Question

Let \[ A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}, B = \begin{pmatrix} 5 & 2 \\ 7 & 4 \end{pmatrix}, C = \begin{pmatrix} 2 & 5 \\ 3 & 8 \end{pmatrix}. \] Then find a matrix D such that \( CD - AB = 0 \).

Hint:

When solving matrix equations like \( CD = AB \), first compute \( AB \), then solve the system of equations formed by multiplying \( C \) with \( D \).

Answer 1

We are asked to find a matrix \( D \) such that: \[ CD - AB = 0 \] This can be rewritten as: \[ CD = AB \] We need to calculate \( AB \) and then solve for \( D \).

Step 1: Calculate \( AB \).
First, compute the product \( AB \): \[ AB = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 5 & 2 \\ 7 & 4 \end{pmatrix} \] Using matrix multiplication: \[ AB = \begin{pmatrix} (2 \times 5 + (-1) \times 7) & (2 \times 2 + (-1) \times 4) \\ (3 \times 5 + 4 \times 7) & (3 \times 2 + 4 \times 4) \end{pmatrix} \] \[ AB = \begin{pmatrix} 10 - 7 & 4 - 4 \\ 15 + 28 & 6 + 16 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 43 & 22 \end{pmatrix} \]

Step 2: Solve for \( D \).
Now, we know \( CD = AB \), so we have the equation: \[ \begin{pmatrix} 2 & 5 \\ 3 & 8 \end{pmatrix} D = \begin{pmatrix} 3 & 0 \\ 43 & 22 \end{pmatrix} \] Let \( D = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \). Now, multiply \( C \) with \( D \): \[ \begin{pmatrix} 2 & 5 \\ 3 & 8 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{pmatrix} \] Equating this with \( AB \), we get the system of equations: \[ 2a + 5c = 3, 2b + 5d = 0 \] \[ 3a + 8c = 43, 3b + 8d = 22 \]

Step 3: Solve the system of equations.
From the first equation, solve for \( a \): \[ 2a + 5c = 3 \implies a = \frac{3 - 5c}{2} \] Substitute \( a = \frac{3 - 5c}{2} \) into the second equation: \[ 3\left(\frac{3 - 5c}{2}\right) + 8c = 43 \] Simplifying: \[ \frac{9 - 15c}{2} + 8c = 43 \] \[ 9 - 15c + 16c = 86 \] \[ 9 + c = 86 \implies c = 77 \] Now substitute \( c = 77 \) into the equation for \( a \): \[ a = \frac{3 - 5(77)}{2} = \frac{3 - 385}{2} = \frac{-382}{2} = -191 \] Now solve for \( b \) and \( d \): From \( 2b + 5d = 0 \): \[ 2b + 5d = 0 \implies b = -\frac{5d}{2} \] Substitute this into the second equation: \[ 3b + 8d = 22 \implies 3\left(-\frac{5d}{2}\right) + 8d = 22 \] Simplifying: \[ -\frac{15d}{2} + 8d = 22 \] \[ -\frac{15d}{2} + \frac{16d}{2} = 22 \implies \frac{d}{2} = 22 \implies d = 44 \] Substitute \( d = 44 \) into the equation for \( b \): \[ b = -\frac{5(44)}{2} = -\frac{220}{2} = -110 \] Thus, the matrix \( D \) is: \[ D = \begin{pmatrix} -191 & -110 \\ 77 & 44 \end{pmatrix} \]

Final Answer: The matrix \( D \) such that \( CD - AB = 0 \) is: \[ \boxed{\begin{pmatrix} -191 & -110 \\ 77 & 44 \end{pmatrix}} \]