This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here's the solution to question 4, parts a) and b): Let t represent the cost of one textbook and e represent the cost of one exercise book. From the previous problem, the system of linear equations is: 25t + 35e = 13500 21t + 38e = 12200 a) Form a matrix equation to represent the above information. Step 1: Identify the coefficient matrix, variable matrix, and constant matrix. The coefficients of t and e form the coefficient matrix. The variables t and e form the variable matrix. The constant terms form the constant matrix. The matrix equation is: 25 & 35 \\ 21 & 38 t \\ e = 13500 \\ 12200 b) Use the matrix method to determine the price of each item. Step 1: Find the determinant of the coefficient matrix. Let A = 25 & 35 \\ 21 & 38 . The determinant of A is det(A) = (25)(38) - (35)(21). det(A) = 950 - 735 det(A) = 215 Step 2: Find the inverse of the coefficient matrix. The inverse of a 2 × 2 matrix a & b \\ c & d is (1)/(ad-bc) d & -b \\ -c & a . A^-1 = (1)/(215) 38 & -35 \\ -21 & 25 Step 3: Multiply the inverse matrix by the constant matrix to find the values of t and e. t \\ e = A^-1 13500 \\ 12200 t \\ e = (1)/(215) 38 & -35 \\ -21 & 25 13500 \\ 12200 Perform the matrix multiplication: t \\ e = (1)/(215) (38)(13500) + (-35)(12200) \\ (-21)(13500) + (25)(12200) t \\ e = (1)/(215) 513000 - 427000 \\ -283500 + 305000 t \\ e = (1)/(215) 86000 \\ 21500 t \\ e = (86000)/(215) \\ (21500)/(215) t \\ e = 400 \\ 100 Step 4: State the price of each item. The cost of one textbook (t) is sh. 400. The cost of one exercise book (e) is sh. 100. The price of one textbook is sh. 400. The price of one exercise book is sh. 100. Last free one today — make it count tomorrow, or type /upgrade for unlimited.