example for each.
b. If \(f:R \rightarrow R\) and \(g:R \rightarrow R\) are defined as \(f(x)=x^{2}\) and \(g(x)=x-3\), find \(f \circ g\) and \(g \circ f\).
c. Given the matrices \(A\) and \(B\), find the product \(A \cdot B\).

\[ A = \begin{bmatrix} 2 & -3 \\ 5 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 6 & -7 \\ 2 & -8 \end{bmatrix} \]
d. Find the minor and co-factors of the elements 2 and 7 in the following matrix:

\[ A = \begin{bmatrix} 2 & -4 & 5 \\ 3 & 1 & 7 \\ 0 & -1 & -2 \end{bmatrix} \]
e. For any three non-empty sets A, B, and C, prove that \(A-(B \cup C)=(A-B) \cap (A-C)\).
f. Find the Trace of the matrix A:

\[ A = \begin{bmatrix} -1 & 0 & 3 \\ 2 & 4 & 5 \\ 7 & 3 & 6 \end{bmatrix} \]
g. If \(P(A)=\frac{2}{3}\), \(P(B)=\frac{3}{4}\), and \(P(A \cup B)=\frac{1}{2}\), find the value of \(P(A/B)\).
h. Define a Partially Ordered Set (POSET).
i. Calculate the median from the following data: - \(x: 1, 2, 3, 4, 5\) - \(y: 8, 5, 6, 2, 1\)
j. What is the empirical relation between mean, median, and mode?

(3) Answer any four from the following¶

Marks: \(5 \times 4 = 20\)

a. For any four non-empty sets A, B, S, and T, prove that \((A \times B) \cap (S \times T) = (A \cap S) \times (B \cap T)\).
b. Using the principle of mathematical induction, prove that \(P(n) = 10^{n} + 3 \cdot 4^{n+2} + 5\) is divisible by 9 for all natural numbers \(n\).
c. If \(f:A \rightarrow B\) and \(g:B \rightarrow C\) are both one-one and onto functions, then prove that:
1. The composite function \(g \circ f\) is also one-one and onto.
2. \((g \circ f)^{-1} = f^{-1} \circ g^{-1}\).
d. Let R be a relation '\(\le\)' (less than or equal to) defined on the set of positive integers. Prove that R is a partial order relation.
e. Find the mean deviation about the median for the given data:

\(x_i\)	5	7	9	10	12	15
\(f_i\)	8	6	2	2	6	6

f. Solve the following system of linear equations using Cramer's rule:

\[ \begin{cases} 2x - y + 3z = 9 \\ x + y + z = 6 \\ x - y + z = 2 \end{cases} \]
g. Find the inverse of the matrix A:

\[ A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \]
h. If A and B are square matrices of the same order, then prove that \(adj(AB) = (adj B) \cdot (adj A)\).

(4) Answer any two from the following¶

Marks: \(10 \times 2 = 20\)

a. Let R be a relation on the set of integers I, defined as \(R = \{(x,y) | (x-y) \text{ is divisible by } 5\}\). Prove that R is an equivalence relation. Also, find the equivalence classes.
b.
1. If \(f:X \rightarrow Y\) is a function and A, B are subsets of Y, then prove that:
  1. - \(f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)\)
  2. - \(f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)\)
2. If R and S are equivalence relations on a set, prove that \(R \cap S\) is also an equivalence relation.
c.
1. State and prove the Principle of Inclusion and Exclusion for two sets.
2. Prove that for any two finite sets A and B, \(n(A \oplus B) = n(A) + n(B) - 2n(A \cap B)\), where \(A \oplus B\) is the symmetric difference of A and B.
d.
1. Solve the following system of equations using the matrix method (inverse matrix method):
  
  \[ \begin{cases} x - y + 2z = 4 \\ 3x + y + 4z = 6 \\ x + y + z = 1 \end{cases} \]
2. For the matrix \(A = \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}\), verify that \(A \cdot (adj A) = (adj A) \cdot A = |A|I\).
e.
1. Write a short note on the limitations of statistics.
2. Find the Arithmetic Mean and Mode from the following frequency distribution:

Class Interval	Frequency
0-10	5
10-20	6
20-30	8
30-40	30
40-50	10

Solutions¶

Section 1: Answer the following¶

a. Define Set.

Ans. A set is a well-defined collection of distinct objects, considered as an object in its own right. The objects that make up the set are called its elements or members.

b. When is a matrix said to be singular?

Ans. A square matrix is said to be singular if the value of its determinant is equal to zero. If the determinant is not zero, the matrix is called non-singular.

c. Give an example of a bijective function.

Ans. An example of a bijective function is \(f: \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = x + 1\). This function is both one-to-one (injective) and onto (surjective).

d. What is the Cardinality of the Set \(A=\{0, 1, -1, 8\}\)?

Ans. The cardinality of a set is the number of elements in the set. The set \(A = \{0, 1, -1, 8\}\) has 4 distinct elements. Therefore, the cardinality of set A, denoted as \(|A|\), is 4.

e. What is Sample Space?

Ans. In probability theory, the sample space of an experiment is the set of all possible outcomes of that experiment. It is usually denoted by the symbol \(S\). For example, when tossing a coin, the sample space is \(S = \{\text{Heads, Tails}\}\).

f. If the probability of an event A is \(P(A)=\frac{2}{3}\), then find \(P(\overline{A})\).

Ans. We know that for any event A, the probability of its complement, \(\overline{A}\), is given by the formula:

\[ P(\overline{A}) = 1 - P(A) \]

Given, \(P(A) = \frac{2}{3}\). Therefore,

\[ P(\overline{A}) = 1 - \frac{2}{3} \]

\[ P(\overline{A}) = \frac{3 - 2}{3} \]

\[ P(\overline{A}) = \frac{1}{3} \]

g. State De Morgan's Law for two sets A and B.

Ans. For any two sets A and B, De Morgan's Laws state: 1. The complement of the union of the two sets is equal to the intersection of their complements: \( (A \cup B)' = A' \cap B' \) 2. The complement of the intersection of the two sets is equal to the union of their complements: \( (A \cap B)' = A' \cup B' \)

h. Find the values of \(x\) and \(y\) if:

\[ \begin{bmatrix}x & 5 \\ 3 & -2 \end{bmatrix} = \begin{bmatrix} -8 & 5 \\ y & -2 \end{bmatrix} \]

Ans. Two matrices are equal if and only if their corresponding elements are equal. By equating the corresponding elements of the given matrices, we get:

Comparing the element in the first row, first column:

\[ x = -8 \]

Comparing the element in the second row, first column:

\[ 3 = y \]

Thus, the values are \(x = -8\) and \(y = 3\).

Section 2: Answer any six from the following¶

a. Define injective and surjective functions with one example for each.

Ans.

Injective Function (One-to-one): A function \(f: A \rightarrow B\) is called injective if distinct elements in the domain \(A\) have distinct images in the codomain \(B\). Mathematically, for all \(a_1, a_2 \in A\), if \(f(a_1) = f(a_2)\), then \(a_1 = a_2\).
- Example: The function \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) defined by \(f(x) = x + 5\) is injective because if \(x_1 + 5 = x_2 + 5\), then \(x_1 = x_2\).
Surjective Function (Onto): A function \(f: A \rightarrow B\) is called surjective if every element in the codomain \(B\) has at least one pre-image in the domain \(A\). Mathematically, for every \(y \in B\), there exists an \(x \in A\) such that \(f(x) = y\).
- Example: The function \(f: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x) = x - 3\) is surjective because for any real number \(y\), we can find \(x = y + 3\) such that \(f(x) = y\).

b. If \(f:R \rightarrow R\) and \(g:R \rightarrow R\) are defined as \(f(x)=x^{2}\) and \(g(x)=x-3\), find \(f \circ g\) and \(g \circ f\).

Ans. Given functions are \(f(x)=x^{2}\) and \(g(x)=x-3\).

To find \(f \circ g\): The composition \(f \circ g\) is defined as \(f(g(x))\).

\[ (f \circ g)(x) = f(g(x)) \]

Substitute the expression for \(g(x)\): [ = f(x-3) ]

Now apply the function \(f\) to \((x-3)\):

\[ = (x-3)^2 \]

\[ = x^2 - 6x + 9 \]

Therefore, \((f \circ g)(x) = x^2 - 6x + 9\).
To find \(g \circ f\): The composition \(g \circ f\) is defined as \(g(f(x))\).

\[ (g \circ f)(x) = g(f(x)) \]

Substitute the expression for \(f(x)\):

\[ = g(x^2) \]

Now apply the function \(g\) to \(x^2\):

\[ = x^2 - 3 \]

Therefore, \((g \circ f)(x) = x^2 - 3\).

c. Given the matrices \(A\) and \(B\), find the product \(A \cdot B\).

\[ A = \begin{bmatrix} 2 & -3 \\ 5 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 6 & -7 \\ 2 & -8 \end{bmatrix} \]

Ans. To find the product \(A \cdot B\), we multiply the rows of matrix \(A\) by the columns of matrix \(B\).

\[ A \cdot B = \begin{bmatrix} 2 & -3 \\ 5 & 4 \end{bmatrix} \begin{bmatrix} 6 & -7 \\ 2 & -8 \end{bmatrix} \]

\[ = \begin{bmatrix} (2)(6)+(-3)(2) & (2)(-7)+(-3)(-8) \\ (5)(6)+(4)(2) & (5)(-7)+(4)(-8) \end{bmatrix} \]

\[ = \begin{bmatrix} 12-6 & -14+24 \\ 30+8 & -35-32 \end{bmatrix} \]

\[ = \begin{bmatrix} 6 & 10 \\ 38 & -67 \end{bmatrix} \]

Thus, the product \(A \cdot B\) is \(\begin{bmatrix} 6 & 10 \\ 38 & -67 \end{bmatrix}\).

d. Find the minor and co-factors of the elements 2 and 7 in the following matrix:

\[ A = \begin{bmatrix} 2 & -4 & 5 \\ 3 & 1 & 7 \\ 0 & -1 & -2 \end{bmatrix} \]

Ans.

For the element 2 (position \(a_{11}\)):

Minor (\(M_{11}\)): The minor is the determinant of the sub-matrix obtained by deleting the 1^st row and 1^st column.

\[ M_{11} = \det \begin{bmatrix} 1 & 7 \\ -1 & -2 \end{bmatrix} = (1)(-2) - (7)(-1) = -2 + 7 = 5 \]
Co-factor (\(C_{11}\)): The co-factor is given by \(C_{ij} = (-1)^{i+j} M_{ij}\).

\[ C_{11} = (-1)^{1+1} M_{11} = (1)(5) = 5 \]

For the element 7 (position \(a_{23}\)):

Minor (\(M_{23}\)): The minor is the determinant of the sub-matrix obtained by deleting the 2^nd row and 3^rd column.

\[ M_{23} = \det \begin{bmatrix} 2 & -4 \\ 0 & -1 \end{bmatrix} = (2)(-1) - (-4)(0) = -2 - 0 = -2 \]
Co-factor (\(C_{23}\)): The co-factor is given by \(C_{ij} = (-1)^{i+j} M_{ij}\).

\[ C_{23} = (-1)^{2+3} M_{23} = (-1)^5(-2) = (-1)(-2) = 2 \]

e. For any three non-empty sets A, B, and C, prove that \(A-(B \cup C)=(A-B) \cap (A-C)\).

Ans. To prove the equality, we will show that any element belonging to the left-hand side (LHS) also belongs to the right-hand side (RHS), and vice-versa.

Let \(x\) be an arbitrary element such that \(x \in A - (B \cup C)\).

\[ x \in A - (B \cup C) \iff x \in A \text{ and } x \notin (B \cup C) \quad \text{(by definition of set difference)} \]

\[ \iff x \in A \text{ and } (x \notin B \text{ and } x \notin C) \quad \text{(by definition of union)} \]

\[ \iff (x \in A \text{ and } x \notin B) \text{ and } (x \in A \text{ and } x \notin C) \quad \text{(Distributive law of logic)} \]

\[ \iff (x \in A-B) \text{ and } (x \in A-C) \quad \text{(by definition of set difference)} \]

\[ \iff x \in (A-B) \cap (A-C) \quad \text{(by definition of intersection)} \]

Since \(x \in A - (B \cup C) \iff x \in (A-B) \cap (A-C)\), the two sets are equal. Hence, \(A-(B \cup C)=(A-B) \cap (A-C)\) is proved.

f. Find the Trace of the matrix A:

\[ A = \begin{bmatrix} -1 & 0 & 3 \\ 2 & 4 & 5 \\ 7 & 3 & 6 \end{bmatrix} \]

Ans. The trace of a square matrix is the sum of the elements on its main diagonal (from the upper left to the lower right). The elements on the main diagonal of matrix A are -1, 4, and 6.

\[ \text{Trace}(A) = -1 + 4 + 6 \]

\[ \text{Trace}(A) = 9 \]

The trace of the matrix A is 9.

g. If \(P(A)=\frac{2}{3}\), \(P(B)=\frac{3}{4}\), and \(P(A \cup B)=\frac{1}{2}\), find the value of \(P(A/B)\).

Ans. The formula for conditional probability is \(P(A/B) = \frac{P(A \cap B)}{P(B)}\). First, we need to find \(P(A \cap B)\) using the addition rule of probability:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Substituting the given values: [ \frac{1}{2} = \frac{2}{3} + \frac{3}{4} - P(A \cap B) ]

\[ P(A \cap B) = \frac{2}{3} + \frac{3}{4} - \frac{1}{2} \]

To add/subtract the fractions, we find a common denominator, which is 12.

\[ P(A \cap B) = \frac{2 \times 4}{12} + \frac{3 \times 3}{12} - \frac{1 \times 6}{12} \]

\[ P(A \cap B) = \frac{8 + 9 - 6}{12} = \frac{11}{12} \]

Now we can find \(P(A/B)\):

\[ P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{11/12}{3/4} = \frac{11}{12} \times \frac{4}{3} = \frac{11 \times 1}{3 \times 3} = \frac{11}{9} \]

Note: The result \(P(A/B) = 11/9\) is greater than 1. This indicates that the initial probability values given in the question are inconsistent, as a probability value can never exceed 1. For example, \(P(A \cap B)\) cannot be greater than \(P(A)\) or \(P(B)\), but here \(\frac{11}{12} > \frac{2}{3}\) and \(\frac{11}{12} > \frac{3}{4}\). However, the calculation based on the given numbers is as shown.

h. Define a Partially Ordered Set (POSET).

Ans.

A Partially Ordered Set (or POSET) consists of a non-empty set \(P\) together with a binary relation \(\preceq\) defined on \(P\), such that the relation satisfies the following three properties for all \(a, b, c \in P\):

Reflexivity: \(a \preceq a\) (Every element is related to itself).
Antisymmetry: If \(a \preceq b\) and \(b \preceq a\), then \(a = b\) (If two elements are related to each other, they must be the same element).
Transitivity: If \(a \preceq b\) and \(b \preceq c\), then \(a \preceq c\) (If a is related to b, and b is related to c, then a is related to c).

The pair \((P, \preceq)\) is called a poset.

i. Calculate the median from the following data: * \(x: 1, 2, 3, 4, 5\) * \(y: 8, 5, 6, 2, 1\)

Ans. Here, \(x\) represents the observations and \(y\) represents their corresponding frequencies (\(f\)). To find the median, we first calculate the cumulative frequency (\(cf\)).

\(x\)	Frequency (\(f\))	Cumulative Frequency (\(cf\))
1	8	8
2	5	8 + 5 = 13
3	6	13 + 6 = 19
4	2	19 + 2 = 21
5	1	21 + 1 = 22

The total number of observations is \(N = \sum f = 22\). Since \(N\) is an even number, the median is the average of the \(\left(\frac{N}{2}\right)^{th}\) and \(\left(\frac{N}{2} + 1\right)^{th}\) observations.

\[ \text{Median} = \frac{\left(\frac{22}{2}\right)^{th} \text{ obs} + \left(\frac{22}{2} + 1\right)^{th} \text{ obs}}{2} = \frac{11^{th} \text{ obs} + 12^{th} \text{ obs}}{2} \]

From the cumulative frequency table: * Observations 1 to 8 have the value 1. * Observations 9 to 13 have the value 2. This means both the \(11^{th}\) and \(12^{th}\) observations fall in the group where \(x=2\).

\[ \text{Median} = \frac{2 + 2}{2} = \frac{4}{2} = 2 \]

The median of the data is 2.

j. What is the empirical relation between mean, median, and mode?

Ans. The empirical relationship between mean, median, and mode describes an approximate relationship between the three measures of central tendency for a unimodal distribution that is moderately skewed. The relation is given by the formula:

\[ \text{Mean} - \text{Mode} \approx 3 (\text{Mean} - \text{Median}) \]

This can be rearranged to express the mode in terms of the mean and median:

\[ \text{Mode} \approx 3 \times \text{Median} - 2 \times \text{Mean} \]

Section 3: Answer any four from the following¶

a. For any four non-empty sets A, B, S, and T, prove that \((A \times B) \cap (S \times T) = (A \cap S) \times (B \cap T)\).

Ans. To prove the set equality, we will show that an element \((x, y)\) belongs to the left-hand side (LHS) if and only if it belongs to the right-hand side (RHS).

Let \((x, y)\) be an arbitrary element of \((A \times B) \cap (S \times T)\).

\[ (x, y) \in (A \times B) \cap (S \times T) \]

By the definition of intersection of sets:

\[ \iff (x, y) \in (A \times B) \quad \text{and} \quad (x, y) \in (S \times T) \]

By the definition of the Cartesian product:

\[ \iff (x \in A \text{ and } y \in B) \quad \text{and} \quad (x \in S \text{ and } y \in T) \]

Rearranging the terms using the commutative and associative properties of logical 'and':

\[ \iff (x \in A \text{ and } x \in S) \quad \text{and} \quad (y \in B \text{ and } y \in T) \]

By the definition of intersection of sets:

\[ \iff x \in (A \cap S) \quad \text{and} \quad y \in (B \cap T) \]

By the definition of the Cartesian product:

\[ \iff (x, y) \in (A \cap S) \times (B \cap T) \]

Since an arbitrary element \((x, y)\) is in the LHS if and only if it is in the RHS, the two sets are equal. Hence, proved.

b. Using the principle of mathematical induction, prove that \(P(n) = 10^{n} + 3 \cdot 4^{n+2} + 5\) is divisible by 9 for all natural numbers \(n\).

Ans. Let the given statement be \(P(n)\): \(10^n + 3 \cdot 4^{n+2} + 5\) is divisible by 9.

Step 1: Base Case For \(n=1\), we have:

\[ P(1) = 10^1 + 3 \cdot 4^{1+2} + 5 \]

\[ = 10 + 3 \cdot 4^3 + 5 \]

\[ = 10 + 3 \cdot 64 + 5 \]

\[ = 10 + 192 + 5 = 207 \]

Since \(207 = 9 \times 23\), \(P(1)\) is divisible by 9. Thus, the base case is true.

Step 2: Inductive Hypothesis Let us assume that \(P(k)\) is true for some natural number \(k\). That is, \(10^k + 3 \cdot 4^{k+2} + 5\) is divisible by 9. We can write this as \(10^k + 3 \cdot 4^{k+2} + 5 = 9m\) for some integer \(m\).

\[ \implies 10^k = 9m - 3 \cdot 4^{k+2} - 5 \quad \dots(i) \]

Step 3: Inductive Step We need to prove that \(P(k+1)\) is also true.

\[ P(k+1) = 10^{k+1} + 3 \cdot 4^{(k+1)+2} + 5 \]

\[ = 10^{k+1} + 3 \cdot 4^{k+3} + 5 \]

\[ = 10 \cdot 10^k + 3 \cdot 4 \cdot 4^{k+2} + 5 \]

\[ = 10 \cdot 10^k + 12 \cdot 4^{k+2} + 5 \]

Substituting the value of \(10^k\) from equation (i):

\[ = 10 (9m - 3 \cdot 4^{k+2} - 5) + 12 \cdot 4^{k+2} + 5 \]

\[ = 90m - 30 \cdot 4^{k+2} - 50 + 12 \cdot 4^{k+2} + 5 \]

\[ = 90m - (30 - 12) \cdot 4^{k+2} - 45 \]

\[ = 90m - 18 \cdot 4^{k+2} - 45 \]

Factoring out 9 from each term:

\[ = 9 (10m - 2 \cdot 4^{k+2} - 5) \]

Since \(m\) and \(k\) are integers, \((10m - 2 \cdot 4^{k+2} - 5)\) is also an integer. Thus, \(P(k+1)\) is a multiple of 9.

By the principle of mathematical induction, \(P(n)\) is true for all natural numbers \(n\). Hence, proved.

e. Find the mean deviation about the median for the given data: | \(x_i\) | 5 | 7 | 9 | 10 | 12 | 15 | | :--- | :-: | :-: | :-: | :--: | :--: | :--: | | \(f_i\) | 8 | 6 | 2 | 2 | 6 | 6 |

Ans. Step 1: Calculate the Median First, we find the cumulative frequency (\(cf\)) to locate the median.

\(x_i\)	Frequency (\(f_i\))	Cumulative Frequency (\(cf\))
5	8	8
7	6	14
9	2	16
10	2	18
12	6	24
15	6	30

The total frequency is \(N = \sum f_i = 30\). Since \(N\) is even, the median is the average of the \(\left(\frac{N}{2}\right)^{th}\) and \(\left(\frac{N}{2} + 1\right)^{th}\) observations. This corresponds to the average of the \(15^{th}\) and \(16^{th}\) observations. From the \(cf\) column, the \(15^{th}\) and \(16^{th}\) observations both fall in the group where \(x_i = 9\). So, Median \(M = 9\).

Step 2: Calculate Mean Deviation about the Median The formula is \(MD_M = \frac{\sum f_i |x_i - M|}{N}\). We create a calculation table:

| \(x_i\) | \(f_i\) | \(|x_i - M| = |x_i - 9|\) | \(f_i |x_i - M|\) | | :---: | :---: | :---: | :---: | | 5 | 8 | 4 | 32 | | 7 | 6 | 2 | 12 | | 9 | 2 | 0 | 0 | | 10 | 2 | 1 | 2 | | 12 | 6 | 3 | 18 | | 15 | 6 | 6 | 36 | | Total | \(N=30\) | | \(\sum f_i |x_i - M| = 100\) |

Now, we compute the mean deviation:

\[ MD_M = \frac{100}{30} = \frac{10}{3} \approx 3.33 \]

The mean deviation about the median is approximately 3.33.

f. Solve the following system of linear equations using Cramer's rule:

\[ \begin{cases} 2x - y + 3z = 9 \\ x + y + z = 6 \\ x - y + z = 2 \end{cases} \]

Ans. The given system of equations can be written in matrix form \(AX=B\), where:

\[ A = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ 6 \\ 2 \end{bmatrix} \]

Step 1: Calculate the determinant D of the coefficient matrix A.

\[ D = \det(A) = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]

\[ = 2(1\cdot1 - 1\cdot(-1)) - (-1)(1\cdot1 - 1\cdot1) + 3(1\cdot(-1) - 1\cdot1) \]

\[ = 2(2) + 1(0) + 3(-2) = 4 - 6 = -2 \]

Since \(D \neq 0\), a unique solution exists.

Step 2: Calculate \(D_x\), \(D_y\), and \(D_z\).

\[ D_x = \begin{vmatrix} 9 & -1 & 3 \\ 6 & 1 & 1 \\ 2 & -1 & 1 \end{vmatrix} = 9(1 - (-1)) - (-1)(6-2) + 3(-6-2) = 18+4-24 = -2 \]

\[ D_y = \begin{vmatrix} 2 & 9 & 3 \\ 1 & 6 & 1 \\ 1 & 2 & 1 \end{vmatrix} = 2(6 - 2) - 9(1-1) + 3(2-6) = 8-0-12 = -4 \]

\[ D_z = \begin{vmatrix} 2 & -1 & 9 \\ 1 & 1 & 6 \\ 1 & -1 & 2 \end{vmatrix} = 2(2 - (-6)) - (-1)(2-6) + 9(-1-1) = 16-4-18 = -6 \]

Step 3: Find x, y, and z. According to Cramer's rule:

\[ x = \frac{D_x}{D} = \frac{-2}{-2} = 1 \]

\[ y = \frac{D_y}{D} = \frac{-4}{-2} = 2 \]

\[ z = \frac{D_z}{D} = \frac{-6}{-2} = 3 \]

The solution is \(x=1, y=2, z=3\).

g. Find the inverse of the matrix A:

\[ A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \]

Ans. We will use the formula \(A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\).

Step 1: Find the determinant of A.

\[ \det(A) = 0(2\cdot1 - 3\cdot1) - 1(1\cdot1 - 3\cdot3) + 2(1\cdot1 - 2\cdot3) \]

\[ = 0 - 1(1-9) + 2(1-6) = -1(-8) + 2(-5) = 8 - 10 = -2 \]

Since \(\det(A) \neq 0\), the inverse exists.

Step 2: Find the Adjoint of A. The adjoint is the transpose of the cofactor matrix. Let's find the cofactors:

\[ C_{11} = (2\cdot1 - 3\cdot1) = -1 \]

\[ C_{12} = -(1\cdot1 - 3\cdot3) = 8 \]

\[ C_{13} = (1\cdot1 - 2\cdot3) = -5 \]

\[ C_{21} = -(1\cdot1 - 2\cdot1) = 1 \]

\[ C_{22} = (0\cdot1 - 2\cdot3) = -6 \]

\[ C_{23} = -(0\cdot1 - 1\cdot3) = 3 \]

\[ C_{31} = (1\cdot3 - 2\cdot2) = -1 \]

\[ C_{32} = -(0\cdot3 - 2\cdot1) = 2 \]

\[ C_{33} = (0\cdot2 - 1\cdot1) = -1 \]

The cofactor matrix is \(C = \begin{bmatrix} -1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1 \end{bmatrix}\). The adjoint of A is the transpose of C:

\[ \text{adj}(A) = C^T = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \]

Step 3: Calculate the inverse.

\[ A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \]

\[ A^{-1} = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{bmatrix} \]

This is the required inverse matrix.