Question: Determine whether the statement below is true or false. Justify the answer.
If $A$ equals \begin{tabular}{|c|c|}\hline a & b \\ \hline c & d \\ \hline \end{tabular} and $ad$ equals $bc$, then $A$ is not invertible.
Choose the correct answer below.
A. The statement is false. If $A$ is invertible, then $ad$ equals $bc$.
B. The statement is true. The matrix $A$ equals \begin{tabular}{|c|c|}\hline a & b \\ \hline c & d \\ \hline \end{tabular} is invertible if and only if $a$ not equals $b$ and $b$ not equals $d$.
C. The statement is false. If $ad$ equals $bc$, then $A$ is invertible.
D. The statement is true. If $ad$ equals $bc$ then $ad - bc$ equals $0$, and $\frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ is undefined.
1. **State the problem:**
We are given a $2 \times 2$ matrix
$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
and the condition $ad = bc$. We need to determine if the statement "If $ad = bc$, then $A$ is not invertible" is true or false.
2. **Recall the formula for invertibility of a $2 \times 2$ matrix:**
A matrix $A$ is invertible if and only if its determinant is nonzero.
The determinant of $A$ is
$$\det(A) = ad - bc.$$
3. **Analyze the given condition:**
If $ad = bc$, then
$$ad - bc = 0.$$
4. **Interpret the determinant:**
Since $\det(A) = 0$, the matrix $A$ is **not invertible**.
5. **Explain why:**
The inverse of $A$ (if it exists) is given by
$$A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$$
If $ad - bc = 0$, then the denominator is zero, making $A^{-1}$ undefined.
6. **Conclusion:**
The statement "If $ad = bc$, then $A$ is not invertible" is **true**.
7. **Check the answer choices:**
- Choice A is false because if $A$ is invertible, then $ad \neq bc$.
- Choice B is false because invertibility depends on $ad - bc \neq 0$, not on $a \neq b$ or $b \neq d$.
- Choice C is false because if $ad = bc$, $A$ is not invertible.
- Choice D is true and correctly explains the reasoning.
**Final answer:** Choice D is correct.