Question

Let $A = \left\{ x \in \mathbb{R} \;\middle|\; -31 < \det \begin{bmatrix} 3x - 1 & 2 \\ -2 & 5 \end{bmatrix} \le 29 \right\}$. Which one of the following statements is TRUE?

• A. $A = (-2, 2]$
• B. $A = (-2, 2)$
• C. $A = [-2, 2)$
• D. $A = [-2, 2]$

Hint:

Pay close attention to the inequality symbols.
The strictly less than symbol ($<$) on the left corresponds to an open boundary (parenthesis '$($'), while the less than or equal to symbol ($\le$) on the right corresponds to a closed boundary (bracket '$]$').

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
The question asks us to find the set of real numbers $A$ defined by a double inequality involving the determinant of a $2 \times 2$ matrix.

Step 2 : Key Formulas and Approach:
For a $2 \times 2$ matrix, the determinant is calculated using the formula:
\[ \det \begin{bmatrix} a & b\\ c & d \end{bmatrix} = ad - bc \]
We will calculate the determinant in terms of $x$, substitute it into the inequality $-31 < \det(\text{matrix}) \le 29$, and solve for $x$.

Step 3 : Detailed Explanation:
Let us calculate the determinant of the given matrix:
\[ D = \det \begin{bmatrix} 3x - 1 & 2 -2 & 5 \end{bmatrix} \]
Applying the formula:
\[ D = (3x - 1)(5) - (2)(-2) \]
\[ D = 15x - 5 + 4 \]
\[ D = 15x - 1 \]
Now, substitute this expression into the inequality:
\[ -31 < 15x - 1 \le 29 \]
Let us add 1 to all parts of the inequality to isolate the term containing $x$:
\[ -31 + 1 < 15x \le 29 + 1 \]
\[ -30 < 15x \le 30 \]
Now, divide all parts of the inequality by 15:
\[ -\frac{30}{15} < x \le \frac{30}{15} \]
\[ -2 < x \le 2 \]
Thus, the set $A$ is the interval $(-2, 2]$.

Step 4 : Final Answer:
The set is $A = (-2, 2]$.
This corresponds to Option (A).