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.
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.
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Since $|R| = 1$, finding the inverse requires no fractional scaling. When solving equations of the form $RS = B$, make sure to pre-multiply by $R^{-1}$ on the left hand side, because matrix multiplication is generally non-commutative ($R^{-1}B \neq BR^{-1}$).
Updated On: Jul 4, 2026
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Solution and Explanation
Concept:
We are given the matrix equation
\[
PQ-RS=O,
\]
where \(O\) is the null matrix.
Rearranging,
\[
RS=PQ.
\]
Since \(R\) is non-singular, pre-multiply both sides by \(R^{-1}\):
\[
R^{-1}RS=R^{-1}PQ
\]
\[
\boxed{S=R^{-1}PQ.}
\]
Step 1: Compute the matrix product \(PQ\).
\[ P= \begin{bmatrix} 2 & -1\\ 3 & 4 \end{bmatrix}, \qquad Q= \begin{bmatrix} 5 & 2\\ 7 & 4 \end{bmatrix}. \] \[ PQ= \begin{bmatrix} 2 & -1\\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2\\ 7 & 4 \end{bmatrix} = \begin{bmatrix} (2)(5)+(-1)(7) & (2)(2)+(-1)(4)\\ (3)(5)+(4)(7) & (3)(2)+(4)(4) \end{bmatrix}. \] \[ PQ= \begin{bmatrix} 10-7 & 4-4\\ 15+28 & 6+16 \end{bmatrix} = \begin{bmatrix} 3 & 0\\ 43 & 22 \end{bmatrix}. \]
Step 2: Find the determinant and verify that \(R\) is invertible.
\[ R= \begin{bmatrix} 2 & 5\\ 3 & 8 \end{bmatrix}. \] Its determinant is \[ |R|=(2)(8)-(5)(3)=16-15=1. \] Since \[ |R|=1\ne0, \] the inverse of \(R\) exists.
Step 3: Find the inverse of \(R\).
Using the inverse formula for a \(2\times2\) matrix, \[ R^{-1} = \frac{1}{|R|} \begin{bmatrix} 8 & -5\\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 8 & -5\\ -3 & 2 \end{bmatrix}. \]
Step 4: Calculate \(S=R^{-1}PQ\).
\[ S= \begin{bmatrix} 8 & -5\\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0\\ 43 & 22 \end{bmatrix}. \] Multiplying, \[ S= \begin{bmatrix} (8)(3)+(-5)(43) & (8)(0)+(-5)(22)\\ (-3)(3)+(2)(43) & (-3)(0)+(2)(22) \end{bmatrix}. \] \[ S= \begin{bmatrix} 24-215 & 0-110\\ -9+86 & 0+44 \end{bmatrix} = \begin{bmatrix} -191 & -110\\ 77 & 44 \end{bmatrix}. \]
Step 1: Compute the matrix product \(PQ\).
\[ P= \begin{bmatrix} 2 & -1\\ 3 & 4 \end{bmatrix}, \qquad Q= \begin{bmatrix} 5 & 2\\ 7 & 4 \end{bmatrix}. \] \[ PQ= \begin{bmatrix} 2 & -1\\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2\\ 7 & 4 \end{bmatrix} = \begin{bmatrix} (2)(5)+(-1)(7) & (2)(2)+(-1)(4)\\ (3)(5)+(4)(7) & (3)(2)+(4)(4) \end{bmatrix}. \] \[ PQ= \begin{bmatrix} 10-7 & 4-4\\ 15+28 & 6+16 \end{bmatrix} = \begin{bmatrix} 3 & 0\\ 43 & 22 \end{bmatrix}. \]
Step 2: Find the determinant and verify that \(R\) is invertible.
\[ R= \begin{bmatrix} 2 & 5\\ 3 & 8 \end{bmatrix}. \] Its determinant is \[ |R|=(2)(8)-(5)(3)=16-15=1. \] Since \[ |R|=1\ne0, \] the inverse of \(R\) exists.
Step 3: Find the inverse of \(R\).
Using the inverse formula for a \(2\times2\) matrix, \[ R^{-1} = \frac{1}{|R|} \begin{bmatrix} 8 & -5\\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 8 & -5\\ -3 & 2 \end{bmatrix}. \]
Step 4: Calculate \(S=R^{-1}PQ\).
\[ S= \begin{bmatrix} 8 & -5\\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0\\ 43 & 22 \end{bmatrix}. \] Multiplying, \[ S= \begin{bmatrix} (8)(3)+(-5)(43) & (8)(0)+(-5)(22)\\ (-3)(3)+(2)(43) & (-3)(0)+(2)(22) \end{bmatrix}. \] \[ S= \begin{bmatrix} 24-215 & 0-110\\ -9+86 & 0+44 \end{bmatrix} = \begin{bmatrix} -191 & -110\\ 77 & 44 \end{bmatrix}. \]
Hence,
\[ \boxed{ S= \begin{bmatrix} -191 & -110\\ 77 & 44 \end{bmatrix} } \]Was this answer helpful?
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