Calculate the range of the data set and the number of ways to put 4 books into 6 bags. Also, define a matrix.

4c) iv) Range

Step 1: Identify the maximum and minimum values in the data set.
The ordered marks are: 60, 65, 65, 70, 70, 70, 75, 80, 85, 90.
Maximum value = 90
Minimum value = 60

Step 2: Calculate the range.

$$\text{Range} = \text{Maximum value} - \text{Minimum value}$$ $$\text{Range} = 90 - 60 = 30$$

The range is $\boxed{30}$.

5a) To find the number of ways 4 books can be put into 6 bags, with not more than one book in each bag, we are essentially selecting 4 bags out of 6 and arranging the 4 distinct books in them. This is a permutation problem.

Step 1: Identify the number of items to arrange ($n$) and the number of positions ($r$).
Number of bags available = 6
Number of books to place = 4
Since each book is distinct and each bag is distinct, and the order of placing books into specific bags matters, we use permutations.

Step 2: Apply the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$.
Here, $n=6$ (bags) and $r=4$ (books).

$$P(6, 4) = \frac{6!}{(6-4)!} = \frac{6!}{2!}$$ $$P(6, 4) = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$$ $$P(6, 4) = 6 \times 5 \times 4 \times 3$$ $$P(6, 4) = 30 \times 12$$ $$P(6, 4) = 360$$

There are $\boxed{360}$ different ways.

5b) A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. It is used to represent linear transformations, systems of linear equations, and data.

5c)

• i) Square Matrix: A square matrix is a matrix that has an equal number of rows and columns. Its dimension is typically denoted as $n \times n$.
• ii) Diagonal Matrix: A diagonal matrix is a square matrix where all the elements outside of the main diagonal are zero. The elements on the main diagonal can be any value.
• iii) Symmetric Matrix: A symmetric matrix is a square matrix that is equal to its own transpose ($A = A^T$). This means that the element in row $i$ and column $j$ is equal to the element in row $j$ and column $i$ ($a_{ij} = a_{ji}$).
• iv) Singular Matrix: A singular matrix is a square matrix whose determinant is equal to zero. A singular matrix does not have an inverse.

5d) Given matrices $A = \begin{bmatrix} 1 & -2 & 3 \\ 4 & 5 & -6 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 0 & 2 \\ -7 & 1 & 8 \end{bmatrix}$.

i) Find $A+B$

Step 1: Add the corresponding elements of matrix A and matrix B.

$$A+B = \begin{bmatrix} 1+3 & -2+0 & 3+2 \\ 4+(-7) & 5+1 & -6+8 \end{bmatrix}$$

Step 2: Perform the additions.

$$A+B = \boxed{\begin{bmatrix} 4 & -2 & 5 \\ -3 & 6 & 2 \end{bmatrix}}$$

ii) Find $A-B$

Step 1: Subtract the corresponding elements of matrix B from matrix A.

$$A-B = \begin{bmatrix} 1-3 & -2-0 & 3-2 \\ 4-(-7) & 5-1 & -6-8 \end{bmatrix}$$

Step 2: Perform the subtractions.

$$A-B = \boxed{\begin{bmatrix} -2 & -2 & 1 \\ 11 & 4 & -14 \end{bmatrix}}$$

6) Find the inverse of Matrix A
Given matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 1 & 5 \\ 6 & 0 & 2 \end{bmatrix}$.
The inverse of a matrix $A$ is given by $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$.

Step 1: Calculate the determinant of A.

$$\det(A) = 1 \begin{vmatrix} 1 & 5 \\ 0 & 2 \end{vmatrix} - 2 \begin{vmatrix} 4 & 5 \\ 6 & 2 \end{vmatrix} + 3 \begin{vmatrix} 4 & 1 \\ 6 & 0 \end{vmatrix}$$ $$\det(A) = 1((1)(2) - (5)(0)) - 2((4)(2) - (5)(6)) + 3((4)(0) - (1)(6))$$ $$\det(A) = 1(2 - 0) - 2(8 - 30) + 3(0 - 6)$$ $$\det(A) = 1(2) - 2(-22) + 3(-6)$$ $$\det(A) = 2 + 44 - 18$$ $$\det(A) = 28$$

Step 2: Calculate the cofactor matrix $C$.

$$C_{11} = \begin{vmatrix} 1 & 5 \\ 0 & 2 \end{vmatrix} = 2 - 0 = 2$$ $$C_{12} = -\begin{vmatrix} 4 & 5 \\ 6 & 2 \end{vmatrix} = -(8 - 30) = -(-22) = 22$$ $$C_{13} = \begin{vmatrix} 4 & 1 \\ 6 & 0 \end{vmatrix} = 0 - 6 = -6$$ $$C_{21} = -\begin{vmatrix} 2 & 3 \\ 0 & 2 \end{vmatrix} = -(4 - 0) = -4$$ $$C_{22} = \begin{vmatrix} 1 & 3 \\ 6 & 2 \end{vmatrix} = 2 - 18 = -16$$ $$C_{23} = -\begin{vmatrix} 1 & 2 \\ 6 & 0 \end{vmatrix} = -(0 - 12) = -(-12) = 12$$ $$C_{31} = \begin{vmatrix} 2 & 3 \\ 1 & 5 \end{vmatrix} = 10 - 3 = 7$$ $$C_{32} = -\begin{vmatrix} 1 & 3 \\ 4 & 5 \end{vmatrix} = -(5 - 12) = -(-7) = 7$$ $$C_{33} = \begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} = 1 - 8 = -7$$

The cofactor matrix is:

$$C = \begin{bmatrix} 2 & 22 & -6 \\ -4 & -16 & 12 \\ 7 & 7 & -7 \end{bmatrix}$$

Step 3: Calculate the adjoint matrix, which is the transpose of the cofactor matrix.

$$\text{adj}(A) = C^T = \begin{bmatrix} 2 & -4 & 7 \\ 22 & -16 & 7 \\ -6 & 12 & -7 \end{bmatrix}$$

Step 4: Calculate the inverse matrix.

$$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$$ $$$$