Calculate the determinant of A.

a) Given matrix $A = \begin{pmatrix} 3 & 4 \\ 2 & 3 \end{pmatrix}$. To find the inverse of a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the formula is $A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. Step 1: Calculate the determinant of $A$. $$ \det(A) = (3)(3) - (4)(2) = 9 - 8 = 1 $$ Step 2: Apply the inverse formula. $$ A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -4 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ -2 & 3 \end{pmatrix} $$ The inverse matrix is $\boxed{\begin{pmatrix} 3 & -4 \\ -2 & 3 \end{pmatrix}}$. b) Let $x$ be the price of one bag of beans and $y$ be the price of one bag of rice. From the given information, we can form a system of linear equations: For School A: $60x + 80y = 240000$ For School B: $150x + 225y = 637500$ Step 1: Write the system of equations in matrix form $M \begin{pmatrix} x \\ y \end{pmatrix} = C$. $$ \begin{pmatrix} 60 & 80 \\ 150 & 225 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 240000 \\ 637500 \end{pmatrix} $$ Step 2: Find the determinant of the coefficient matrix $M = \begin{pmatrix} 60 & 80 \\ 150 & 225 \end{pmatrix}$. $$ \det(M) = (60)(225) - (80)(150) = 13500 - 12000 = 1500 $$ Step 3: Find the inverse of the coefficient matrix $M^{-1}$. $$ M^{-1} = \frac{1}{1500} \begin{pmatrix} 225 & -80 \\ -150 & 60 \end{pmatrix} $$ Step 4: Multiply $M^{-1}$ by the constant matrix $C$ to find $\begin{pmatrix} x \\ y \end{pmatrix}$. $$ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{1500} \begin{pmatrix} 225 & -80 \\ -150 & 60 \end{pmatrix} \begin{pmatrix} 240000 \\ 637500 \end{pmatrix} $$ $$ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{1500} \begin{pmatrix} (225)(240000) + (-80)(637500) \\ (-150)(240000) + (60)(637500) \end{pmatrix} $$ $$ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{1500} \begin{pmatrix} 54000000 - 51000000 \\ -36000000 + 38250000 \end{pmatrix} $$ $$ \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{1500} \begin{pmatrix} 3000000 \\ 2250000 \end{pmatrix} $$ $$ x = \frac{3000000}{1500} = 2000 $$ $$ y = \frac{2250000}{1500} = 1500 $$ The price of one bag of beans is Kshs. 2000 and the price of one bag of rice is Kshs. 1500. c) Step 1: Calculate the new price of beans. The price of beans decreased in the ratio 4:5. New price of beans = $\frac{4}{5} \times 2000 = 4 \times 400 = 1600$ Kshs. Step 2: Calculate the new price of rice. The price of rice increased by 20%. New price of rice = $1500 \times (1 + 0.20) = 1500 \times 1.20 = 1800$ Kshs. Step 3: Calculate the total amount paid by the businessman. The businessman bought 20 bags of beans and 30 bags of rice. Cost of beans = $20 \times 1600 = 32000$ Kshs. Cost of rice = $30 \times 1800 = 54000$ Kshs. Total amount paid = $32000 + 54000 = 86000$ Kshs. The businessman paid Kshs. 86,000. Send me the next one 📸