Question

If \( \alpha \) is a real root of the equation \( x^3 + 6x^2 + 5x - 42 = 0 \), then the determinant of the matrix \[ \begin{bmatrix} \alpha - 1 & \alpha + 1 & \alpha + 2 \\ \alpha - 2 & \alpha + 3 & \alpha - 3 \\ \alpha + 4 & \alpha - 4 & \alpha + 5 \end{bmatrix} \] is Options:

• A. 90
• B. 120
• C. -105
• D. -135

Hint:

When computing determinants, use row operations to simplify the matrix, and double-check with multiple methods (direct formula and cofactor expansion) to ensure accuracy. If the answer doesn’t match the expected option, consider potential errors in the problem statement.

Answer 1

The Correct Option is B

Step 1: Use the given equation to find \( \alpha \).
The equation is \( x^3 + 6x^2 + 5x - 42 = 0 \). Since \( \alpha \) is a real root, we can use the Rational Root Theorem to test possible rational roots. Possible factors of \(-42\) over factors of \(1\) are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42 \).
Testing \( x = 2 \):
\( 2^3 + 6(2^2) + 5(2) - 42 = 8 + 24 + 10 - 42 = 0 \).
So, \( \alpha = 2 \) is a root.
Step 2: Substitute \( \alpha = 2 \) into the matrix.
The matrix becomes:
\[ \begin{bmatrix} 2 - 1 & 2 + 1 & 2 + 2 \\ 2 - 2 & 2 + 3 & 2 - 3 \\ 2 + 4 & 2 - 4 & 2 + 5 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 4 \\ 0 & 5 & -1 \\ 6 & -2 & 7 \end{bmatrix} \] Step 3: Compute the determinant of the matrix.
For a 3\(\times\)3 matrix \( \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \), the determinant is:
\( a(ei - fh) - b(di - fg) + c(dh - eg) \).
Here, the matrix is \( \begin{bmatrix} 1 & 3 & 4 \\ 0 & 5 & -1 \\ 6 & -2 & 7 \end{bmatrix} \).
Using the formula:
\( a = 1 \), \( b = 3 \), \( c = 4 \), \( d = 0 \), \( e = 5 \), \( f = -1 \), \( g = 6 \), \( h = -2 \), \( i = 7 \).
First term: \( 1 \cdot (5 \cdot 7 - (-1) \cdot (-2)) = 1 \cdot (35 - 2) = 1 \cdot 33 = 33 \).
Second term: \( -3 \cdot (0 \cdot 7 - (-1) \cdot 6) = -3 \cdot (0 + 6) = -3 \cdot 6 = -18 \).
Third term: \( 4 \cdot (0 \cdot (-2) - 5 \cdot 6) = 4 \cdot (0 - 30) = 4 \cdot (-30) = -120 \).
Total determinant:
\( 33 - 18 - 120 = 33 - 138 = -105 \).
Step 4: Recompute for accuracy.
Use cofactor expansion along the first row:
\( 1 \cdot \det \begin{bmatrix} 5 & -1 \\ -2 & 7 \end{bmatrix} = 1 \cdot (5 \cdot 7 - (-1) \cdot (-2)) = 1 \cdot (35 - 2) = 33 \).
\( -3 \cdot \det \begin{bmatrix} 0 & -1 \\ 6 & 7 \end{bmatrix} = -3 \cdot (0 \cdot 7 - (-1) \cdot 6) = -3 \cdot (0 + 6) = -18 \).
\( 4 \cdot \det \begin{bmatrix} 0 & 5 \\ 6 & -2 \end{bmatrix} = 4 \cdot (0 \cdot (-2) - 5 \cdot 6) = 4 \cdot (0 - 30) = -120 \).
Determinant: \( 33 - 18 - 120 = -105 \). Matches the previous calculation.
Step 5: Explore the possibility of an error.
The determinant consistently computes to \( -105 \), which is option (3), but the given correct answer is (2) 120. Simplify the matrix using row operations:
Row 2: \( R_2 - R_1 \):
\( [0 - 1, 5 - 3, -1 - 4] = [-1, 2, -5] \).
Row 3: \( R_3 - R_1 \):
\( [6 - 1, -2 - 3, 7 - 4] = [5, -5, 3] \).
New matrix:
\[ \begin{bmatrix} 1 & 3 & 4 \\ -1 & 2 & -5 \\ 5 & -5 & 3 \end{bmatrix} \] Recalculate the determinant:
\( 1 \cdot (2 \cdot 3 - (-5) \cdot (-5)) = 1 \cdot (6 - 25) = 1 \cdot (-19) = -19 \).
\( -3 \cdot ((-1) \cdot 3 - (-5) \cdot 5) = -3 \cdot (-3 + 25) = -3 \cdot 22 = -66 \).
\( 4 \cdot ((-1) \cdot (-5) - 2 \cdot 5) = 4 \cdot (5 - 10) = 4 \cdot (-5) = -20 \).
Determinant: \( -19 - 66 - 20 = -105 \). Consistent result.
Step 6: Consider the given answer and problem context.
The determinant is \( -105 \), but the correct answer is given as 120. This suggests a potential error in the problem statement or answer key. Since \( -105 \) is an option, it’s likely the intended answer, and 120 may be a typo in the answer key.
Final Answer:
The determinant is \( -105 \), but since the given correct answer is 120, there may be an error in the problem setup or answer key. Based on calculation, the answer should be (3) \( -105 \) (in magnitude, but \( -105 \) as computed).
\[ \boxed{-105} \]