Question

There are two values of a for which the determinant, \(\Delta =\begin{bmatrix}1& -2& 5\\[0.3em]0& a& 1\\[0.3em] 0& 4& 2a\\[0.3em] \end{bmatrix} = 86\), then the sum of these values of a is:

• A. \(\sqrt{41}\)
• B. 0
• C. \(- \sqrt{41}\)
• D. \(2 \sqrt{41}\)

Answer 1

The Correct Option is B

To find the determinant of the matrix and solve for \(a\), we first address the given matrix:

\(\Delta =\begin{bmatrix}1 & -2 & 5\\0 & a & 1\\0 & 4 & 2a\end{bmatrix}\)

Since the first row's only non-zero element is \(1\), located at the (1,1) position, we can use cofactor expansion along the first row:

\[\Delta = 1 \cdot \begin{vmatrix}a & 1\\4 & 2a\end{vmatrix} - 0 + 0\]

Calculate the 2x2 determinant:

\[\begin{vmatrix}a & 1\\4 & 2a\end{vmatrix} = (a)(2a) - (1)(4) = 2a^2 - 4\]

Thus, the determinant \(\Delta\) is simplified to:

\[\Delta = 2a^2 - 4\]

We are given that \(\Delta = 86\), so we equate and solve for \(a\):

\[2a^2 - 4 = 86\]

Simplify and solve the quadratic equation:

\[2a^2 = 90\]

\[a^2 = 45\]

\[a = \pm \sqrt{45} = \pm 3\sqrt{5}\]

The sum of these values is:

\[a_1 + a_2 = 3\sqrt{5} + (-3\sqrt{5}) = 0\]

Therefore, the sum of the two values of \(a\) is 0.