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

To solve this problem, we need to determine the range of the set \( A \) defined by the inequality involving the determinant of the given matrix.

The matrix provided is:

\(\begin{bmatrix} 3x - 1 & 2 \\ -2 & 5 \end{bmatrix}\)

To find the determinant of this matrix, we use the formula for a 2x2 matrix:

\(\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc\)

Applying this to our matrix, we have:

\(\det\begin{bmatrix} 3x - 1 & 2 \\ -2 & 5 \end{bmatrix} = (3x - 1) \cdot 5 - (2) \cdot (-2)\)

Simplifying the expression:

• First term: \(5 \cdot (3x - 1) = 15x - 5\)
• Second term: \(2 \cdot (-2) = -4\)

Combining these, the determinant becomes:

\(15x - 5 + 4 = 15x - 1\)

We are given that this determinant is constrained as:

\(-31 < 15x - 1 \le 29\)

We solve these linear inequalities separately:

1. For \(-31 < 15x - 1\):

Add 1 to both sides:

\(-30 < 15x\)

Divide by 15:

\(-\frac{30}{15} < x\)

Simplifying gives:

\(-2 < x\)

1. For \(15x - 1 \le 29\):

Add 1 to both sides:

\(15x \le 30\)

Divide by 15:

\(x \le \frac{30}{15}\)

Simplifying gives:

\(x \le 2\)

Combining these results, we find:

\(-2 < x \le 2\)

Thus, the set \( A \) is:

\(A = (-2, 2]\)

This corresponds to the given option:

A = (-2, 2]