(b) Given that $$P = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}, Q = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} \text{ and } R = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix},$$ find a matrix $S$ such that $PQ - RS$ is a null matrix. (matrix, null matrix, matrix multiplication, matrix subtraction)

Step 1: Set up the equation

We are given that the difference between the product of matrices $P$ and $Q$, and the product of matrices $R$ and $S$ is a null matrix. A null matrix, denoted by $O$, is a matrix where all elements are zero. We can rearrange this equation to solve for $S$ after calculating $PQ$ and finding the inverse of $R$.

Step 2: Calculate the product PQ

To find the product of two matrices, we multiply the rows of the first matrix by the columns of the second matrix. Each element in the resulting matrix is the sum of the products of the corresponding elements from the row and column. For example, the element in the first row and first column of $PQ$ is $(2)(5) + (-1)(7) = 10 - 7 = 3$.

Step 3: Find the determinant of R

To find the inverse of matrix $R$, we first need to calculate its determinant. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is calculated as $ad - bc$. We substitute the values from matrix $R$ to find its determinant.

Step 4: Find the inverse of R (R⁻¹)

The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. Since the determinant of $R$ is $1$, its inverse is simply the adjugate matrix.

Step 5: Solve for S

To find matrix $S$, we pre-multiply both sides of the equation $RS = PQ$ by $R^{-1}$. Since $R^{-1}R = I$ (the identity matrix) and $IS = S$, we get $S = R^{-1}PQ$. Then we perform the matrix multiplication of $R^{-1}$ and $PQ$ to find the elements of $S$.

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