Question

The product of all eigenvalues of the matrix \[ \begin{bmatrix} 1 & 2 & 3
4 & 5 & 6
7 & 8 & 9 \end{bmatrix} is: \]

• A. \( -1 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

The determinant of a matrix is a quick way to find the product of all eigenvalues. If the determinant is zero, at least one eigenvalue is zero.

Answer 1

The Correct Option is B

Step 1: Determine the product of eigenvalues. The product of all eigenvalues of a square matrix is equal to the determinant of the matrix. Step 2: Calculate the determinant. \[ \det\left(\begin{bmatrix} 1 & 2 & 3
4 & 5 & 6
7 & 8 & 9 \end{bmatrix}\right) = 1\begin{vmatrix} 5 & 6
8 & 9 \end{vmatrix} - 2\begin{vmatrix} 4 & 6
7 & 9 \end{vmatrix} + 3\begin{vmatrix} 4 & 5
7 & 8 \end{vmatrix}. \]
Evaluate each minor determinant: \[ \begin{vmatrix} 5 & 6
8 & 9 \end{vmatrix} = (5)(9) - (6)(8) = 45 - 48 = -3, \] \[ \begin{vmatrix} 4 & 6
7 & 9 \end{vmatrix} = (4)(9) - (6)(7) = 36 - 42 = -6, \] \[ \begin{vmatrix} 4 & 5
7 & 8 \end{vmatrix} = (4)(8) - (5)(7) = 32 - 35 = -3. \]
Substitute back: \[ \det = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0. \] Step 3: Conclusion. The determinant is \( 0 \), so the product of eigenvalues is \( 0 \).