Question

Suppose that \(A=\begin{bmatrix}2 & -1\\3 & 4\end{bmatrix}\), \(B=\begin{bmatrix}5 & 2\\7 & 4\end{bmatrix}\), \(C=\begin{bmatrix}2 & 5\\3 & 8\end{bmatrix}\). Find the matrix \(D\) such that \(CD-AB=0\).

Hint:

From CD = AB, find D by multiplying the inverse of C with the product AB.

Answer 1

Step 1: Understanding the Concept:
The equation \(CD-AB=0\) means \(CD=AB\).
Since \(C\) is a square matrix, if \(C\) is invertible, then \(D=C^{-1}(AB)\).

Step 2: Computing the product AB:
\[ AB=\begin{bmatrix}2 & -1\\3 & 4\end{bmatrix}\begin{bmatrix}5 & 2\\7 & 4\end{bmatrix}=\begin{bmatrix}10-7 & 4-4\\15+28 & 6+16\end{bmatrix}=\begin{bmatrix}3 & 0\\43 & 22\end{bmatrix} \]

Step 3: Finding the inverse of C:
\(\det C=2\times8-5\times3=16-15=1\), so \(C\) is invertible.
Adjoint of \(C\) is obtained by swapping the diagonal entries and changing the sign of the off-diagonal entries.
\[ C^{-1}=\dfrac{1}{\det C}\,\text{adj}(C)=\begin{bmatrix}8 & -5\\-3 & 2\end{bmatrix} \]

Step 4: Computing D:
\[ D=C^{-1}(AB)=\begin{bmatrix}8 & -5\\-3 & 2\end{bmatrix}\begin{bmatrix}3 & 0\\43 & 22\end{bmatrix}=\begin{bmatrix}24-215 & 0-110\\-9+86 & 0+44\end{bmatrix}=\begin{bmatrix}-191 & -110\\77 & 44\end{bmatrix} \]

Final Answer:
This D satisfies CD=AB, so CD-AB=0.
\[ \boxed{D=\begin{bmatrix}-191 & -110\\77 & 44\end{bmatrix}} \]