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: Setting up unknown entries:
Let $D=\begin{bmatrix}a & b\\c & d\end{bmatrix}$. From $CD=AB$, first compute $AB=\begin{bmatrix}3 & 0\\43 & 22\end{bmatrix}$ as before.

Step 2: Writing CD in terms of a, b, c, d:
\[ CD=\begin{bmatrix}2 & 5\\3 & 8\end{bmatrix}\begin{bmatrix}a & b\\c & d\end{bmatrix}=\begin{bmatrix}2a+5c & 2b+5d\\3a+8c & 3b+8d\end{bmatrix} \]
Equating $CD$ to $AB$ gives four equations: $2a+5c=3$, $3a+8c=43$, $2b+5d=0$, $3b+8d=22$.

Step 3: Solving for a and c:
Multiply the first equation by 8 and the second by 5, then subtract.
\[ 16a+40c=24,\quad 15a+40c=215 \implies a=24-215=-191 \]
Put back: $2(-191)+5c=3 \implies c=\dfrac{3+382}{5}=77$.

Step 4: Solving for b and d:
Multiply the third equation by 8 and the fourth by 5, then subtract.
\[ 16b+40d=0,\quad 15b+40d=110 \implies b=0-110=-110 \]
Put back: $2(-110)+5d=0 \implies d=44$.

Final Answer:
Solving the equations directly gives the same matrix as the inverse method.
\[ \boxed{D=\begin{bmatrix}-191 & -110\\77 & 44\end{bmatrix}} \]