Question

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

Hint:

When solving matrix equations of the form $RS = T$, remember that multiplication is not commutative. You must pre-multiply by $R^{-1}$ on both sides. If the equation was $SR = T$, you would post-multiply by $R^{-1}$ on both sides.

Answer 1

Concept:
Given: \[ PQ-RS=O \] we get: \[ RS=PQ \] To find \(S\), pre-multiply both sides by \(R^{-1}\): \[ S=R^{-1}PQ \] 
Step 1: Calculate \(PQ\)
Given: \[ P= \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}, \qquad Q= \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} \] Therefore: \[ PQ= \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} \] \[ PQ= \begin{bmatrix} 2(5)+(-1)(7) & 2(2)+(-1)(4) \\ 3(5)+4(7) & 3(2)+4(4) \end{bmatrix} \] \[ PQ= \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} \] 
Step 2: Find \(R^{-1}\)
Given: \[ R= \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \] Its determinant is: \[ |R|=(2)(8)-(5)(3) \] \[ |R|=16-15=1 \] Therefore: \[ R^{-1} = \frac{1}{|R|} \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} \] \[ R^{-1} = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} \] 
Step 3: Calculate \(S\)
Using: \[ S=R^{-1}PQ \] \[ S= \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} \] \[ S= \begin{bmatrix} 8(3)+(-5)(43) & 8(0)+(-5)(22) \\ (-3)(3)+2(43) & (-3)(0)+2(22) \end{bmatrix} \] \[ S= \begin{bmatrix} 24-215 & -110 \\ -9+86 & 44 \end{bmatrix} \] \[ S= \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix} \] 
Final Answer:
Therefore, \[ \boxed{ S= \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix} } \]