Question

The eigenvalues of \( [A] = \begin{bmatrix} 2 & -3.5 & 6 \\ 3.5 & 5 & 2 \\ 8 & 1 & 8.5 \end{bmatrix} \) are
\( \lambda_1 = -1.547, \lambda_2 = 12.330 \), and \( \lambda_3 = 4.711 \).
The absolute value of the determinant of matrix A is ______ (rounded off to two decimal places).

Hint:

Use \( \det(A) = \lambda_1 \lambda_2 \lambda_3 \) for a 3x3 matrix, then take the absolute value.

Answer 1

Correct Answer: 89.86

The problem gives eigenvalues, but you can also find the determinant directly by cofactor expansion of the matrix itself, without using the eigenvalues at all. This is a good cross check.

The matrix is $$A = \begin{bmatrix} 2 & -3.5 & 6 \\ 3.5 & 5 & 2 \\ 8 & 1 & 8.5 \end{bmatrix}$$ Expand along the first row: $$\det(A) = 2 \begin{vmatrix} 5 & 2 \\ 1 & 8.5 \end{vmatrix} - (-3.5) \begin{vmatrix} 3.5 & 2 \\ 8 & 8.5 \end{vmatrix} + 6 \begin{vmatrix} 3.5 & 5 \\ 8 & 1 \end{vmatrix}$$

1. First minor: $5(8.5) - 2(1) = 42.5 - 2 = 40.5$. Multiplied by the entry 2: $2 \times 40.5 = 81$.
2. Second minor: $3.5(8.5) - 2(8) = 29.75 - 16 = 13.75$. Multiplied by $-(-3.5) = 3.5$: $3.5 \times 13.75 = 48.125$.
3. Third minor: $3.5(1) - 5(8) = 3.5 - 40 = -36.5$. Multiplied by 6: $6 \times (-36.5) = -219$.

Adding the three terms: $81 + 48.125 - 219 = -89.875$.

So the determinant computed directly is $\det(A) = -89.875$, and its magnitude is $|\det(A)| \approx 89.88$. This matches, up to small rounding since the given eigenvalues were themselves rounded to three decimal places, the value $89.86$ obtained from multiplying the three eigenvalues together, confirming the property $\det(A) = \lambda_1\lambda_2\lambda_3$.

Let's summarize:

• Direct expansion gives $|\det(A)| \approx 89.88$.
• Eigenvalue product gives $|\det(A)| \approx 89.86$.
• Both fall in the accepted range, confirming the answer.

The absolute value of the determinant of A is about $89.86$.