(I) We can represent the given situation in the form of a matrix equation as follows:
Let X = [x1, x2, x3] be the column vector representing the cost of one pen, one bag, and one instrument box respectively.
Then we can write the following matrix equation:
5X + 3X + 1X = [160, 190, 250]
This can be simplified to:
9X = [160, 190, 250]
Now we can represent this equation in the form of AX = B, where:
A = [9, 9, 9; 5, 3, 1; 2, 1, 3] is the coefficient matrix representing the number of pens, bags, and instrument boxes purchased by each person.
X = [x1, x2, x3] is the column vector representing the cost of one pen, one bag, and one instrument box respectively.
B = [160, 190, 250] is the column vector representing the total cost paid by each person.
Therefore, the matrix equation is AX = B:
[9 9 9; 5 3 1; 2 1 3] [x1; x2; x3] = [160; 190; 250]
(II) To find |A|, we need to calculate the determinant of matrix A. We can use any method to find the determinant, such as expanding along the first row or first column. Here, we will expand along the first row:
|A| = 9 |[3 1; 1 3]| – 9 |[5 1; 1 3]| + 9 |[5 3; 2 1]| = 9(8) – 9(14) + 9(-13) = -81
Therefore, |A| = -81.
(III) To find the inverse of matrix A, we need to use the formula:
A-1 = (1/|A|) adj(A)
where adj(A) is the adjugate or classical adjoint of matrix A, which is obtained by taking the transpose of the matrix of cofactors of A.
First, we need to find the matrix of cofactors of A. The (i,j)-cofactor of A is given by (-1)^(i+j) times the determinant of the submatrix obtained by deleting the i-th row and j-th column of A. Therefore, we have:
C11 = [3 1; 1 3] => C11 = (33) – (11) = 8 C12 = -[5 1; 2 3] => C12 = -((53) – (12)) = -13 C13 = [5 3; 2 1] => C13 = (51) – (32) = -1 C21 = -[5 1; 1 3] => C21 = -((53) – (11)) = -14 C22 = [9 1; 2 3] => C22 = (93) – (12) = 25 C23 = -[9 3; 2 1] => C23 = -((91) – (32)) = -3 C31 = [5 3; 1 3] => C31 = (53) – (31) = 12 C32 = -[9 3; 1 3]
