Samacheer Kalvi Class 12 Maths Solutions Chapter 12 Discrete Mathematics Exercise 12.1

Question 1. Determine whether * is a binary operation on the sets-given below
(i) a * b – a. |b| on R
(ii) a * b = min (a, b) on A = {1, 2, 3, 4, 5}
(iii) (a * b) = avb is binary on R
Answer:
(i) Yes, the operation `*` defined as \(a * b = a - |b|\) is a binary operation on R.
If we take any two real numbers \(a\) and \(b\), then \(|b|\) is also a real number. When you subtract \(|b|\) from \(a\), the result \(a - |b|\) will also be a real number. This means the result stays within the set R.
In simple words: For any two real numbers, if you calculate \(a\) minus the absolute value of \(b\), the answer will always be another real number. This makes it a binary operation.

🎯 Exam Tip: To check if an operation is binary, test if the result of the operation on any two elements from the set always stays within the same set. If it does, it's binary; otherwise, it's not.

Answer:
(ii) Yes, the operation `*` defined as \(a * b = \min(a, b)\) is a binary operation on \(A = \{1, 2, 3, 4, 5\}\).
When you pick any two numbers from the set \(A\), the minimum of those two numbers will always be one of the numbers already present in the set \(A\). For example, \(\min(2, 5) = 2\), which is in \(A\). This means the operation keeps the result within the set.
In simple words: When you find the smaller of any two numbers in the set \(\{1, 2, 3, 4, 5\}\), the answer is always one of those numbers. So, it's a binary operation.

🎯 Exam Tip: For finite sets, test a few examples to confirm closure. The minimum of two elements from a set will always be an element of that set, ensuring closure.

Answer:
(iii) No, the operation `*` defined as \(a * b = a\sqrt{b}\) is not a binary operation on R.
For an operation to be binary on R, the result must always be a real number for any real numbers \(a\) and \(b\). If we choose a negative real number for \(b\), for example, \(b = -4\), then \(\sqrt{b} = \sqrt{-4} = 2i\), which is an imaginary number and not a real number. Therefore, \(a\sqrt{b}\) would not be a real number in this case, meaning the operation is not closed on R.
In simple words: If you take a real number for \(b\) that is negative, like -4, its square root is not a real number. So, \(a\sqrt{b}\) won't always be a real number, meaning this operation is not binary on R.

🎯 Exam Tip: To prove an operation is *not* binary, a single counterexample (one case where the result falls outside the set) is sufficient. Always consider edge cases like negative numbers or zero.

Question 2. On Z, define \( \otimes \) by \( (m \otimes n) = m^n + n^m : \forall m, n \in Z \). Is \( \otimes \) binary on Z?
Answer: No, the operation \( \otimes \) is not a binary operation on Z.
A binary operation on Z must always produce an integer result when applied to any two integers. However, if we take negative integers for \(n\) (for example, \(n = -k\) where \(k\) is a positive integer), then \(m^n\) becomes \(m^{-k}\), which is equal to \( \frac{1}{m^k} \). This result is often a fraction and not an integer, unless \(m\) is 1 or -1.
For example, if we take \(m = 2\) and \(n = -1\):
\( (2 \otimes -1) = 2^{-1} + (-1)^2 \)
\( = \frac{1}{2^1} + 1 \)
\( = \frac{1}{2} + 1 \)
\( = \frac{3}{2} \)
The result \( \frac{3}{2} \) is not an integer (it's not in Z). Since we found one case where the operation does not produce an integer, it is not a binary operation on Z.
In simple words: If you use negative numbers for \(n\), \(m^n\) can become a fraction. Since fractions are not integers, the operation \( \otimes \) does not always give an integer result, so it's not a binary operation on integers.

🎯 Exam Tip: When checking for binary operations on integers involving powers, always consider negative exponents, as they often lead to fractions which are not integers.

Question 3. Let * be defined on R by \( (a * b) = a + b + ab - 7 \). Is * binary on R? If so, find \( 3 * \left(\frac{-7}{15}\right) \)
Answer: Yes, the operation `*` is a binary operation on R.
Given \( (a * b) = a + b + ab - 7 \). If \(a\) and \(b\) are any two real numbers (R), then \(a+b\), \(ab\), and \(a+b+ab-7\) are all also real numbers. This means the operation is closed within the set of real numbers.
Now, let's find \( 3 * \left(\frac{-7}{15}\right) \):
Using the definition \( (a * b) = a + b + ab - 7 \), with \(a = 3\) and \(b = \frac{-7}{15}\):
\( 3 * \left(\frac{-7}{15}\right) = 3 + \left(\frac{-7}{15}\right) + 3 \left(\frac{-7}{15}\right) - 7 \)
\( = 3 - \frac{7}{15} - \frac{21}{15} - 7 \)
First, combine the whole numbers:
\( = (3 - 7) - \left(\frac{7}{15} + \frac{21}{15}\right) \)
\( = -4 - \frac{28}{15} \)
To subtract, find a common denominator, which is 15:
\( = \frac{-4 \times 15}{1 \times 15} - \frac{28}{15} \)
\( = \frac{-60}{15} - \frac{28}{15} \)
\( = \frac{-60 - 28}{15} \)
\( = \frac{-88}{15} \)
Therefore, \( 3 * \left(\frac{-7}{15}\right) = \frac{-88}{15} \).
In simple words: Since adding, subtracting, and multiplying real numbers always gives a real number, this operation is binary on R. To find the value, simply put the numbers into the given formula and do the math.

🎯 Exam Tip: Clearly state if the operation is binary first, then show the step-by-step calculation. Always double-check your arithmetic, especially with fractions and negative signs.

Question 4. Let \( A = \{a + \sqrt{5} b: a, b \in Z\} \). Check whether the usual multiplication is a binary operation on A.
Answer: Yes, the usual multiplication is a binary operation on \(A\).
For multiplication to be a binary operation on \(A\), if we multiply any two elements from \(A\), the result must also be an element of \(A\).
Let \(X = a + \sqrt{5}b\) and \(Y = c + \sqrt{5}d\) be two elements in \(A\), where \(a, b, c, d\) are integers (Z).
Now, let's multiply \(X\) and \(Y\):
\( X \times Y = (a + \sqrt{5}b)(c + \sqrt{5}d) \)
\( = ac + a(\sqrt{5}d) + (\sqrt{5}b)c + (\sqrt{5}b)(\sqrt{5}d) \)
\( = ac + \sqrt{5}ad + \sqrt{5}bc + 5bd \)
Group the terms without \(\sqrt{5}\) and terms with \(\sqrt{5}\):
\( = (ac + 5bd) + \sqrt{5}(ad + bc) \)
Since \(a, b, c, d\) are integers, their products (\(ac, bd, ad, bc\)) and sums (\(ac + 5bd\), \(ad + bc\)) are also integers. Let \(P = ac + 5bd\) and \(Q = ad + bc\). Both \(P\) and \(Q\) are integers.
So, the product \(X \times Y\) can be written in the form \(P + \sqrt{5}Q\), where \(P, Q \in Z\). This matches the definition of an element in set \(A\). Therefore, multiplication is a binary operation on \(A\).
In simple words: When you multiply two numbers from set \(A\), the answer can also be written in the same special form of \( \text{integer} + \sqrt{5} \times \text{integer} \). This means multiplication is a binary operation for this set.

🎯 Exam Tip: For sets involving irrational numbers like \(\sqrt{5}\), always expand the product carefully and group terms to ensure the result fits the required form for set membership.

Question 5. (i) Define an operation * on Q as follows: \( a * b = \left(\frac{a+b}{2}\right) \); a, b \( \in \) Q. Examine the closure, commutative and associate properties satisfied by * on Q.
(ii) Define an operation * on Q as follows: \( a * b = \left(\frac{a+b}{2}\right) \); a, b \( \in \) Q. Examine the existence of identity and the existence of inverse for the operation * on Q.
Answer:
(i) Let the operation `*` be defined on the set of rational numbers Q by \( a * b = \frac{a+b}{2} \).
1. **Closure Property:**
If \(a\) and \(b\) are rational numbers, then \(a+b\) is also a rational number. Dividing a rational number by 2 (which is a non-zero rational number) also results in a rational number. So, \( \frac{a+b}{2} \in Q \). Thus, `*` is closed on Q. It means the result of the operation always stays within the set Q.
2. **Commutative Property:**
An operation is commutative if \( a * b = b * a \) for all \(a, b \in Q\).
\( a * b = \frac{a+b}{2} \)
\( b * a = \frac{b+a}{2} \)
Since \( a+b = b+a \) (addition of rational numbers is commutative), we have \( \frac{a+b}{2} = \frac{b+a}{2} \).
So, \( a * b = b * a \). Thus, `*` is commutative on Q. The order of numbers does not change the outcome.
3. **Associative Property:**
An operation is associative if \( a * (b * c) = (a * b) * c \) for all \(a, b, c \in Q\).
First, calculate \( a * (b * c) \):
\( b * c = \frac{b+c}{2} \)
\( a * (b * c) = a * \left(\frac{b+c}{2}\right) \)
\( = \frac{a + \left(\frac{b+c}{2}\right)}{2} \)
\( = \frac{\frac{2a + b + c}{2}}{2} \)
\( = \frac{2a+b+c}{4} \)
Next, calculate \( (a * b) * c \):
\( a * b = \frac{a+b}{2} \)
\( (a * b) * c = \left(\frac{a+b}{2}\right) * c \)
\( = \frac{\left(\frac{a+b}{2}\right) + c}{2} \)
\( = \frac{\frac{a+b+2c}{2}}{2} \)
\( = \frac{a+b+2c}{4} \)
Since \( \frac{2a+b+c}{4} \neq \frac{a+b+2c}{4} \) (unless \(a=c\)), the operation `*` is not associative on Q. The way numbers are grouped affects the result.
In simple words: This operation always gives a rational number (closed) and the order doesn't matter (commutative). But, how you group three numbers changes the answer (not associative).

🎯 Exam Tip: When checking associativity, carefully calculate both sides, \(a * (b * c)\) and \((a * b) * c\), and compare the final expressions. A single counterexample is enough to prove non-associativity.

Answer:
(ii) Let the operation `*` be defined on the set of rational numbers Q by \( a * b = \frac{a+b}{2} \).
1. **Existence of Identity Element:**
An identity element \(e\) exists if \( a * e = e * a = a \) for all \(a \in Q\).
Using \( a * e = a \):
\( \frac{a+e}{2} = a \)
\( a+e = 2a \)
\( e = 2a - a \)
\( e = a \)
This shows that the identity element \(e\) depends on \(a\), which means there isn't a single identity element that works for all rational numbers. For an identity element to exist, it must be unique and constant for the entire set. Therefore, a unique identity element does not exist for this operation on Q.
2. **Existence of Inverse Element:**
Since a unique identity element does not exist, an inverse element cannot exist either, as the definition of an inverse requires an identity element. For an inverse \(a'\) to exist, we need \( a * a' = a' * a = e \), where \(e\) is the identity element. As there is no such \(e\), there are no inverse elements.
In simple words: There is no special number that leaves other numbers unchanged when you apply this operation (no identity element). Because there's no identity, there can't be an inverse number that "undoes" the operation.

🎯 Exam Tip: To find an identity element, set \(a * e = a\) (or \(e * a = a\)) and solve for \(e\). If \(e\) depends on \(a\) or is not in the set, a unique identity element does not exist. The existence of an inverse directly depends on the existence of a unique identity.

Question 6. Fill in the following table so that the binary operation * on \( A = \{a, b, c\} \) is commutative.
Answer: To make the binary operation `*` commutative, we must ensure that for any elements \(x, y\) in set \(A\), \(x * y = y * x\). We will fill the table based on this rule.

Explanation of how the table is filled:
Given elements:
- \(a * a = b\)
- \(b * b = b\)
- \(c * c = c\)
Using the commutative property \(x * y = y * x\):
- Since \(b * a = c\) is given (from row b, column a), then \(a * b\) must also be \(c\).
- Since \(c * a = a\) is given (from row c, column a), then \(a * c\) must also be \(a\).
- Since \(b * c = a\) is given (from row b, column c), then \(c * b\) must also be \(a\).
The completed table shows these values, ensuring the operation is commutative.
In simple words: To make the table work for a commutative operation, just make sure that if \( \text{element1} * \text{element2} \) gives an answer, then \( \text{element2} * \text{element1} \) must give the exact same answer. Fill in the missing spots using this rule.

🎯 Exam Tip: For a commutative table, the entries are symmetrical across the main diagonal (from top-left to bottom-right). If you have \(i * j\), its value should be the same as \(j * i\).

Question 7. Consider the binary operation * defined on the set \( A = \{a, b, c, d\} \) by the following table:
Answer: Let's examine the properties of the binary operation `*` on set \( A = \{a, b, c, d\} \) using the given table.

**Check for Commutativity:**
An operation is commutative if \(x * y = y * x\) for all \(x, y\) in the set.
Let's check \(a * b\) and \(b * a\):
From the table, \(a * b = c\).
From the table, \(b * a = d\).
Since \(c \neq d\), the operation is not commutative. The order of elements matters here.

**Check for Associativity:**
An operation is associative if \(x * (y * z) = (x * y) * z\) for all \(x, y, z\) in the set.
Let's test with \(x=a, y=b, z=c\):
First, calculate \( (a * b) * c \):
\( (a * b) = c \) (from table)
So, \( (a * b) * c = c * c \)
From the table, \( c * c = a \).
Thus, \( (a * b) * c = a \).
Next, calculate \( a * (b * c) \):
\( (b * c) = b \) (from table)
So, \( a * (b * c) = a * b \)
From the table, \( a * b = c \).
Thus, \( a * (b * c) = c \).
Since \( (a * b) * c = a \) and \( a * (b * c) = c \), and \( a \neq c \), the operation is not associative. The grouping of elements changes the outcome.
In simple words: This operation is not commutative because changing the order of \(a\) and \(b\) gives different results. It's also not associative because grouping \(a, b, c\) in different ways gives different final answers.

🎯 Exam Tip: For large tables, use specific examples to quickly disprove commutativity (check \(a*b\) vs \(b*a\)) or associativity (check \((a*b)*c\) vs \(a*(b*c)\)). You only need one counterexample for each property.

Question 8. Let
\[ A = \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}, \quad C = \begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{pmatrix} \]
be any three boolean matrices of the same type. Find
(i) \( A \lor B \)
(ii) \( A \land B \)
(iii) \( (A \lor B) \land C \)
(iv) \( (A \land B) \lor C \).
Answer: We will perform the boolean matrix operations as requested.

(i) To find \( A \lor B \), we apply the logical OR operation element-wise:
\[ A \lor B = \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix} \lor \begin{pmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 1 \lor 0 & 0 \lor 1 & 1 \lor 0 & 0 \lor 1 \\ 0 \lor 1 & 1 \lor 0 & 0 \lor 1 & 1 \lor 0 \\ 1 \lor 1 & 0 \lor 0 & 0 \lor 0 & 1 \lor 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix} \]

(ii) To find \( A \land B \), we apply the logical AND operation element-wise:
\[ A \land B = \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix} \land \begin{pmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 1 \land 0 & 0 \land 1 & 1 \land 0 & 0 \land 1 \\ 0 \land 1 & 1 \land 0 & 0 \land 1 & 1 \land 0 \\ 1 \land 1 & 0 \land 0 & 0 \land 0 & 1 \land 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} \]

(iii) To find \( (A \lor B) \land C \), we use the result from part (i) and apply logical AND with \(C\):
\[ (A \lor B) \land C = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix} \land \begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 1 \land 1 & 1 \land 1 & 1 \land 0 & 1 \land 1 \\ 1 \land 0 & 1 \land 1 & 1 \land 1 & 1 \land 0 \\ 1 \land 1 & 0 \land 1 & 0 \land 1 & 1 \land 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} \]

(iv) To find \( (A \land B) \lor C \), we use the result from part (ii) and apply logical OR with \(C\):
\[ (A \land B) \lor C = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} \lor \begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 0 \lor 1 & 0 \lor 1 & 0 \lor 0 & 0 \lor 1 \\ 0 \lor 0 & 0 \lor 1 & 0 \lor 1 & 0 \lor 0 \\ 1 \lor 1 & 0 \lor 1 & 0 \lor 1 & 1 \lor 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{pmatrix} \]
In simple words: For boolean matrices, 'OR' means if at least one cell has a '1', the result is '1'. 'AND' means both cells must have a '1' for the result to be '1'. Just apply these rules for each matching position in the matrices.

🎯 Exam Tip: Remember the basic rules for boolean operations: for OR (\(\lor\)), \(0 \lor 0 = 0\), all others are 1. For AND (\(\land\)), \(1 \land 1 = 1\), all others are 0. Work element-by-element for each matrix position.

Question 9. (i) Let \( M = \left\{\begin{pmatrix} x & x \\ x & x \end{pmatrix} : x \in R-\{0\}\right\} \) and let * be the matrix multiplication. Determine whether M is closed under *. If so, examine the commutative and associative properties satisfied by * on M.
(ii) Let \( M = \left\{\begin{pmatrix} x & x \\ x & x \end{pmatrix} : x \in R-\{0\}\right\} \) and let * be the matrix multiplication. Determine whether M is closed under *. If so, examine the existence of identity, existence of inverse properties for the operation * on M
Answer:
(i) Given set \( M = \left\{\begin{pmatrix} x & x \\ x & x \end{pmatrix} : x \in R-\{0\}\right\} \) and the operation `*` is matrix multiplication.
1. **Closure Property:**
Let \( A = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \) and \( B = \begin{pmatrix} y & y \\ y & y \end{pmatrix} \) be two elements in \(M\), where \(x, y \in R-\{0\}\) (meaning \(x\) and \(y\) are non-zero real numbers).
Now, let's find the product \( A * B \):
\[ A * B = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \begin{pmatrix} y & y \\ y & y \end{pmatrix} = \begin{pmatrix} (x \cdot y + x \cdot y) & (x \cdot y + x \cdot y) \\ (x \cdot y + x \cdot y) & (x \cdot y + x \cdot y) \end{pmatrix} \]
\[ = \begin{pmatrix} 2xy & 2xy \\ 2xy & 2xy \end{pmatrix} \]
Since \(x \in R-\{0\}\) and \(y \in R-\{0\}\), their product \(xy\) is also a non-zero real number. Therefore, \(2xy\) is also a non-zero real number. The resulting matrix \( \begin{pmatrix} 2xy & 2xy \\ 2xy & 2xy \end{pmatrix} \) is of the form \( \begin{pmatrix} z & z \\ z & z \end{pmatrix} \) where \(z = 2xy \in R-\{0\}\). So, \(A * B \in M\). Thus, `*` is closed on \(M\). The product matrix remains in the set.

2. **Commutative Property:**
For commutativity, we need \( A * B = B * A \). We already found \( A * B = \begin{pmatrix} 2xy & 2xy \\ 2xy & 2xy \end{pmatrix} \).
Now let's find \( B * A \):
\[ B * A = \begin{pmatrix} y & y \\ y & y \end{pmatrix} \begin{pmatrix} x & x \\ x & x \end{pmatrix} = \begin{pmatrix} (y \cdot x + y \cdot x) & (y \cdot x + y \cdot x) \\ (y \cdot x + y \cdot x) & (y \cdot x + y \cdot x) \end{pmatrix} \]
\[ = \begin{pmatrix} 2yx & 2yx \\ 2yx & 2yx \end{pmatrix} \]
Since \(xy = yx\) for real numbers, \(2xy = 2yx\). Therefore, \( A * B = B * A \). Thus, `*` is commutative on \(M\). The order of multiplication does not change the result.

3. **Associative Property:**
Matrix multiplication is generally associative. For any three matrices \(A, B, C\) in \(M\), we know that \( (A * B) * C = A * (B * C) \). This property holds true for elements of \(M\) as well because matrix multiplication itself is an associative operation.
Thus, `*` is associative on \(M\). Grouping of matrices in multiplication does not affect the final product.
In simple words: This matrix set is closed under multiplication because the result stays in the set. It's also commutative because the order of multiplication doesn't change the answer. And matrix multiplication is always associative, so this set follows that rule too.

🎯 Exam Tip: When proving closure for matrices, perform the multiplication and ensure each element of the resulting matrix satisfies the conditions for membership in the set. For commutativity, calculate both \(AB\) and \(BA\). Associativity of matrix multiplication is a standard property that usually holds.

Answer:
(ii) Given set \( M = \left\{\begin{pmatrix} x & x \\ x & x \end{pmatrix} : x \in R-\{0\}\right\} \) and the operation `*` is matrix multiplication.
1. **Existence of Identity Element:**
An identity element \(E \in M\) must satisfy \( A * E = E * A = A \) for all \( A \in M \).
Let \( E = \begin{pmatrix} e & e \\ e & e \end{pmatrix} \) where \( e \in R-\{0\} \).
We need \( A * E = A \):
\[ \begin{pmatrix} x & x \\ x & x \end{pmatrix} \begin{pmatrix} e & e \\ e & e \end{pmatrix} = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \]
\[ \begin{pmatrix} 2xe & 2xe \\ 2xe & 2xe \end{pmatrix} = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \]
Comparing the elements, we get \( 2xe = x \). Since \(x \in R-\{0\}\), we can divide by \(x\):
\( 2e = 1 \)
\( e = \frac{1}{2} \)
Since \( \frac{1}{2} \in R-\{0\} \), an identity element exists and is \( E = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix} \). This matrix is indeed in \(M\).

2. **Existence of Inverse Element:**
An inverse element \(B \in M\) for \(A \in M\) must satisfy \( A * B = B * A = E \), where \(E\) is the identity element found above.
Let \( A = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \) and its inverse be \( B = \begin{pmatrix} y & y \\ y & y \end{pmatrix} \), where \(y \in R-\{0\}\).
We need \( A * B = E \):
\[ \begin{pmatrix} x & x \\ x & x \end{pmatrix} \begin{pmatrix} y & y \\ y & y \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix} \]
\[ \begin{pmatrix} 2xy & 2xy \\ 2xy & 2xy \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix} \]
Comparing the elements, we get \( 2xy = \frac{1}{2} \).
\( y = \frac{1}{4x} \)
Since \(x \in R-\{0\}\), \(4x\) is also a non-zero real number, so \( \frac{1}{4x} \) is a non-zero real number. Thus, \(y = \frac{1}{4x} \in R-\{0\}\).
Therefore, the inverse element for \( A = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \) is \( B = \begin{pmatrix} 1/(4x) & 1/(4x) \\ 1/(4x) & 1/(4x) \end{pmatrix} \), which exists and is in \(M\).
In simple words: This set has a special identity matrix that doesn't change other matrices when multiplied. Also, every matrix in the set has an inverse matrix that, when multiplied, gives back the identity matrix.

🎯 Exam Tip: To find the identity matrix, assume a generic identity matrix of the required form, multiply it with a generic element from the set, and equate it to the generic element. Solve for the parameters of the identity matrix. Do the same for the inverse, equating the product to the identity matrix.

Question 10. (i) Let A be Q\{1} Define * on A by \( x * y = x + y - xy \). Is * binary on A ? If so, examine the commutative and associative properties satisfied by * on A.
(ii) Let A be Q\{1}. Define * on A by \( x * y = x + y - xy \). Is * binary on A ? If so, examine the existence of identity, the existence of inverse properties for the operation * on A.
Answer:
(i) Given set \( A = Q-\{1\} \) (rational numbers excluding 1) and the operation \( x * y = x + y - xy \).
1. **Closure Property:**
For `*` to be binary on \(A\), if \(x, y \in A\), then \(x * y\) must also be in \(A\). This means \(x * y \neq 1\).
Assume for a moment that \(x * y = 1\):
\( x + y - xy = 1 \)
Rearrange the terms:
\( xy - x - y + 1 = 0 \)
This equation can be factored as:
\( (x-1)(y-1) = 0 \)
This implies either \(x-1 = 0\) or \(y-1 = 0\), which means \(x = 1\) or \(y = 1\).
However, we are given that \(x, y \in Q-\{1\}\), meaning \(x \neq 1\) and \(y \neq 1\). Therefore, \( (x-1)(y-1) \neq 0 \), which implies \( x * y \neq 1 \).
So, if \(x, y \in A\), then \(x * y \in A\). Thus, `*` is closed on \(A\). The result of the operation never becomes 1.

2. **Commutative Property:**
An operation is commutative if \( x * y = y * x \) for all \(x, y \in A\).
\( x * y = x + y - xy \)
\( y * x = y + x - yx \)
Since addition and multiplication of rational numbers are commutative (\(x+y = y+x\) and \(xy = yx\)), we have \( x + y - xy = y + x - yx \).
Therefore, \( x * y = y * x \). Thus, `*` is commutative on \(A\). The order of the numbers does not matter.

3. **Associative Property:**
An operation is associative if \( x * (y * z) = (x * y) * z \) for all \(x, y, z \in A\).
First, calculate \( x * (y * z) \):
\( y * z = y + z - yz \)
Now, substitute this into the expression for \( x * (y * z) \):
\( x * (y * z) = x + (y+z-yz) - x(y+z-yz) \)
\( = x + y + z - yz - xy - xz + xyz \) (Equation 1)
Next, calculate \( (x * y) * z \):
\( x * y = x + y - xy \)
Now, substitute this into the expression for \( (x * y) * z \):
\( (x * y) * z = (x+y-xy) + z - (x+y-xy)z \)
\( = x + y - xy + z - xz - yz + xyz \) (Equation 2)
Comparing Equation 1 and Equation 2, we see that they are identical.
Therefore, \( x * (y * z) = (x * y) * z \). Thus, `*` is associative on \(A\). Grouping does not affect the result.
In simple words: This operation always gives a number that is not 1 (closed). The order of numbers does not change the answer (commutative). Also, how you group three numbers when doing the operation doesn't change the final answer (associative).

🎯 Exam Tip: For closure, factor the expression to show that \(x*y=1\) only if \(x=1\) or \(y=1\), which are excluded from the set. For associativity, carefully expand both sides of the equation and ensure all terms match perfectly.

Answer:
(ii) Given set \( A = Q-\{1\} \) and the operation \( x * y = x + y - xy \).
1. **Existence of Identity Element:**
An identity element \(e \in A\) must satisfy \( x * e = e * x = x \) for all \(x \in A\).
Using \( x * e = x \):
\( x + e - xe = x \)
Subtract \(x\) from both sides:
\( e - xe = 0 \)
Factor out \(e\):
\( e(1 - x) = 0 \)
Since \(x \in Q-\{1\}\), we know that \(x \neq 1\), so \( (1 - x) \neq 0 \). Therefore, we can divide by \( (1-x) \):
\( e = \frac{0}{1-x} \)
\( e = 0 \)
The identity element is \(e = 0\). Since \(0\) is a rational number and \(0 \neq 1\), \(0 \in A\). Thus, a unique identity element exists.

2. **Existence of Inverse Element:**
An inverse element \(x' \in A\) for \(x \in A\) must satisfy \( x * x' = x' * x = e \), where \(e=0\) is the identity element.
Using \( x * x' = 0 \):
\( x + x' - xx' = 0 \)
Factor out \(x'\) from the terms involving \(x'\):
\( x + x'(1 - x) = 0 \)
Subtract \(x\) from both sides:
\( x'(1 - x) = -x \)
Since \(x \in Q-\{1\}\), \( (1 - x) \neq 0 \), so we can divide by \( (1-x) \):
\( x' = \frac{-x}{1-x} \)
This can also be written as \( x' = \frac{x}{x-1} \).
We need to check if \(x' \in A\), which means \(x'\) must be a rational number and \(x' \neq 1\).
Since \(x\) is a rational number and \(x \neq 1\), \(x-1\) is also a non-zero rational number. Therefore, \( \frac{x}{x-1} \) is a rational number.
Now, let's check if \(x' = 1\):
If \( \frac{x}{x-1} = 1 \), then \( x = x-1 \), which simplifies to \( 0 = -1 \). This is false.
Therefore, \(x' \neq 1\). So, for every \(x \in A\), its inverse \(x' = \frac{x}{x-1}\) exists and is also in \(A\).
In simple words: For this operation, 0 is the special number that acts as the identity, meaning \(x * 0 = x\). Also, every number \(x\) in the set has an inverse number \(x'\) such that \(x * x' = 0\), and this inverse number is never 1.

🎯 Exam Tip: To find the identity element, set the operation to equal the original element (\(x*e=x\)). For the inverse, set the operation to equal the identity element (\(x*x'=e\)). Always verify that the found identity and inverse elements belong to the defined set \(A\).