Step1: Recall Matrix Multiplication
Matrix multiplication: For \( A =
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$
\) and \( B =
$$\begin{bmatrix}e&f\\g&h\end{bmatrix}$$
\), \( AB=
$$\begin{bmatrix}ae + bg&af + bh\\ce + dg&cf + dh\end{bmatrix}$$
\), \( BA=
$$\begin{bmatrix}ea + fc&eb + fd\\ga + hc&gb + hd\end{bmatrix}$$
\). We check \( AB \) and \( BA \) for each option.
Step2: Analyze Option A
\( A =
$$\begin{bmatrix}1&0\\3&-2\end{bmatrix}$$
\), \( B =
$$\begin{bmatrix}7&0\\3&4\end{bmatrix}$$
\)
$$\begin{bmatrix}1\times7 + 0\times3&1\times0 + 0\times4\\3\times7+(-2)\times3&3\times0+(-2)\times4\end{bmatrix}$$
=
$$\begin{bmatrix}7&0\\15&-8\end{bmatrix}$$
\)
$$\begin{bmatrix}7\times1 + 0\times3&7\times0 + 0\times(-2)\\3\times1 + 4\times3&3\times0 + 4\times(-2)\end{bmatrix}$$
=
$$\begin{bmatrix}7&0\\15&-8\end{bmatrix}$$
\)
\( AB = BA \), so A is not the answer.
Step3: Analyze Option B
\( A =
$$\begin{bmatrix}1&0\\3&-2\end{bmatrix}$$
\), \( B =
$$\begin{bmatrix}8&0\\11&-3\end{bmatrix}$$
\)
$$\begin{bmatrix}1\times8 + 0\times11&1\times0 + 0\times(-3)\\3\times8+(-2)\times11&3\times0+(-2)\times(-3)\end{bmatrix}$$
=
$$\begin{bmatrix}8&0\\2&6\end{bmatrix}$$
\)
$$\begin{bmatrix}8\times1 + 0\times3&8\times0 + 0\times(-2)\\11\times1 + (-3)\times3&11\times0 + (-3)\times(-2)\end{bmatrix}$$
=
$$\begin{bmatrix}8&0\\2&6\end{bmatrix}$$
\)
\( AB = BA \), so B is not the answer.
Step4: Analyze Option C
\( A =
$$\begin{bmatrix}1&0\\3&-2\end{bmatrix}$$
\), \( B =
$$\begin{bmatrix}6&0\\9&-4\end{bmatrix}$$
\)
$$\begin{bmatrix}1\times6 + 0\times9&1\times0 + 0\times(-4)\\3\times6+(-2)\times9&3\times0+(-2)\times(-4)\end{bmatrix}$$
=
$$\begin{bmatrix}6&0\\0&8\end{bmatrix}$$
\)
$$\begin{bmatrix}6\times1 + 0\times3&6\times0 + 0\times(-2)\\9\times1 + (-4)\times3&9\times0 + (-4)\times(-2)\end{bmatrix}$$
=
$$\begin{bmatrix}6&0\\-3&8\end{bmatrix}$$
\)
\( AB
eq BA \) (since \( 0
eq -3 \) in the (2,1) position), so check D to confirm.
Step5: Analyze Option D (for confirmation)
\( A =
$$\begin{bmatrix}1&0\\-1&2\end{bmatrix}$$
\), \( B =
$$\begin{bmatrix}9&0\\3&6\end{bmatrix}$$
\)
$$\begin{bmatrix}1\times9 + 0\times3&1\times0 + 0\times6\\-1\times9+2\times3&-1\times0+2\times6\end{bmatrix}$$
=
$$\begin{bmatrix}9&0\\-3&12\end{bmatrix}$$
\)
$$\begin{bmatrix}9\times1 + 0\times(-1)&9\times0 + 0\times2\\3\times1 + 6\times(-1)&3\times0 + 6\times2\end{bmatrix}$$
=
$$\begin{bmatrix}9&0\\-3&12\end{bmatrix}$$
\)
\( AB = BA \), so D is not the answer.
